Easier way to find eigenvalues of Matrices?












1












$begingroup$


I am trying to find eigenvalues for this matrix,

A =
$begin{bmatrix}
3 & 2 & -3 \
-3 & -4 & 9 \
-1 & -2 & 5 \
end{bmatrix}$



I find the characteristic equation here:
$(lambda I - A)
=
begin{bmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{bmatrix}$



The difficult part I am having is finding the determinant of the characteristic equation, mainly that it becomes insanely difficult for me to keep track of the factoring to get the eigenvalues...here I use rule of Sarrus to try to calculate the eigenvalues.
$begin{vmatrix}
lambda I - A \
end{vmatrix}
=
begin{bmatrix}
lambda - 3 & -2 & 3 & lambda -3 & -2 \
3 & lambda + 4 & -9 & 3 & lambda + 4 \
1 & 2 & lambda - 5 & 1 & 2 \
end{bmatrix}$

= $(1)(lambda + 4)(3)$
$+ (2)(-9)(lambda - 3)$
$+ (lambda - 5)(3)(-2)$
$- (lambda - 3)(lambda + 4)(lambda - 5)$
$- (-2)(-9)(1)$
$- (3)(3)(2)$



= $(3)(lambda + 4) + (-18)(lambda - 3) + (-6)(lambda -5) - (lambda - 3)(lambda + 4)(lambda - 5) - 18 - 18$

= $(3)(lambda + 4) + (-18)(lambda - 3) + (-6)(lambda -5) - (lambda - 3)(lambda + 4)(lambda - 5) - 36$

= $((lambda + 4)(3) -(lambda -3)(lambda -5)) + (-18)(lambda -3) + (-6)(lambda - 5) - 36$

= $(lambda - 3)((lambda + 4)(3) -(lambda - 5)) -18 + (-6)(lambda - 5) - 36$

= $(lambda - 5)(lambda -3)((lambda + 4)(3) - 1)) - 18 + (-6)(1) - 36$

= $(lambda - 5)(lambda -3)((lambda + 4)(3) - 1)) - 60$

= $(lambda - 5)(lambda - 3)(3lambda + 12 -1 - 60)$

= $(lambda - 5)(lambda - 3)(3lambda - 49)$



I end up with 5, 3, and 16.3 as the eigenvalues (16.3 seems off). Obviously that was a ridiculous amount of simplification I had to do just to get eigenvalues and when I feel like I messed up (like here), it is pretty impossible to check my work in an effective manner. The amount of time it takes for me to calculate eigenvalues is unacceptable for my upcoming final exam. Do you guys have any tips or tricks that makes this process easier?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    $16.3 neq frac{49}{3}.$ Please just use fractions. Also, this is the way it's done - not much shortcuts are around.
    $endgroup$
    – Sean Roberson
    Dec 6 '18 at 18:42






  • 1




    $begingroup$
    If you compute the determinant of $A$, you will get zero, hence $A$ is singular, hence $0$ is an eigenvalue. So your characteristic polynomial can't be right. You can plug in $lambda = 0$ starting with your first expression to find out where you went wrong.
    $endgroup$
    – Bungo
    Dec 6 '18 at 18:55










  • $begingroup$
    Many of the steps in your calculation are not things you're allowed to do to simplify an expression. For example, in the next-to-last step, you go from $(lambda-3)(lambda-5)(3lambda+11) - 60$ to $(lambda-3)(lambda-5)(3lambda+11-60)$, which ignores order of operations.
    $endgroup$
    – Misha Lavrov
    Dec 6 '18 at 19:02












  • $begingroup$
    Evan, it is easier and more reliable to separately work out the coefficients. Carrying around polynomials inside a determinant is a way to cause errors... Anyway, I put an answer. The same scheme, in an analogous manner works for larger size square matrices as well,,,
    $endgroup$
    – Will Jagy
    Dec 6 '18 at 19:13
















1












$begingroup$


I am trying to find eigenvalues for this matrix,

A =
$begin{bmatrix}
3 & 2 & -3 \
-3 & -4 & 9 \
-1 & -2 & 5 \
end{bmatrix}$



I find the characteristic equation here:
$(lambda I - A)
=
begin{bmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{bmatrix}$



The difficult part I am having is finding the determinant of the characteristic equation, mainly that it becomes insanely difficult for me to keep track of the factoring to get the eigenvalues...here I use rule of Sarrus to try to calculate the eigenvalues.
$begin{vmatrix}
lambda I - A \
end{vmatrix}
=
begin{bmatrix}
lambda - 3 & -2 & 3 & lambda -3 & -2 \
3 & lambda + 4 & -9 & 3 & lambda + 4 \
1 & 2 & lambda - 5 & 1 & 2 \
end{bmatrix}$

= $(1)(lambda + 4)(3)$
$+ (2)(-9)(lambda - 3)$
$+ (lambda - 5)(3)(-2)$
$- (lambda - 3)(lambda + 4)(lambda - 5)$
$- (-2)(-9)(1)$
$- (3)(3)(2)$



= $(3)(lambda + 4) + (-18)(lambda - 3) + (-6)(lambda -5) - (lambda - 3)(lambda + 4)(lambda - 5) - 18 - 18$

= $(3)(lambda + 4) + (-18)(lambda - 3) + (-6)(lambda -5) - (lambda - 3)(lambda + 4)(lambda - 5) - 36$

= $((lambda + 4)(3) -(lambda -3)(lambda -5)) + (-18)(lambda -3) + (-6)(lambda - 5) - 36$

= $(lambda - 3)((lambda + 4)(3) -(lambda - 5)) -18 + (-6)(lambda - 5) - 36$

= $(lambda - 5)(lambda -3)((lambda + 4)(3) - 1)) - 18 + (-6)(1) - 36$

= $(lambda - 5)(lambda -3)((lambda + 4)(3) - 1)) - 60$

= $(lambda - 5)(lambda - 3)(3lambda + 12 -1 - 60)$

= $(lambda - 5)(lambda - 3)(3lambda - 49)$



I end up with 5, 3, and 16.3 as the eigenvalues (16.3 seems off). Obviously that was a ridiculous amount of simplification I had to do just to get eigenvalues and when I feel like I messed up (like here), it is pretty impossible to check my work in an effective manner. The amount of time it takes for me to calculate eigenvalues is unacceptable for my upcoming final exam. Do you guys have any tips or tricks that makes this process easier?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    $16.3 neq frac{49}{3}.$ Please just use fractions. Also, this is the way it's done - not much shortcuts are around.
    $endgroup$
    – Sean Roberson
    Dec 6 '18 at 18:42






  • 1




    $begingroup$
    If you compute the determinant of $A$, you will get zero, hence $A$ is singular, hence $0$ is an eigenvalue. So your characteristic polynomial can't be right. You can plug in $lambda = 0$ starting with your first expression to find out where you went wrong.
    $endgroup$
    – Bungo
    Dec 6 '18 at 18:55










  • $begingroup$
    Many of the steps in your calculation are not things you're allowed to do to simplify an expression. For example, in the next-to-last step, you go from $(lambda-3)(lambda-5)(3lambda+11) - 60$ to $(lambda-3)(lambda-5)(3lambda+11-60)$, which ignores order of operations.
    $endgroup$
    – Misha Lavrov
    Dec 6 '18 at 19:02












  • $begingroup$
    Evan, it is easier and more reliable to separately work out the coefficients. Carrying around polynomials inside a determinant is a way to cause errors... Anyway, I put an answer. The same scheme, in an analogous manner works for larger size square matrices as well,,,
    $endgroup$
    – Will Jagy
    Dec 6 '18 at 19:13














1












1








1


1



$begingroup$


I am trying to find eigenvalues for this matrix,

A =
$begin{bmatrix}
3 & 2 & -3 \
-3 & -4 & 9 \
-1 & -2 & 5 \
end{bmatrix}$



I find the characteristic equation here:
$(lambda I - A)
=
begin{bmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{bmatrix}$



The difficult part I am having is finding the determinant of the characteristic equation, mainly that it becomes insanely difficult for me to keep track of the factoring to get the eigenvalues...here I use rule of Sarrus to try to calculate the eigenvalues.
$begin{vmatrix}
lambda I - A \
end{vmatrix}
=
begin{bmatrix}
lambda - 3 & -2 & 3 & lambda -3 & -2 \
3 & lambda + 4 & -9 & 3 & lambda + 4 \
1 & 2 & lambda - 5 & 1 & 2 \
end{bmatrix}$

= $(1)(lambda + 4)(3)$
$+ (2)(-9)(lambda - 3)$
$+ (lambda - 5)(3)(-2)$
$- (lambda - 3)(lambda + 4)(lambda - 5)$
$- (-2)(-9)(1)$
$- (3)(3)(2)$



= $(3)(lambda + 4) + (-18)(lambda - 3) + (-6)(lambda -5) - (lambda - 3)(lambda + 4)(lambda - 5) - 18 - 18$

= $(3)(lambda + 4) + (-18)(lambda - 3) + (-6)(lambda -5) - (lambda - 3)(lambda + 4)(lambda - 5) - 36$

= $((lambda + 4)(3) -(lambda -3)(lambda -5)) + (-18)(lambda -3) + (-6)(lambda - 5) - 36$

= $(lambda - 3)((lambda + 4)(3) -(lambda - 5)) -18 + (-6)(lambda - 5) - 36$

= $(lambda - 5)(lambda -3)((lambda + 4)(3) - 1)) - 18 + (-6)(1) - 36$

= $(lambda - 5)(lambda -3)((lambda + 4)(3) - 1)) - 60$

= $(lambda - 5)(lambda - 3)(3lambda + 12 -1 - 60)$

= $(lambda - 5)(lambda - 3)(3lambda - 49)$



I end up with 5, 3, and 16.3 as the eigenvalues (16.3 seems off). Obviously that was a ridiculous amount of simplification I had to do just to get eigenvalues and when I feel like I messed up (like here), it is pretty impossible to check my work in an effective manner. The amount of time it takes for me to calculate eigenvalues is unacceptable for my upcoming final exam. Do you guys have any tips or tricks that makes this process easier?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




I am trying to find eigenvalues for this matrix,

A =
$begin{bmatrix}
3 & 2 & -3 \
-3 & -4 & 9 \
-1 & -2 & 5 \
end{bmatrix}$



I find the characteristic equation here:
$(lambda I - A)
=
begin{bmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{bmatrix}$



The difficult part I am having is finding the determinant of the characteristic equation, mainly that it becomes insanely difficult for me to keep track of the factoring to get the eigenvalues...here I use rule of Sarrus to try to calculate the eigenvalues.
$begin{vmatrix}
lambda I - A \
end{vmatrix}
=
begin{bmatrix}
lambda - 3 & -2 & 3 & lambda -3 & -2 \
3 & lambda + 4 & -9 & 3 & lambda + 4 \
1 & 2 & lambda - 5 & 1 & 2 \
end{bmatrix}$

= $(1)(lambda + 4)(3)$
$+ (2)(-9)(lambda - 3)$
$+ (lambda - 5)(3)(-2)$
$- (lambda - 3)(lambda + 4)(lambda - 5)$
$- (-2)(-9)(1)$
$- (3)(3)(2)$



= $(3)(lambda + 4) + (-18)(lambda - 3) + (-6)(lambda -5) - (lambda - 3)(lambda + 4)(lambda - 5) - 18 - 18$

= $(3)(lambda + 4) + (-18)(lambda - 3) + (-6)(lambda -5) - (lambda - 3)(lambda + 4)(lambda - 5) - 36$

= $((lambda + 4)(3) -(lambda -3)(lambda -5)) + (-18)(lambda -3) + (-6)(lambda - 5) - 36$

= $(lambda - 3)((lambda + 4)(3) -(lambda - 5)) -18 + (-6)(lambda - 5) - 36$

= $(lambda - 5)(lambda -3)((lambda + 4)(3) - 1)) - 18 + (-6)(1) - 36$

= $(lambda - 5)(lambda -3)((lambda + 4)(3) - 1)) - 60$

= $(lambda - 5)(lambda - 3)(3lambda + 12 -1 - 60)$

= $(lambda - 5)(lambda - 3)(3lambda - 49)$



I end up with 5, 3, and 16.3 as the eigenvalues (16.3 seems off). Obviously that was a ridiculous amount of simplification I had to do just to get eigenvalues and when I feel like I messed up (like here), it is pretty impossible to check my work in an effective manner. The amount of time it takes for me to calculate eigenvalues is unacceptable for my upcoming final exam. Do you guys have any tips or tricks that makes this process easier?






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 18:45









Moo

5,63131020




5,63131020










asked Dec 6 '18 at 18:34









Evan KimEvan Kim

5059




5059












  • $begingroup$
    $16.3 neq frac{49}{3}.$ Please just use fractions. Also, this is the way it's done - not much shortcuts are around.
    $endgroup$
    – Sean Roberson
    Dec 6 '18 at 18:42






  • 1




    $begingroup$
    If you compute the determinant of $A$, you will get zero, hence $A$ is singular, hence $0$ is an eigenvalue. So your characteristic polynomial can't be right. You can plug in $lambda = 0$ starting with your first expression to find out where you went wrong.
    $endgroup$
    – Bungo
    Dec 6 '18 at 18:55










  • $begingroup$
    Many of the steps in your calculation are not things you're allowed to do to simplify an expression. For example, in the next-to-last step, you go from $(lambda-3)(lambda-5)(3lambda+11) - 60$ to $(lambda-3)(lambda-5)(3lambda+11-60)$, which ignores order of operations.
    $endgroup$
    – Misha Lavrov
    Dec 6 '18 at 19:02












  • $begingroup$
    Evan, it is easier and more reliable to separately work out the coefficients. Carrying around polynomials inside a determinant is a way to cause errors... Anyway, I put an answer. The same scheme, in an analogous manner works for larger size square matrices as well,,,
    $endgroup$
    – Will Jagy
    Dec 6 '18 at 19:13


















  • $begingroup$
    $16.3 neq frac{49}{3}.$ Please just use fractions. Also, this is the way it's done - not much shortcuts are around.
    $endgroup$
    – Sean Roberson
    Dec 6 '18 at 18:42






  • 1




    $begingroup$
    If you compute the determinant of $A$, you will get zero, hence $A$ is singular, hence $0$ is an eigenvalue. So your characteristic polynomial can't be right. You can plug in $lambda = 0$ starting with your first expression to find out where you went wrong.
    $endgroup$
    – Bungo
    Dec 6 '18 at 18:55










  • $begingroup$
    Many of the steps in your calculation are not things you're allowed to do to simplify an expression. For example, in the next-to-last step, you go from $(lambda-3)(lambda-5)(3lambda+11) - 60$ to $(lambda-3)(lambda-5)(3lambda+11-60)$, which ignores order of operations.
    $endgroup$
    – Misha Lavrov
    Dec 6 '18 at 19:02












  • $begingroup$
    Evan, it is easier and more reliable to separately work out the coefficients. Carrying around polynomials inside a determinant is a way to cause errors... Anyway, I put an answer. The same scheme, in an analogous manner works for larger size square matrices as well,,,
    $endgroup$
    – Will Jagy
    Dec 6 '18 at 19:13
















$begingroup$
$16.3 neq frac{49}{3}.$ Please just use fractions. Also, this is the way it's done - not much shortcuts are around.
$endgroup$
– Sean Roberson
Dec 6 '18 at 18:42




$begingroup$
$16.3 neq frac{49}{3}.$ Please just use fractions. Also, this is the way it's done - not much shortcuts are around.
$endgroup$
– Sean Roberson
Dec 6 '18 at 18:42




1




1




$begingroup$
If you compute the determinant of $A$, you will get zero, hence $A$ is singular, hence $0$ is an eigenvalue. So your characteristic polynomial can't be right. You can plug in $lambda = 0$ starting with your first expression to find out where you went wrong.
$endgroup$
– Bungo
Dec 6 '18 at 18:55




$begingroup$
If you compute the determinant of $A$, you will get zero, hence $A$ is singular, hence $0$ is an eigenvalue. So your characteristic polynomial can't be right. You can plug in $lambda = 0$ starting with your first expression to find out where you went wrong.
$endgroup$
– Bungo
Dec 6 '18 at 18:55












$begingroup$
Many of the steps in your calculation are not things you're allowed to do to simplify an expression. For example, in the next-to-last step, you go from $(lambda-3)(lambda-5)(3lambda+11) - 60$ to $(lambda-3)(lambda-5)(3lambda+11-60)$, which ignores order of operations.
$endgroup$
– Misha Lavrov
Dec 6 '18 at 19:02






$begingroup$
Many of the steps in your calculation are not things you're allowed to do to simplify an expression. For example, in the next-to-last step, you go from $(lambda-3)(lambda-5)(3lambda+11) - 60$ to $(lambda-3)(lambda-5)(3lambda+11-60)$, which ignores order of operations.
$endgroup$
– Misha Lavrov
Dec 6 '18 at 19:02














$begingroup$
Evan, it is easier and more reliable to separately work out the coefficients. Carrying around polynomials inside a determinant is a way to cause errors... Anyway, I put an answer. The same scheme, in an analogous manner works for larger size square matrices as well,,,
$endgroup$
– Will Jagy
Dec 6 '18 at 19:13




$begingroup$
Evan, it is easier and more reliable to separately work out the coefficients. Carrying around polynomials inside a determinant is a way to cause errors... Anyway, I put an answer. The same scheme, in an analogous manner works for larger size square matrices as well,,,
$endgroup$
– Will Jagy
Dec 6 '18 at 19:13










3 Answers
3






active

oldest

votes


















2












$begingroup$

Here is a simple computation:
begin{align}
det(lambda I - A)& = begin{vmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=begin{vmatrix}
lambda - 2 & 0 & lambda - 2 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}\& =(lambda - 2 )begin{vmatrix}
1 & 0 & 1 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=(lambda - 2 )biggl(begin{vmatrix}
lambda + 4 & -9 \
2 & lambda - 5 \
end{vmatrix}+begin{vmatrix}
3 & lambda + 4 \
1 & 2 \
end{vmatrix}biggl)\
&=(lambda - 2 )Bigl(bigl(
(lambda + 4)(lambda - 5)+18bigr)
+ (6 -lambda - 4) Bigl)=(lambda - 2 )(lambda^2-2lambda)\
&=lambda(lambda - 2)^2.
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from 3 to $lambda - 2$ in the top right corner, $A_13$?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:10








  • 1




    $begingroup$
    I added the third row to the first.
    $endgroup$
    – Bernard
    Dec 6 '18 at 21:19










  • $begingroup$
    So adding rows to each other will not change characteristic equation, but scaling rows will?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 23:34










  • $begingroup$
    It doesn't change any determinant, and scalind a row or a column scales the determinant, since a determinant is linear w.r.t. each row or column.
    $endgroup$
    – Bernard
    Dec 6 '18 at 23:37



















0












$begingroup$

I would do what you did: to apply the rule of Sarrus. But the polynomial that you got cannot be correct. First of all, it is not monic. On the other hand, if you sum all entries of each row of the matrix, you get $2$. But that means that $2$ is an eigenvalue and that $(1,1,1)$ is an eigenvector of $A$. However, $2$ is not a root of your polynomial.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What do you mean by "if you sum all entries of each line on the matrix, you get 2." Is each line referencing the row or the column?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:04










  • $begingroup$
    I meant row. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 20:06










  • $begingroup$
    @JoséCarlos, some students really liked the recipe in my answer, although this OP seems less than impressed. Just a way to reduce errors that I have not usually seen in textbooks.
    $endgroup$
    – Will Jagy
    Dec 7 '18 at 1:00



















0












$begingroup$

For 2 by 2 and 3 by 3 there are worthwhile recipes for the characteristic polynomial. For 2 by 2, it is $lambda^2 - T lambda + D,$ where $T$ is the trace and $D$ is the determinant.



more in a few minutes.



For 3 by 3, it is $$ lambda^3 - sigma_1 lambda^2 + sigma_2 lambda - sigma_3 $$
Here $sigma_1$ is the trace and $sigma_3$ is the determinant. The middle one, $sigma_2$ is the sum of the three two by two mini-determinants with both diagonal entries on the main diagonal.
With your original matrix, we have
$$
sigma_2 =
left|
begin{array}{rr}
3 & 2 \
-3 & -4 \
end{array}
right| +
left|
begin{array}{rr}
3 & -3 \
-1 & 5 \
end{array}
right| +
left|
begin{array}{rr}
-4 & 9 \
-2 & 5 \
end{array}
right| = -6 + 12 -2 = 4
$$



We have the familiar trace $sigma_1 = 4,$ and the determinant $sigma_3 = 0$ by that Rule of Sarrus.
$$ lambda^3 - 4 lambda^2 + 4 lambda - 0 = lambda (lambda- 2 )^2 $$



=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=



Note that the same method, done carefully, works in any dimension. In dimension 4, we get $$ lambda^4 - sigma_1 lambda^3 + sigma_2 lambda^2 - sigma_3 lambda + sigma_4. $$
Once again $sigma_1$ is the trace and $sigma_4$ the determinant. This time we have two middle terms. Now $sigma_2$ is the sum of (determinants of) six little 2 by 2 matrices along the diagonal, while $sigma_3$ is the sum of (determinants of) four little 3 by 3 matrices along the diagonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what do you refer to as a "Trace"? Why do you also alterante the (-) and (+) signs? My book does not appear to do that.
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:53













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Here is a simple computation:
begin{align}
det(lambda I - A)& = begin{vmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=begin{vmatrix}
lambda - 2 & 0 & lambda - 2 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}\& =(lambda - 2 )begin{vmatrix}
1 & 0 & 1 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=(lambda - 2 )biggl(begin{vmatrix}
lambda + 4 & -9 \
2 & lambda - 5 \
end{vmatrix}+begin{vmatrix}
3 & lambda + 4 \
1 & 2 \
end{vmatrix}biggl)\
&=(lambda - 2 )Bigl(bigl(
(lambda + 4)(lambda - 5)+18bigr)
+ (6 -lambda - 4) Bigl)=(lambda - 2 )(lambda^2-2lambda)\
&=lambda(lambda - 2)^2.
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from 3 to $lambda - 2$ in the top right corner, $A_13$?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:10








  • 1




    $begingroup$
    I added the third row to the first.
    $endgroup$
    – Bernard
    Dec 6 '18 at 21:19










  • $begingroup$
    So adding rows to each other will not change characteristic equation, but scaling rows will?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 23:34










  • $begingroup$
    It doesn't change any determinant, and scalind a row or a column scales the determinant, since a determinant is linear w.r.t. each row or column.
    $endgroup$
    – Bernard
    Dec 6 '18 at 23:37
















2












$begingroup$

Here is a simple computation:
begin{align}
det(lambda I - A)& = begin{vmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=begin{vmatrix}
lambda - 2 & 0 & lambda - 2 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}\& =(lambda - 2 )begin{vmatrix}
1 & 0 & 1 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=(lambda - 2 )biggl(begin{vmatrix}
lambda + 4 & -9 \
2 & lambda - 5 \
end{vmatrix}+begin{vmatrix}
3 & lambda + 4 \
1 & 2 \
end{vmatrix}biggl)\
&=(lambda - 2 )Bigl(bigl(
(lambda + 4)(lambda - 5)+18bigr)
+ (6 -lambda - 4) Bigl)=(lambda - 2 )(lambda^2-2lambda)\
&=lambda(lambda - 2)^2.
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from 3 to $lambda - 2$ in the top right corner, $A_13$?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:10








  • 1




    $begingroup$
    I added the third row to the first.
    $endgroup$
    – Bernard
    Dec 6 '18 at 21:19










  • $begingroup$
    So adding rows to each other will not change characteristic equation, but scaling rows will?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 23:34










  • $begingroup$
    It doesn't change any determinant, and scalind a row or a column scales the determinant, since a determinant is linear w.r.t. each row or column.
    $endgroup$
    – Bernard
    Dec 6 '18 at 23:37














2












2








2





$begingroup$

Here is a simple computation:
begin{align}
det(lambda I - A)& = begin{vmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=begin{vmatrix}
lambda - 2 & 0 & lambda - 2 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}\& =(lambda - 2 )begin{vmatrix}
1 & 0 & 1 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=(lambda - 2 )biggl(begin{vmatrix}
lambda + 4 & -9 \
2 & lambda - 5 \
end{vmatrix}+begin{vmatrix}
3 & lambda + 4 \
1 & 2 \
end{vmatrix}biggl)\
&=(lambda - 2 )Bigl(bigl(
(lambda + 4)(lambda - 5)+18bigr)
+ (6 -lambda - 4) Bigl)=(lambda - 2 )(lambda^2-2lambda)\
&=lambda(lambda - 2)^2.
end{align}






share|cite|improve this answer









$endgroup$



Here is a simple computation:
begin{align}
det(lambda I - A)& = begin{vmatrix}
lambda - 3 & -2 & 3 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=begin{vmatrix}
lambda - 2 & 0 & lambda - 2 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}\& =(lambda - 2 )begin{vmatrix}
1 & 0 & 1 \
3 & lambda + 4 & -9 \
1 & 2 & lambda - 5 \
end{vmatrix}=(lambda - 2 )biggl(begin{vmatrix}
lambda + 4 & -9 \
2 & lambda - 5 \
end{vmatrix}+begin{vmatrix}
3 & lambda + 4 \
1 & 2 \
end{vmatrix}biggl)\
&=(lambda - 2 )Bigl(bigl(
(lambda + 4)(lambda - 5)+18bigr)
+ (6 -lambda - 4) Bigl)=(lambda - 2 )(lambda^2-2lambda)\
&=lambda(lambda - 2)^2.
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 19:33









BernardBernard

122k741116




122k741116












  • $begingroup$
    How do you go from 3 to $lambda - 2$ in the top right corner, $A_13$?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:10








  • 1




    $begingroup$
    I added the third row to the first.
    $endgroup$
    – Bernard
    Dec 6 '18 at 21:19










  • $begingroup$
    So adding rows to each other will not change characteristic equation, but scaling rows will?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 23:34










  • $begingroup$
    It doesn't change any determinant, and scalind a row or a column scales the determinant, since a determinant is linear w.r.t. each row or column.
    $endgroup$
    – Bernard
    Dec 6 '18 at 23:37


















  • $begingroup$
    How do you go from 3 to $lambda - 2$ in the top right corner, $A_13$?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:10








  • 1




    $begingroup$
    I added the third row to the first.
    $endgroup$
    – Bernard
    Dec 6 '18 at 21:19










  • $begingroup$
    So adding rows to each other will not change characteristic equation, but scaling rows will?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 23:34










  • $begingroup$
    It doesn't change any determinant, and scalind a row or a column scales the determinant, since a determinant is linear w.r.t. each row or column.
    $endgroup$
    – Bernard
    Dec 6 '18 at 23:37
















$begingroup$
How do you go from 3 to $lambda - 2$ in the top right corner, $A_13$?
$endgroup$
– Evan Kim
Dec 6 '18 at 20:10






$begingroup$
How do you go from 3 to $lambda - 2$ in the top right corner, $A_13$?
$endgroup$
– Evan Kim
Dec 6 '18 at 20:10






1




1




$begingroup$
I added the third row to the first.
$endgroup$
– Bernard
Dec 6 '18 at 21:19




$begingroup$
I added the third row to the first.
$endgroup$
– Bernard
Dec 6 '18 at 21:19












$begingroup$
So adding rows to each other will not change characteristic equation, but scaling rows will?
$endgroup$
– Evan Kim
Dec 6 '18 at 23:34




$begingroup$
So adding rows to each other will not change characteristic equation, but scaling rows will?
$endgroup$
– Evan Kim
Dec 6 '18 at 23:34












$begingroup$
It doesn't change any determinant, and scalind a row or a column scales the determinant, since a determinant is linear w.r.t. each row or column.
$endgroup$
– Bernard
Dec 6 '18 at 23:37




$begingroup$
It doesn't change any determinant, and scalind a row or a column scales the determinant, since a determinant is linear w.r.t. each row or column.
$endgroup$
– Bernard
Dec 6 '18 at 23:37











0












$begingroup$

I would do what you did: to apply the rule of Sarrus. But the polynomial that you got cannot be correct. First of all, it is not monic. On the other hand, if you sum all entries of each row of the matrix, you get $2$. But that means that $2$ is an eigenvalue and that $(1,1,1)$ is an eigenvector of $A$. However, $2$ is not a root of your polynomial.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What do you mean by "if you sum all entries of each line on the matrix, you get 2." Is each line referencing the row or the column?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:04










  • $begingroup$
    I meant row. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 20:06










  • $begingroup$
    @JoséCarlos, some students really liked the recipe in my answer, although this OP seems less than impressed. Just a way to reduce errors that I have not usually seen in textbooks.
    $endgroup$
    – Will Jagy
    Dec 7 '18 at 1:00
















0












$begingroup$

I would do what you did: to apply the rule of Sarrus. But the polynomial that you got cannot be correct. First of all, it is not monic. On the other hand, if you sum all entries of each row of the matrix, you get $2$. But that means that $2$ is an eigenvalue and that $(1,1,1)$ is an eigenvector of $A$. However, $2$ is not a root of your polynomial.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What do you mean by "if you sum all entries of each line on the matrix, you get 2." Is each line referencing the row or the column?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:04










  • $begingroup$
    I meant row. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 20:06










  • $begingroup$
    @JoséCarlos, some students really liked the recipe in my answer, although this OP seems less than impressed. Just a way to reduce errors that I have not usually seen in textbooks.
    $endgroup$
    – Will Jagy
    Dec 7 '18 at 1:00














0












0








0





$begingroup$

I would do what you did: to apply the rule of Sarrus. But the polynomial that you got cannot be correct. First of all, it is not monic. On the other hand, if you sum all entries of each row of the matrix, you get $2$. But that means that $2$ is an eigenvalue and that $(1,1,1)$ is an eigenvector of $A$. However, $2$ is not a root of your polynomial.






share|cite|improve this answer











$endgroup$



I would do what you did: to apply the rule of Sarrus. But the polynomial that you got cannot be correct. First of all, it is not monic. On the other hand, if you sum all entries of each row of the matrix, you get $2$. But that means that $2$ is an eigenvalue and that $(1,1,1)$ is an eigenvector of $A$. However, $2$ is not a root of your polynomial.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 20:06

























answered Dec 6 '18 at 18:53









José Carlos SantosJosé Carlos Santos

166k22132235




166k22132235












  • $begingroup$
    What do you mean by "if you sum all entries of each line on the matrix, you get 2." Is each line referencing the row or the column?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:04










  • $begingroup$
    I meant row. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 20:06










  • $begingroup$
    @JoséCarlos, some students really liked the recipe in my answer, although this OP seems less than impressed. Just a way to reduce errors that I have not usually seen in textbooks.
    $endgroup$
    – Will Jagy
    Dec 7 '18 at 1:00


















  • $begingroup$
    What do you mean by "if you sum all entries of each line on the matrix, you get 2." Is each line referencing the row or the column?
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:04










  • $begingroup$
    I meant row. I've edited my answer.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 20:06










  • $begingroup$
    @JoséCarlos, some students really liked the recipe in my answer, although this OP seems less than impressed. Just a way to reduce errors that I have not usually seen in textbooks.
    $endgroup$
    – Will Jagy
    Dec 7 '18 at 1:00
















$begingroup$
What do you mean by "if you sum all entries of each line on the matrix, you get 2." Is each line referencing the row or the column?
$endgroup$
– Evan Kim
Dec 6 '18 at 20:04




$begingroup$
What do you mean by "if you sum all entries of each line on the matrix, you get 2." Is each line referencing the row or the column?
$endgroup$
– Evan Kim
Dec 6 '18 at 20:04












$begingroup$
I meant row. I've edited my answer.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 20:06




$begingroup$
I meant row. I've edited my answer.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 20:06












$begingroup$
@JoséCarlos, some students really liked the recipe in my answer, although this OP seems less than impressed. Just a way to reduce errors that I have not usually seen in textbooks.
$endgroup$
– Will Jagy
Dec 7 '18 at 1:00




$begingroup$
@JoséCarlos, some students really liked the recipe in my answer, although this OP seems less than impressed. Just a way to reduce errors that I have not usually seen in textbooks.
$endgroup$
– Will Jagy
Dec 7 '18 at 1:00











0












$begingroup$

For 2 by 2 and 3 by 3 there are worthwhile recipes for the characteristic polynomial. For 2 by 2, it is $lambda^2 - T lambda + D,$ where $T$ is the trace and $D$ is the determinant.



more in a few minutes.



For 3 by 3, it is $$ lambda^3 - sigma_1 lambda^2 + sigma_2 lambda - sigma_3 $$
Here $sigma_1$ is the trace and $sigma_3$ is the determinant. The middle one, $sigma_2$ is the sum of the three two by two mini-determinants with both diagonal entries on the main diagonal.
With your original matrix, we have
$$
sigma_2 =
left|
begin{array}{rr}
3 & 2 \
-3 & -4 \
end{array}
right| +
left|
begin{array}{rr}
3 & -3 \
-1 & 5 \
end{array}
right| +
left|
begin{array}{rr}
-4 & 9 \
-2 & 5 \
end{array}
right| = -6 + 12 -2 = 4
$$



We have the familiar trace $sigma_1 = 4,$ and the determinant $sigma_3 = 0$ by that Rule of Sarrus.
$$ lambda^3 - 4 lambda^2 + 4 lambda - 0 = lambda (lambda- 2 )^2 $$



=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=



Note that the same method, done carefully, works in any dimension. In dimension 4, we get $$ lambda^4 - sigma_1 lambda^3 + sigma_2 lambda^2 - sigma_3 lambda + sigma_4. $$
Once again $sigma_1$ is the trace and $sigma_4$ the determinant. This time we have two middle terms. Now $sigma_2$ is the sum of (determinants of) six little 2 by 2 matrices along the diagonal, while $sigma_3$ is the sum of (determinants of) four little 3 by 3 matrices along the diagonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what do you refer to as a "Trace"? Why do you also alterante the (-) and (+) signs? My book does not appear to do that.
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:53


















0












$begingroup$

For 2 by 2 and 3 by 3 there are worthwhile recipes for the characteristic polynomial. For 2 by 2, it is $lambda^2 - T lambda + D,$ where $T$ is the trace and $D$ is the determinant.



more in a few minutes.



For 3 by 3, it is $$ lambda^3 - sigma_1 lambda^2 + sigma_2 lambda - sigma_3 $$
Here $sigma_1$ is the trace and $sigma_3$ is the determinant. The middle one, $sigma_2$ is the sum of the three two by two mini-determinants with both diagonal entries on the main diagonal.
With your original matrix, we have
$$
sigma_2 =
left|
begin{array}{rr}
3 & 2 \
-3 & -4 \
end{array}
right| +
left|
begin{array}{rr}
3 & -3 \
-1 & 5 \
end{array}
right| +
left|
begin{array}{rr}
-4 & 9 \
-2 & 5 \
end{array}
right| = -6 + 12 -2 = 4
$$



We have the familiar trace $sigma_1 = 4,$ and the determinant $sigma_3 = 0$ by that Rule of Sarrus.
$$ lambda^3 - 4 lambda^2 + 4 lambda - 0 = lambda (lambda- 2 )^2 $$



=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=



Note that the same method, done carefully, works in any dimension. In dimension 4, we get $$ lambda^4 - sigma_1 lambda^3 + sigma_2 lambda^2 - sigma_3 lambda + sigma_4. $$
Once again $sigma_1$ is the trace and $sigma_4$ the determinant. This time we have two middle terms. Now $sigma_2$ is the sum of (determinants of) six little 2 by 2 matrices along the diagonal, while $sigma_3$ is the sum of (determinants of) four little 3 by 3 matrices along the diagonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what do you refer to as a "Trace"? Why do you also alterante the (-) and (+) signs? My book does not appear to do that.
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:53
















0












0








0





$begingroup$

For 2 by 2 and 3 by 3 there are worthwhile recipes for the characteristic polynomial. For 2 by 2, it is $lambda^2 - T lambda + D,$ where $T$ is the trace and $D$ is the determinant.



more in a few minutes.



For 3 by 3, it is $$ lambda^3 - sigma_1 lambda^2 + sigma_2 lambda - sigma_3 $$
Here $sigma_1$ is the trace and $sigma_3$ is the determinant. The middle one, $sigma_2$ is the sum of the three two by two mini-determinants with both diagonal entries on the main diagonal.
With your original matrix, we have
$$
sigma_2 =
left|
begin{array}{rr}
3 & 2 \
-3 & -4 \
end{array}
right| +
left|
begin{array}{rr}
3 & -3 \
-1 & 5 \
end{array}
right| +
left|
begin{array}{rr}
-4 & 9 \
-2 & 5 \
end{array}
right| = -6 + 12 -2 = 4
$$



We have the familiar trace $sigma_1 = 4,$ and the determinant $sigma_3 = 0$ by that Rule of Sarrus.
$$ lambda^3 - 4 lambda^2 + 4 lambda - 0 = lambda (lambda- 2 )^2 $$



=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=



Note that the same method, done carefully, works in any dimension. In dimension 4, we get $$ lambda^4 - sigma_1 lambda^3 + sigma_2 lambda^2 - sigma_3 lambda + sigma_4. $$
Once again $sigma_1$ is the trace and $sigma_4$ the determinant. This time we have two middle terms. Now $sigma_2$ is the sum of (determinants of) six little 2 by 2 matrices along the diagonal, while $sigma_3$ is the sum of (determinants of) four little 3 by 3 matrices along the diagonal.






share|cite|improve this answer











$endgroup$



For 2 by 2 and 3 by 3 there are worthwhile recipes for the characteristic polynomial. For 2 by 2, it is $lambda^2 - T lambda + D,$ where $T$ is the trace and $D$ is the determinant.



more in a few minutes.



For 3 by 3, it is $$ lambda^3 - sigma_1 lambda^2 + sigma_2 lambda - sigma_3 $$
Here $sigma_1$ is the trace and $sigma_3$ is the determinant. The middle one, $sigma_2$ is the sum of the three two by two mini-determinants with both diagonal entries on the main diagonal.
With your original matrix, we have
$$
sigma_2 =
left|
begin{array}{rr}
3 & 2 \
-3 & -4 \
end{array}
right| +
left|
begin{array}{rr}
3 & -3 \
-1 & 5 \
end{array}
right| +
left|
begin{array}{rr}
-4 & 9 \
-2 & 5 \
end{array}
right| = -6 + 12 -2 = 4
$$



We have the familiar trace $sigma_1 = 4,$ and the determinant $sigma_3 = 0$ by that Rule of Sarrus.
$$ lambda^3 - 4 lambda^2 + 4 lambda - 0 = lambda (lambda- 2 )^2 $$



=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=



Note that the same method, done carefully, works in any dimension. In dimension 4, we get $$ lambda^4 - sigma_1 lambda^3 + sigma_2 lambda^2 - sigma_3 lambda + sigma_4. $$
Once again $sigma_1$ is the trace and $sigma_4$ the determinant. This time we have two middle terms. Now $sigma_2$ is the sum of (determinants of) six little 2 by 2 matrices along the diagonal, while $sigma_3$ is the sum of (determinants of) four little 3 by 3 matrices along the diagonal.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 20:06

























answered Dec 6 '18 at 18:48









Will JagyWill Jagy

104k5102201




104k5102201












  • $begingroup$
    what do you refer to as a "Trace"? Why do you also alterante the (-) and (+) signs? My book does not appear to do that.
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:53




















  • $begingroup$
    what do you refer to as a "Trace"? Why do you also alterante the (-) and (+) signs? My book does not appear to do that.
    $endgroup$
    – Evan Kim
    Dec 6 '18 at 20:53


















$begingroup$
what do you refer to as a "Trace"? Why do you also alterante the (-) and (+) signs? My book does not appear to do that.
$endgroup$
– Evan Kim
Dec 6 '18 at 20:53






$begingroup$
what do you refer to as a "Trace"? Why do you also alterante the (-) and (+) signs? My book does not appear to do that.
$endgroup$
– Evan Kim
Dec 6 '18 at 20:53




















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