Find the values for which the matrix is diagonalizable












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I have this matrix A= $$begin{bmatrix} 2&0&3\0&L&0\1&0&4end{bmatrix}$$ and for find characteristic matrix I do $lambda I - A$ so I've got $$begin{bmatrix} lambda - 2&0&-3\0&lambda-L&0\-1&0&lambda-4end {bmatrix}$$ and here I'm in stuck, I think it's something simple but I can't to calcolate this: $(lambda-2)(lambda-L)(lambda-4)-(3lambda-L)=lambda^3-6lambda^2-Llambda^2+6lambda+6Llambda-8L$ and it's not correct. Is here someone who can explain to me whot I'm doing wrong or whot rule or theorem I must to study. Here I used Sarrus for find determinant of the matrix.






matrix-calculus







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  • $begingroup$
    you miscalculated you nebative diagonal, it should be $(-1)(-3)(lambda - L)=3(lambda - L) = (3 lambda - 3 L)$
    $endgroup$
    – Enkidu
    Dec 7 '18 at 11:33


















0












$begingroup$


I have this matrix A= $$begin{bmatrix} 2&0&3\0&L&0\1&0&4end{bmatrix}$$ and for find characteristic matrix I do $lambda I - A$ so I've got $$begin{bmatrix} lambda - 2&0&-3\0&lambda-L&0\-1&0&lambda-4end {bmatrix}$$ and here I'm in stuck, I think it's something simple but I can't to calcolate this: $(lambda-2)(lambda-L)(lambda-4)-(3lambda-L)=lambda^3-6lambda^2-Llambda^2+6lambda+6Llambda-8L$ and it's not correct. Is here someone who can explain to me whot I'm doing wrong or whot rule or theorem I must to study. Here I used Sarrus for find determinant of the matrix.






matrix-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    you miscalculated you nebative diagonal, it should be $(-1)(-3)(lambda - L)=3(lambda - L) = (3 lambda - 3 L)$
    $endgroup$
    – Enkidu
    Dec 7 '18 at 11:33
















0












0








0





$begingroup$


I have this matrix A= $$begin{bmatrix} 2&0&3\0&L&0\1&0&4end{bmatrix}$$ and for find characteristic matrix I do $lambda I - A$ so I've got $$begin{bmatrix} lambda - 2&0&-3\0&lambda-L&0\-1&0&lambda-4end {bmatrix}$$ and here I'm in stuck, I think it's something simple but I can't to calcolate this: $(lambda-2)(lambda-L)(lambda-4)-(3lambda-L)=lambda^3-6lambda^2-Llambda^2+6lambda+6Llambda-8L$ and it's not correct. Is here someone who can explain to me whot I'm doing wrong or whot rule or theorem I must to study. Here I used Sarrus for find determinant of the matrix.






matrix-calculus







share|cite|improve this question











$endgroup$




I have this matrix A= $$begin{bmatrix} 2&0&3\0&L&0\1&0&4end{bmatrix}$$ and for find characteristic matrix I do $lambda I - A$ so I've got $$begin{bmatrix} lambda - 2&0&-3\0&lambda-L&0\-1&0&lambda-4end {bmatrix}$$ and here I'm in stuck, I think it's something simple but I can't to calcolate this: $(lambda-2)(lambda-L)(lambda-4)-(3lambda-L)=lambda^3-6lambda^2-Llambda^2+6lambda+6Llambda-8L$ and it's not correct. Is here someone who can explain to me whot I'm doing wrong or whot rule or theorem I must to study. Here I used Sarrus for find determinant of the matrix.






matrix-calculus




matrix-calculus






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edited Dec 7 '18 at 11:27









José Carlos Santos

155k22124227




155k22124227










asked Dec 7 '18 at 11:25









CiaoCiao

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  • $begingroup$
    you miscalculated you nebative diagonal, it should be $(-1)(-3)(lambda - L)=3(lambda - L) = (3 lambda - 3 L)$
    $endgroup$
    – Enkidu
    Dec 7 '18 at 11:33




















  • $begingroup$
    you miscalculated you nebative diagonal, it should be $(-1)(-3)(lambda - L)=3(lambda - L) = (3 lambda - 3 L)$
    $endgroup$
    – Enkidu
    Dec 7 '18 at 11:33


















$begingroup$
you miscalculated you nebative diagonal, it should be $(-1)(-3)(lambda - L)=3(lambda - L) = (3 lambda - 3 L)$
$endgroup$
– Enkidu
Dec 7 '18 at 11:33






$begingroup$
you miscalculated you nebative diagonal, it should be $(-1)(-3)(lambda - L)=3(lambda - L) = (3 lambda - 3 L)$
$endgroup$
– Enkidu
Dec 7 '18 at 11:33












2 Answers
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Just use the fact that the middle column has two zeros:$$begin{vmatrix}lambda-2&0&-3\0&lambda-L&0\-1&0&lambda-4end{vmatrix}=(lambda-L)begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}.$$So, the eigenvalues are $L$ and the roots of the quadratic equation$$begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}=0.$$






share|cite|improve this answer









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    1












    $begingroup$

    You can find the eigenvalues of this matrix by inspection by looking for simple eigenvectors. The second column only has a non-zero entry in its second element, so $(0,1,0)^T$ is an eigenvector with eigenvalue $L$. Adding up the first and third columns gives you $(5,0,5)^T = 5(1,0,1)^T$, but this column addition is equivalent to multiplying by $(1,0,1)^T$, so that’s an eigenvector with eigenvalue $5$. You get the last eigenvalue “for free” by using the trace: $(L+6)-L-5=1$. The characteristic polynomial is therefore $(lambda-L)(lambda-5)(lambda-1)$, so you’ll need to investigate what happens when $L=5$ or $L=1$.






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      2 Answers
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      2 Answers
      2






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      active

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      1












      $begingroup$

      Just use the fact that the middle column has two zeros:$$begin{vmatrix}lambda-2&0&-3\0&lambda-L&0\-1&0&lambda-4end{vmatrix}=(lambda-L)begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}.$$So, the eigenvalues are $L$ and the roots of the quadratic equation$$begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}=0.$$






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        1












        $begingroup$

        Just use the fact that the middle column has two zeros:$$begin{vmatrix}lambda-2&0&-3\0&lambda-L&0\-1&0&lambda-4end{vmatrix}=(lambda-L)begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}.$$So, the eigenvalues are $L$ and the roots of the quadratic equation$$begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}=0.$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Just use the fact that the middle column has two zeros:$$begin{vmatrix}lambda-2&0&-3\0&lambda-L&0\-1&0&lambda-4end{vmatrix}=(lambda-L)begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}.$$So, the eigenvalues are $L$ and the roots of the quadratic equation$$begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}=0.$$






          share|cite|improve this answer









          $endgroup$



          Just use the fact that the middle column has two zeros:$$begin{vmatrix}lambda-2&0&-3\0&lambda-L&0\-1&0&lambda-4end{vmatrix}=(lambda-L)begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}.$$So, the eigenvalues are $L$ and the roots of the quadratic equation$$begin{vmatrix}lambda-2&-3\-1&lambda-4end{vmatrix}=0.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 11:31









          José Carlos SantosJosé Carlos Santos

          155k22124227




          155k22124227























              1












              $begingroup$

              You can find the eigenvalues of this matrix by inspection by looking for simple eigenvectors. The second column only has a non-zero entry in its second element, so $(0,1,0)^T$ is an eigenvector with eigenvalue $L$. Adding up the first and third columns gives you $(5,0,5)^T = 5(1,0,1)^T$, but this column addition is equivalent to multiplying by $(1,0,1)^T$, so that’s an eigenvector with eigenvalue $5$. You get the last eigenvalue “for free” by using the trace: $(L+6)-L-5=1$. The characteristic polynomial is therefore $(lambda-L)(lambda-5)(lambda-1)$, so you’ll need to investigate what happens when $L=5$ or $L=1$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                You can find the eigenvalues of this matrix by inspection by looking for simple eigenvectors. The second column only has a non-zero entry in its second element, so $(0,1,0)^T$ is an eigenvector with eigenvalue $L$. Adding up the first and third columns gives you $(5,0,5)^T = 5(1,0,1)^T$, but this column addition is equivalent to multiplying by $(1,0,1)^T$, so that’s an eigenvector with eigenvalue $5$. You get the last eigenvalue “for free” by using the trace: $(L+6)-L-5=1$. The characteristic polynomial is therefore $(lambda-L)(lambda-5)(lambda-1)$, so you’ll need to investigate what happens when $L=5$ or $L=1$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You can find the eigenvalues of this matrix by inspection by looking for simple eigenvectors. The second column only has a non-zero entry in its second element, so $(0,1,0)^T$ is an eigenvector with eigenvalue $L$. Adding up the first and third columns gives you $(5,0,5)^T = 5(1,0,1)^T$, but this column addition is equivalent to multiplying by $(1,0,1)^T$, so that’s an eigenvector with eigenvalue $5$. You get the last eigenvalue “for free” by using the trace: $(L+6)-L-5=1$. The characteristic polynomial is therefore $(lambda-L)(lambda-5)(lambda-1)$, so you’ll need to investigate what happens when $L=5$ or $L=1$.






                  share|cite|improve this answer









                  $endgroup$



                  You can find the eigenvalues of this matrix by inspection by looking for simple eigenvectors. The second column only has a non-zero entry in its second element, so $(0,1,0)^T$ is an eigenvector with eigenvalue $L$. Adding up the first and third columns gives you $(5,0,5)^T = 5(1,0,1)^T$, but this column addition is equivalent to multiplying by $(1,0,1)^T$, so that’s an eigenvector with eigenvalue $5$. You get the last eigenvalue “for free” by using the trace: $(L+6)-L-5=1$. The characteristic polynomial is therefore $(lambda-L)(lambda-5)(lambda-1)$, so you’ll need to investigate what happens when $L=5$ or $L=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 7 '18 at 21:09









                  amdamd

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                  29.6k21050






























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