Compute the characteristic equation 3x3 matrix
$begingroup$
Can someone help me explain how to solve this problem below:
Something like this?
But I'm not sure.
linear-algebra
asked Mar 31 '16 at 12:58
AdiTAdiT
148119
$endgroup$
add a comment |
$begingroup$
Can someone help me explain how to solve this problem below:
Something like this?
But I'm not sure.
linear-algebra
asked Mar 31 '16 at 12:58
AdiTAdiT
148119
$endgroup$
$begingroup$
Yes, that is correct.
$endgroup$
– Inazuma
Mar 31 '16 at 12:59
2
$begingroup$
Do it like in this MSE-question.
$endgroup$
– Dietrich Burde
Mar 31 '16 at 13:05
add a comment |
$begingroup$
Can someone help me explain how to solve this problem below:
Something like this?
But I'm not sure.
linear-algebra
asked Mar 31 '16 at 12:58
AdiTAdiT
148119
$endgroup$
Can someone help me explain how to solve this problem below:
Something like this?
But I'm not sure.
linear-algebra
linear-algebra
asked Mar 31 '16 at 12:58
AdiTAdiT
148119
asked Mar 31 '16 at 12:58
AdiTAdiT
148119
asked Mar 31 '16 at 12:58
AdiTAdiT
148119
asked Mar 31 '16 at 12:58
AdiTAdiT
148119
asked Mar 31 '16 at 12:58
AdiTAdiT
148119
148119
$begingroup$
Yes, that is correct.
$endgroup$
– Inazuma
Mar 31 '16 at 12:59
2
$begingroup$
Do it like in this MSE-question.
$endgroup$
– Dietrich Burde
Mar 31 '16 at 13:05
add a comment |
$begingroup$
Yes, that is correct.
$endgroup$
– Inazuma
Mar 31 '16 at 12:59
2
$begingroup$
Do it like in this MSE-question.
$endgroup$
– Dietrich Burde
Mar 31 '16 at 13:05
$begingroup$
Yes, that is correct.
$endgroup$
– Inazuma
Mar 31 '16 at 12:59
$begingroup$
Yes, that is correct.
$endgroup$
– Inazuma
Mar 31 '16 at 12:59
2
2
$begingroup$
Do it like in this MSE-question.
$endgroup$
– Dietrich Burde
Mar 31 '16 at 13:05
$begingroup$
Do it like in this MSE-question.
$endgroup$
– Dietrich Burde
Mar 31 '16 at 13:05
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The easy and quick way to compute the characteristic equation of 3x3 matrix is to use the formulae
$$x^3-tr(A)x^2+(A_{11}+A_{22}+A_{33})x-det(A)=0$$
For given matrix $$tr(A)=4, A_{11}(cofa_{11})=3, A_{22}(cofa_{22})=1, A_{33}(cofa_{33})=1, det(A)=2$$
so the char equation will be
$x^3-4x^2+5x-2=0$
answered Mar 31 '16 at 13:07
Rayees AhmadRayees Ahmad
884413
$endgroup$
$begingroup$
Your formula is incorrect: $A_{11} + A_{22} + A_{33}$ is the same thing as the trace, which is not the linear coefficient. The correct formula is given here.
$endgroup$
– André 3000
Mar 31 '16 at 13:15
2
$begingroup$
Sir these are cofactors of diogonal elements not the elements of matrix
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 13:32
1
$begingroup$
Oh, my mistake. You should probably change your notation: for a matrix $A$, $A_{ij}$ typically denotes the $i,j$ entry.
$endgroup$
– André 3000
Mar 31 '16 at 15:58
2
$begingroup$
For elements of matrix we use small alphabets/
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 16:51
1
$begingroup$
In any case, I think it is a good idea to include what the $A_{ij}$ denote in your answer.
$endgroup$
– André 3000
Mar 31 '16 at 17:04
|
show 1 more comment
$begingroup$
Quick way: Note that sum of each column of $A$ is $2$, hence $lambda_1=2$ is an eigenvalue of $A$.Let $lambda_1$ and $lambda_2$ be other eigenvalues,using trace and determinant $lambda_1 +lambda_2=2$ and $lambda_1 lambda_2=1$,hence $lambda_1=lambda_2=1$
Hence Characteristic polynomial $p(x)$ is $$p(x)=(x-1)^2(x-2)$$
answered Mar 31 '16 at 13:11
Arpit KansalArpit Kansal
6,97911135
$endgroup$
1
$begingroup$
But you don't show how to compute the characteristic equation?
$endgroup$
– AdiT
Mar 31 '16 at 14:01
$begingroup$
@AdiT Once you know your the eigenvalues, you know the factorization of the characteristic polynomial.
$endgroup$
– André 3000
Mar 31 '16 at 17:02
$begingroup$
Could you elaborate on this method? I've never heard of any result that relates the sum of the columns to an eigenvalue. For the rest, I think I follow your reasoning for the trace ($2 + lambda_1 + lambda_2 = 4 ; Leftrightarrow ; lambda_1 + lambda_2 = 2$), but I don't see how you identified the determinant without computation. Most curious about the "sum of each column" part, though.
$endgroup$
– Rax Adaam
Nov 25 '18 at 3:52
1
$begingroup$
@RaxAdaam: Consider transpose of the matrix and multiply this transposed matrix with the vector $(1,1,1,...1)$. Finally recall that $A$ and $A^T$ have same eigenvalues
$endgroup$
– Arpit Kansal
Dec 3 '18 at 10:41
add a comment |
$begingroup$
The characteristic equation is used to find the eigenvalues of a square matrix A.
First: Know that an eigenvector of some square matrix A is a non-zero vector x such that Ax = λx.
Second: Through standard mathematical operations we can go from this: Ax = λx, to this: (A - λI)x = 0
The solutions to the equation det(A - λI) = 0 will yield your eigenvalues.
The previously mentioned equation is the characteristic equation.
-- SKIP TO THIS POINT FOR THE SHORT ANSWER --
Given some square matrix A, the characteristic equation is det(A - λI) = 0.
Example:
$$ A = begin{bmatrix}
a & b \
c & d \
end{bmatrix}$$$$ λI = begin{bmatrix}
λ & 0 \
0 & λ \
end{bmatrix}$$
$$ A - λI = begin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix}$$
$$detbegin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix} = 0$$ Characteristic Equation: $$(a-λ)(d-λ) - (c)(b) = 0$$
Source Cited: Lay, David C., et al. Linear Algebra and Its Applications, Fifth Edition. Pearson, 2016.
answered Dec 5 '18 at 21:53
sethledsethled
113
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add a comment |
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That is incorrect. The characteristic equation comes from computing the $lambda$ values necessary to give you a matrix determinant equal to zero.
$$begin{vmatrix}
-lambda & -1 & -1 \ 1 & 2 - lambda & 1 \ 1 & 1 & 2- lambda
end{vmatrix} = 0
$$
Notice that for $lambda = 1$ all three columns become identical and we can come up with two eigenvectors. Coming up with the last eigenvalue should be easy if we use the fact that the trace of the matrix is the sum of the eigenvalues.
The eigenvalues for this matrix are thus $lambda_1 = 1, lambda_2 = 1, lambda_3 = 2$ as they make the matrix singular.
answered Mar 31 '16 at 13:04
Klint QinamiKlint Qinami
1,137510
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The easy and quick way to compute the characteristic equation of 3x3 matrix is to use the formulae
$$x^3-tr(A)x^2+(A_{11}+A_{22}+A_{33})x-det(A)=0$$
For given matrix $$tr(A)=4, A_{11}(cofa_{11})=3, A_{22}(cofa_{22})=1, A_{33}(cofa_{33})=1, det(A)=2$$
so the char equation will be
$x^3-4x^2+5x-2=0$
answered Mar 31 '16 at 13:07
Rayees AhmadRayees Ahmad
884413
$endgroup$
$begingroup$
Your formula is incorrect: $A_{11} + A_{22} + A_{33}$ is the same thing as the trace, which is not the linear coefficient. The correct formula is given here.
$endgroup$
– André 3000
Mar 31 '16 at 13:15
2
$begingroup$
Sir these are cofactors of diogonal elements not the elements of matrix
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 13:32
1
$begingroup$
Oh, my mistake. You should probably change your notation: for a matrix $A$, $A_{ij}$ typically denotes the $i,j$ entry.
$endgroup$
– André 3000
Mar 31 '16 at 15:58
2
$begingroup$
For elements of matrix we use small alphabets/
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 16:51
1
$begingroup$
In any case, I think it is a good idea to include what the $A_{ij}$ denote in your answer.
$endgroup$
– André 3000
Mar 31 '16 at 17:04
|
show 1 more comment
$begingroup$
The easy and quick way to compute the characteristic equation of 3x3 matrix is to use the formulae
$$x^3-tr(A)x^2+(A_{11}+A_{22}+A_{33})x-det(A)=0$$
For given matrix $$tr(A)=4, A_{11}(cofa_{11})=3, A_{22}(cofa_{22})=1, A_{33}(cofa_{33})=1, det(A)=2$$
so the char equation will be
$x^3-4x^2+5x-2=0$
answered Mar 31 '16 at 13:07
Rayees AhmadRayees Ahmad
884413
$endgroup$
$begingroup$
Your formula is incorrect: $A_{11} + A_{22} + A_{33}$ is the same thing as the trace, which is not the linear coefficient. The correct formula is given here.
$endgroup$
– André 3000
Mar 31 '16 at 13:15
2
$begingroup$
Sir these are cofactors of diogonal elements not the elements of matrix
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 13:32
1
$begingroup$
Oh, my mistake. You should probably change your notation: for a matrix $A$, $A_{ij}$ typically denotes the $i,j$ entry.
$endgroup$
– André 3000
Mar 31 '16 at 15:58
2
$begingroup$
For elements of matrix we use small alphabets/
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 16:51
1
$begingroup$
In any case, I think it is a good idea to include what the $A_{ij}$ denote in your answer.
$endgroup$
– André 3000
Mar 31 '16 at 17:04
|
show 1 more comment
$begingroup$
The easy and quick way to compute the characteristic equation of 3x3 matrix is to use the formulae
$$x^3-tr(A)x^2+(A_{11}+A_{22}+A_{33})x-det(A)=0$$
For given matrix $$tr(A)=4, A_{11}(cofa_{11})=3, A_{22}(cofa_{22})=1, A_{33}(cofa_{33})=1, det(A)=2$$
so the char equation will be
$x^3-4x^2+5x-2=0$
answered Mar 31 '16 at 13:07
Rayees AhmadRayees Ahmad
884413
$endgroup$
The easy and quick way to compute the characteristic equation of 3x3 matrix is to use the formulae
$$x^3-tr(A)x^2+(A_{11}+A_{22}+A_{33})x-det(A)=0$$
For given matrix $$tr(A)=4, A_{11}(cofa_{11})=3, A_{22}(cofa_{22})=1, A_{33}(cofa_{33})=1, det(A)=2$$
so the char equation will be
$x^3-4x^2+5x-2=0$
answered Mar 31 '16 at 13:07
Rayees AhmadRayees Ahmad
884413
edited Mar 31 '16 at 17:13
answered Mar 31 '16 at 13:07
Rayees AhmadRayees Ahmad
884413
answered Mar 31 '16 at 13:07
Rayees AhmadRayees Ahmad
884413
answered Mar 31 '16 at 13:07
Rayees AhmadRayees Ahmad
884413
884413
$begingroup$
Your formula is incorrect: $A_{11} + A_{22} + A_{33}$ is the same thing as the trace, which is not the linear coefficient. The correct formula is given here.
$endgroup$
– André 3000
Mar 31 '16 at 13:15
2
$begingroup$
Sir these are cofactors of diogonal elements not the elements of matrix
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 13:32
1
$begingroup$
Oh, my mistake. You should probably change your notation: for a matrix $A$, $A_{ij}$ typically denotes the $i,j$ entry.
$endgroup$
– André 3000
Mar 31 '16 at 15:58
2
$begingroup$
For elements of matrix we use small alphabets/
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 16:51
1
$begingroup$
In any case, I think it is a good idea to include what the $A_{ij}$ denote in your answer.
$endgroup$
– André 3000
Mar 31 '16 at 17:04
|
show 1 more comment
$begingroup$
Your formula is incorrect: $A_{11} + A_{22} + A_{33}$ is the same thing as the trace, which is not the linear coefficient. The correct formula is given here.
$endgroup$
– André 3000
Mar 31 '16 at 13:15
2
$begingroup$
Sir these are cofactors of diogonal elements not the elements of matrix
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 13:32
1
$begingroup$
Oh, my mistake. You should probably change your notation: for a matrix $A$, $A_{ij}$ typically denotes the $i,j$ entry.
$endgroup$
– André 3000
Mar 31 '16 at 15:58
2
$begingroup$
For elements of matrix we use small alphabets/
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 16:51
1
$begingroup$
In any case, I think it is a good idea to include what the $A_{ij}$ denote in your answer.
$endgroup$
– André 3000
Mar 31 '16 at 17:04
$begingroup$
Your formula is incorrect: $A_{11} + A_{22} + A_{33}$ is the same thing as the trace, which is not the linear coefficient. The correct formula is given here.
$endgroup$
– André 3000
Mar 31 '16 at 13:15
$begingroup$
Your formula is incorrect: $A_{11} + A_{22} + A_{33}$ is the same thing as the trace, which is not the linear coefficient. The correct formula is given here.
$endgroup$
– André 3000
Mar 31 '16 at 13:15
2
2
$begingroup$
Sir these are cofactors of diogonal elements not the elements of matrix
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 13:32
$begingroup$
Sir these are cofactors of diogonal elements not the elements of matrix
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 13:32
1
1
$begingroup$
Oh, my mistake. You should probably change your notation: for a matrix $A$, $A_{ij}$ typically denotes the $i,j$ entry.
$endgroup$
– André 3000
Mar 31 '16 at 15:58
$begingroup$
Oh, my mistake. You should probably change your notation: for a matrix $A$, $A_{ij}$ typically denotes the $i,j$ entry.
$endgroup$
– André 3000
Mar 31 '16 at 15:58
2
2
$begingroup$
For elements of matrix we use small alphabets/
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 16:51
$begingroup$
For elements of matrix we use small alphabets/
$endgroup$
– Rayees Ahmad
Mar 31 '16 at 16:51
1
1
$begingroup$
In any case, I think it is a good idea to include what the $A_{ij}$ denote in your answer.
$endgroup$
– André 3000
Mar 31 '16 at 17:04
$begingroup$
In any case, I think it is a good idea to include what the $A_{ij}$ denote in your answer.
$endgroup$
– André 3000
Mar 31 '16 at 17:04
|
show 1 more comment
$begingroup$
Quick way: Note that sum of each column of $A$ is $2$, hence $lambda_1=2$ is an eigenvalue of $A$.Let $lambda_1$ and $lambda_2$ be other eigenvalues,using trace and determinant $lambda_1 +lambda_2=2$ and $lambda_1 lambda_2=1$,hence $lambda_1=lambda_2=1$
Hence Characteristic polynomial $p(x)$ is $$p(x)=(x-1)^2(x-2)$$
answered Mar 31 '16 at 13:11
Arpit KansalArpit Kansal
6,97911135
$endgroup$
1
$begingroup$
But you don't show how to compute the characteristic equation?
$endgroup$
– AdiT
Mar 31 '16 at 14:01
$begingroup$
@AdiT Once you know your the eigenvalues, you know the factorization of the characteristic polynomial.
$endgroup$
– André 3000
Mar 31 '16 at 17:02
$begingroup$
Could you elaborate on this method? I've never heard of any result that relates the sum of the columns to an eigenvalue. For the rest, I think I follow your reasoning for the trace ($2 + lambda_1 + lambda_2 = 4 ; Leftrightarrow ; lambda_1 + lambda_2 = 2$), but I don't see how you identified the determinant without computation. Most curious about the "sum of each column" part, though.
$endgroup$
– Rax Adaam
Nov 25 '18 at 3:52
1
$begingroup$
@RaxAdaam: Consider transpose of the matrix and multiply this transposed matrix with the vector $(1,1,1,...1)$. Finally recall that $A$ and $A^T$ have same eigenvalues
$endgroup$
– Arpit Kansal
Dec 3 '18 at 10:41
add a comment |
$begingroup$
Quick way: Note that sum of each column of $A$ is $2$, hence $lambda_1=2$ is an eigenvalue of $A$.Let $lambda_1$ and $lambda_2$ be other eigenvalues,using trace and determinant $lambda_1 +lambda_2=2$ and $lambda_1 lambda_2=1$,hence $lambda_1=lambda_2=1$
Hence Characteristic polynomial $p(x)$ is $$p(x)=(x-1)^2(x-2)$$
answered Mar 31 '16 at 13:11
Arpit KansalArpit Kansal
6,97911135
$endgroup$
1
$begingroup$
But you don't show how to compute the characteristic equation?
$endgroup$
– AdiT
Mar 31 '16 at 14:01
$begingroup$
@AdiT Once you know your the eigenvalues, you know the factorization of the characteristic polynomial.
$endgroup$
– André 3000
Mar 31 '16 at 17:02
$begingroup$
Could you elaborate on this method? I've never heard of any result that relates the sum of the columns to an eigenvalue. For the rest, I think I follow your reasoning for the trace ($2 + lambda_1 + lambda_2 = 4 ; Leftrightarrow ; lambda_1 + lambda_2 = 2$), but I don't see how you identified the determinant without computation. Most curious about the "sum of each column" part, though.
$endgroup$
– Rax Adaam
Nov 25 '18 at 3:52
1
$begingroup$
@RaxAdaam: Consider transpose of the matrix and multiply this transposed matrix with the vector $(1,1,1,...1)$. Finally recall that $A$ and $A^T$ have same eigenvalues
$endgroup$
– Arpit Kansal
Dec 3 '18 at 10:41
add a comment |
$begingroup$
Quick way: Note that sum of each column of $A$ is $2$, hence $lambda_1=2$ is an eigenvalue of $A$.Let $lambda_1$ and $lambda_2$ be other eigenvalues,using trace and determinant $lambda_1 +lambda_2=2$ and $lambda_1 lambda_2=1$,hence $lambda_1=lambda_2=1$
Hence Characteristic polynomial $p(x)$ is $$p(x)=(x-1)^2(x-2)$$
answered Mar 31 '16 at 13:11
Arpit KansalArpit Kansal
6,97911135
$endgroup$
Quick way: Note that sum of each column of $A$ is $2$, hence $lambda_1=2$ is an eigenvalue of $A$.Let $lambda_1$ and $lambda_2$ be other eigenvalues,using trace and determinant $lambda_1 +lambda_2=2$ and $lambda_1 lambda_2=1$,hence $lambda_1=lambda_2=1$
Hence Characteristic polynomial $p(x)$ is $$p(x)=(x-1)^2(x-2)$$
answered Mar 31 '16 at 13:11
Arpit KansalArpit Kansal
6,97911135
edited Mar 31 '16 at 17:37
answered Mar 31 '16 at 13:11
Arpit KansalArpit Kansal
6,97911135
answered Mar 31 '16 at 13:11
Arpit KansalArpit Kansal
6,97911135
answered Mar 31 '16 at 13:11
Arpit KansalArpit Kansal
6,97911135
6,97911135
1
$begingroup$
But you don't show how to compute the characteristic equation?
$endgroup$
– AdiT
Mar 31 '16 at 14:01
$begingroup$
@AdiT Once you know your the eigenvalues, you know the factorization of the characteristic polynomial.
$endgroup$
– André 3000
Mar 31 '16 at 17:02
$begingroup$
Could you elaborate on this method? I've never heard of any result that relates the sum of the columns to an eigenvalue. For the rest, I think I follow your reasoning for the trace ($2 + lambda_1 + lambda_2 = 4 ; Leftrightarrow ; lambda_1 + lambda_2 = 2$), but I don't see how you identified the determinant without computation. Most curious about the "sum of each column" part, though.
$endgroup$
– Rax Adaam
Nov 25 '18 at 3:52
1
$begingroup$
@RaxAdaam: Consider transpose of the matrix and multiply this transposed matrix with the vector $(1,1,1,...1)$. Finally recall that $A$ and $A^T$ have same eigenvalues
$endgroup$
– Arpit Kansal
Dec 3 '18 at 10:41
add a comment |
1
$begingroup$
But you don't show how to compute the characteristic equation?
$endgroup$
– AdiT
Mar 31 '16 at 14:01
$begingroup$
@AdiT Once you know your the eigenvalues, you know the factorization of the characteristic polynomial.
$endgroup$
– André 3000
Mar 31 '16 at 17:02
$begingroup$
Could you elaborate on this method? I've never heard of any result that relates the sum of the columns to an eigenvalue. For the rest, I think I follow your reasoning for the trace ($2 + lambda_1 + lambda_2 = 4 ; Leftrightarrow ; lambda_1 + lambda_2 = 2$), but I don't see how you identified the determinant without computation. Most curious about the "sum of each column" part, though.
$endgroup$
– Rax Adaam
Nov 25 '18 at 3:52
1
$begingroup$
@RaxAdaam: Consider transpose of the matrix and multiply this transposed matrix with the vector $(1,1,1,...1)$. Finally recall that $A$ and $A^T$ have same eigenvalues
$endgroup$
– Arpit Kansal
Dec 3 '18 at 10:41
1
1
$begingroup$
But you don't show how to compute the characteristic equation?
$endgroup$
– AdiT
Mar 31 '16 at 14:01
$begingroup$
But you don't show how to compute the characteristic equation?
$endgroup$
– AdiT
Mar 31 '16 at 14:01
$begingroup$
@AdiT Once you know your the eigenvalues, you know the factorization of the characteristic polynomial.
$endgroup$
– André 3000
Mar 31 '16 at 17:02
$begingroup$
@AdiT Once you know your the eigenvalues, you know the factorization of the characteristic polynomial.
$endgroup$
– André 3000
Mar 31 '16 at 17:02
$begingroup$
Could you elaborate on this method? I've never heard of any result that relates the sum of the columns to an eigenvalue. For the rest, I think I follow your reasoning for the trace ($2 + lambda_1 + lambda_2 = 4 ; Leftrightarrow ; lambda_1 + lambda_2 = 2$), but I don't see how you identified the determinant without computation. Most curious about the "sum of each column" part, though.
$endgroup$
– Rax Adaam
Nov 25 '18 at 3:52
$begingroup$
Could you elaborate on this method? I've never heard of any result that relates the sum of the columns to an eigenvalue. For the rest, I think I follow your reasoning for the trace ($2 + lambda_1 + lambda_2 = 4 ; Leftrightarrow ; lambda_1 + lambda_2 = 2$), but I don't see how you identified the determinant without computation. Most curious about the "sum of each column" part, though.
$endgroup$
– Rax Adaam
Nov 25 '18 at 3:52
1
1
$begingroup$
@RaxAdaam: Consider transpose of the matrix and multiply this transposed matrix with the vector $(1,1,1,...1)$. Finally recall that $A$ and $A^T$ have same eigenvalues
$endgroup$
– Arpit Kansal
Dec 3 '18 at 10:41
$begingroup$
@RaxAdaam: Consider transpose of the matrix and multiply this transposed matrix with the vector $(1,1,1,...1)$. Finally recall that $A$ and $A^T$ have same eigenvalues
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– Arpit Kansal
Dec 3 '18 at 10:41
add a comment |
$begingroup$
The characteristic equation is used to find the eigenvalues of a square matrix A.
First: Know that an eigenvector of some square matrix A is a non-zero vector x such that Ax = λx.
Second: Through standard mathematical operations we can go from this: Ax = λx, to this: (A - λI)x = 0
The solutions to the equation det(A - λI) = 0 will yield your eigenvalues.
The previously mentioned equation is the characteristic equation.
-- SKIP TO THIS POINT FOR THE SHORT ANSWER --
Given some square matrix A, the characteristic equation is det(A - λI) = 0.
Example:
$$ A = begin{bmatrix}
a & b \
c & d \
end{bmatrix}$$$$ λI = begin{bmatrix}
λ & 0 \
0 & λ \
end{bmatrix}$$
$$ A - λI = begin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix}$$
$$detbegin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix} = 0$$ Characteristic Equation: $$(a-λ)(d-λ) - (c)(b) = 0$$
Source Cited: Lay, David C., et al. Linear Algebra and Its Applications, Fifth Edition. Pearson, 2016.
answered Dec 5 '18 at 21:53
sethledsethled
113
$endgroup$
add a comment |
$begingroup$
The characteristic equation is used to find the eigenvalues of a square matrix A.
First: Know that an eigenvector of some square matrix A is a non-zero vector x such that Ax = λx.
Second: Through standard mathematical operations we can go from this: Ax = λx, to this: (A - λI)x = 0
The solutions to the equation det(A - λI) = 0 will yield your eigenvalues.
The previously mentioned equation is the characteristic equation.
-- SKIP TO THIS POINT FOR THE SHORT ANSWER --
Given some square matrix A, the characteristic equation is det(A - λI) = 0.
Example:
$$ A = begin{bmatrix}
a & b \
c & d \
end{bmatrix}$$$$ λI = begin{bmatrix}
λ & 0 \
0 & λ \
end{bmatrix}$$
$$ A - λI = begin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix}$$
$$detbegin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix} = 0$$ Characteristic Equation: $$(a-λ)(d-λ) - (c)(b) = 0$$
Source Cited: Lay, David C., et al. Linear Algebra and Its Applications, Fifth Edition. Pearson, 2016.
answered Dec 5 '18 at 21:53
sethledsethled
113
$endgroup$
add a comment |
$begingroup$
The characteristic equation is used to find the eigenvalues of a square matrix A.
First: Know that an eigenvector of some square matrix A is a non-zero vector x such that Ax = λx.
Second: Through standard mathematical operations we can go from this: Ax = λx, to this: (A - λI)x = 0
The solutions to the equation det(A - λI) = 0 will yield your eigenvalues.
The previously mentioned equation is the characteristic equation.
-- SKIP TO THIS POINT FOR THE SHORT ANSWER --
Given some square matrix A, the characteristic equation is det(A - λI) = 0.
Example:
$$ A = begin{bmatrix}
a & b \
c & d \
end{bmatrix}$$$$ λI = begin{bmatrix}
λ & 0 \
0 & λ \
end{bmatrix}$$
$$ A - λI = begin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix}$$
$$detbegin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix} = 0$$ Characteristic Equation: $$(a-λ)(d-λ) - (c)(b) = 0$$
Source Cited: Lay, David C., et al. Linear Algebra and Its Applications, Fifth Edition. Pearson, 2016.
answered Dec 5 '18 at 21:53
sethledsethled
113
$endgroup$
The characteristic equation is used to find the eigenvalues of a square matrix A.
First: Know that an eigenvector of some square matrix A is a non-zero vector x such that Ax = λx.
Second: Through standard mathematical operations we can go from this: Ax = λx, to this: (A - λI)x = 0
The solutions to the equation det(A - λI) = 0 will yield your eigenvalues.
The previously mentioned equation is the characteristic equation.
-- SKIP TO THIS POINT FOR THE SHORT ANSWER --
Given some square matrix A, the characteristic equation is det(A - λI) = 0.
Example:
$$ A = begin{bmatrix}
a & b \
c & d \
end{bmatrix}$$$$ λI = begin{bmatrix}
λ & 0 \
0 & λ \
end{bmatrix}$$
$$ A - λI = begin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix}$$
$$detbegin{bmatrix}
a-λ & b \
c & d-λ \
end{bmatrix} = 0$$ Characteristic Equation: $$(a-λ)(d-λ) - (c)(b) = 0$$
Source Cited: Lay, David C., et al. Linear Algebra and Its Applications, Fifth Edition. Pearson, 2016.
answered Dec 5 '18 at 21:53
sethledsethled
113
answered Dec 5 '18 at 21:53
sethledsethled
113
answered Dec 5 '18 at 21:53
sethledsethled
113
answered Dec 5 '18 at 21:53
sethledsethled
113
113
add a comment |
add a comment |
$begingroup$
That is incorrect. The characteristic equation comes from computing the $lambda$ values necessary to give you a matrix determinant equal to zero.
$$begin{vmatrix}
-lambda & -1 & -1 \ 1 & 2 - lambda & 1 \ 1 & 1 & 2- lambda
end{vmatrix} = 0
$$
Notice that for $lambda = 1$ all three columns become identical and we can come up with two eigenvectors. Coming up with the last eigenvalue should be easy if we use the fact that the trace of the matrix is the sum of the eigenvalues.
The eigenvalues for this matrix are thus $lambda_1 = 1, lambda_2 = 1, lambda_3 = 2$ as they make the matrix singular.
answered Mar 31 '16 at 13:04
Klint QinamiKlint Qinami
1,137510
$endgroup$
add a comment |
$begingroup$
That is incorrect. The characteristic equation comes from computing the $lambda$ values necessary to give you a matrix determinant equal to zero.
$$begin{vmatrix}
-lambda & -1 & -1 \ 1 & 2 - lambda & 1 \ 1 & 1 & 2- lambda
end{vmatrix} = 0
$$
Notice that for $lambda = 1$ all three columns become identical and we can come up with two eigenvectors. Coming up with the last eigenvalue should be easy if we use the fact that the trace of the matrix is the sum of the eigenvalues.
The eigenvalues for this matrix are thus $lambda_1 = 1, lambda_2 = 1, lambda_3 = 2$ as they make the matrix singular.
answered Mar 31 '16 at 13:04
Klint QinamiKlint Qinami
1,137510
$endgroup$
add a comment |
$begingroup$
That is incorrect. The characteristic equation comes from computing the $lambda$ values necessary to give you a matrix determinant equal to zero.
$$begin{vmatrix}
-lambda & -1 & -1 \ 1 & 2 - lambda & 1 \ 1 & 1 & 2- lambda
end{vmatrix} = 0
$$
Notice that for $lambda = 1$ all three columns become identical and we can come up with two eigenvectors. Coming up with the last eigenvalue should be easy if we use the fact that the trace of the matrix is the sum of the eigenvalues.
The eigenvalues for this matrix are thus $lambda_1 = 1, lambda_2 = 1, lambda_3 = 2$ as they make the matrix singular.
answered Mar 31 '16 at 13:04
Klint QinamiKlint Qinami
1,137510
$endgroup$
That is incorrect. The characteristic equation comes from computing the $lambda$ values necessary to give you a matrix determinant equal to zero.
$$begin{vmatrix}
-lambda & -1 & -1 \ 1 & 2 - lambda & 1 \ 1 & 1 & 2- lambda
end{vmatrix} = 0
$$
Notice that for $lambda = 1$ all three columns become identical and we can come up with two eigenvectors. Coming up with the last eigenvalue should be easy if we use the fact that the trace of the matrix is the sum of the eigenvalues.
The eigenvalues for this matrix are thus $lambda_1 = 1, lambda_2 = 1, lambda_3 = 2$ as they make the matrix singular.
answered Mar 31 '16 at 13:04
Klint QinamiKlint Qinami
1,137510
edited Mar 31 '16 at 13:28
answered Mar 31 '16 at 13:04
Klint QinamiKlint Qinami
1,137510
answered Mar 31 '16 at 13:04
Klint QinamiKlint Qinami
1,137510
answered Mar 31 '16 at 13:04
Klint QinamiKlint Qinami
1,137510
1,137510
add a comment |
add a comment |
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Yes, that is correct.
$endgroup$
– Inazuma
Mar 31 '16 at 12:59
2
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Do it like in this MSE-question.
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– Dietrich Burde
Mar 31 '16 at 13:05