Compute the characteristic equation 3x3 matrix












1












$begingroup$


Can someone help me explain how to solve this problem below:



enter image description here



Something like this?



enter image description here



But I'm not sure.






linear-algebra







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  • $begingroup$
    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


















1












$begingroup$


Can someone help me explain how to solve this problem below:



enter image description here



Something like this?



enter image description here



But I'm not sure.






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












1








1


3



$begingroup$


Can someone help me explain how to solve this problem below:



enter image description here



Something like this?



enter image description here



But I'm not sure.






linear-algebra







share|cite|improve this question









$endgroup$




Can someone help me explain how to solve this problem below:



enter image description here



Something like this?



enter image description here



But I'm not sure.






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    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












4 Answers
4






active

oldest

votes


















5












$begingroup$

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
$x^3-4x^2+5x-2=0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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 $$p(x)=(x-1)^2(x-2)$$






share|cite|improve this answer











$endgroup$









  • 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





















1












$begingroup$

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.






share|cite|improve this answer









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    0












    $begingroup$

    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.






    share|cite|improve this answer











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      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
      $x^3-4x^2+5x-2=0$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        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
















      5












      $begingroup$

      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
      $x^3-4x^2+5x-2=0$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        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














      5












      5








      5





      $begingroup$

      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
      $x^3-4x^2+5x-2=0$






      share|cite|improve this answer











      $endgroup$



      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
      $x^3-4x^2+5x-2=0$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 31 '16 at 17:13

























      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


















      • $begingroup$
        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











      1












      $begingroup$

      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 $$p(x)=(x-1)^2(x-2)$$






      share|cite|improve this answer











      $endgroup$









      • 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


















      1












      $begingroup$

      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 $$p(x)=(x-1)^2(x-2)$$






      share|cite|improve this answer











      $endgroup$









      • 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
















      1












      1








      1





      $begingroup$

      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 $$p(x)=(x-1)^2(x-2)$$






      share|cite|improve this answer











      $endgroup$



      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 $$p(x)=(x-1)^2(x-2)$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 31 '16 at 17:37

























      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
















      • 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
      $endgroup$
      – Arpit Kansal
      Dec 3 '18 at 10:41













      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 21:53









          sethledsethled

          113




          113























              0












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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.






                  share|cite|improve this answer











                  $endgroup$



                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 31 '16 at 13:28

























                  answered Mar 31 '16 at 13:04









                  Klint QinamiKlint Qinami

                  1,137510




                  1,137510






























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