Finding the eigenvalues - missing a trick?












0












$begingroup$


I can compute the eigenvalues of the product of these matrices $X_1 X_2$ by multiplying out the matrices, but I'm certain I am missing a trick that will make this computation less tedious. Any hints?



$X_1$:
$$
begin{matrix}
1 & 2 \
0 & 2 \
1 & 0 \
2 & 0 \
end{matrix}
$$



$X_2$:



$$
begin{matrix}
0 & 2 & 3 & 1\
2 & 1 & 3 & -1 \
end{matrix}
$$



The product of these matrices being:



begin{matrix}
4 & 6 & 8 & -1\
4 & 6 & 12 & 0 \
0 & 2 & 3 & 1 \
0 & 4 & 6 & 2 \
end{matrix}



$$
$$






linear-algebra







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$endgroup$












  • $begingroup$
    "I'm certain I am missing a trick that will make this computation less tedious" Why?
    $endgroup$
    – user587192
    Dec 16 '18 at 1:22










  • $begingroup$
    Mainly because this was a question on a previous exam - I doubt the instructor would expect us to find the eigenvalues for a 4x4 matrix.
    $endgroup$
    – user3424575
    Dec 16 '18 at 1:25










  • $begingroup$
    What would a rank of the resulting matrix be?
    $endgroup$
    – user58697
    Dec 16 '18 at 1:36






  • 1




    $begingroup$
    The first two rows of the product matrix are incorrect. I'm not sure if this helps at all with the question though.
    $endgroup$
    – MAX
    Dec 16 '18 at 1:43












  • $begingroup$
    The only thing special I can see is that row 4 is twice row 3, so the matrix is not invertible, making $lambda = 0$ one of the eigenvalues. Also, if you calculate $det(A-lambda I) = 0$, using the zeros in the first column to your advantage should make the determinant a lot less tedious.
    $endgroup$
    – Matthias
    Dec 16 '18 at 1:45


















0












$begingroup$


I can compute the eigenvalues of the product of these matrices $X_1 X_2$ by multiplying out the matrices, but I'm certain I am missing a trick that will make this computation less tedious. Any hints?



$X_1$:
$$
begin{matrix}
1 & 2 \
0 & 2 \
1 & 0 \
2 & 0 \
end{matrix}
$$



$X_2$:



$$
begin{matrix}
0 & 2 & 3 & 1\
2 & 1 & 3 & -1 \
end{matrix}
$$



The product of these matrices being:



begin{matrix}
4 & 6 & 8 & -1\
4 & 6 & 12 & 0 \
0 & 2 & 3 & 1 \
0 & 4 & 6 & 2 \
end{matrix}



$$
$$






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    "I'm certain I am missing a trick that will make this computation less tedious" Why?
    $endgroup$
    – user587192
    Dec 16 '18 at 1:22










  • $begingroup$
    Mainly because this was a question on a previous exam - I doubt the instructor would expect us to find the eigenvalues for a 4x4 matrix.
    $endgroup$
    – user3424575
    Dec 16 '18 at 1:25










  • $begingroup$
    What would a rank of the resulting matrix be?
    $endgroup$
    – user58697
    Dec 16 '18 at 1:36






  • 1




    $begingroup$
    The first two rows of the product matrix are incorrect. I'm not sure if this helps at all with the question though.
    $endgroup$
    – MAX
    Dec 16 '18 at 1:43












  • $begingroup$
    The only thing special I can see is that row 4 is twice row 3, so the matrix is not invertible, making $lambda = 0$ one of the eigenvalues. Also, if you calculate $det(A-lambda I) = 0$, using the zeros in the first column to your advantage should make the determinant a lot less tedious.
    $endgroup$
    – Matthias
    Dec 16 '18 at 1:45
















0












0








0





$begingroup$


I can compute the eigenvalues of the product of these matrices $X_1 X_2$ by multiplying out the matrices, but I'm certain I am missing a trick that will make this computation less tedious. Any hints?



$X_1$:
$$
begin{matrix}
1 & 2 \
0 & 2 \
1 & 0 \
2 & 0 \
end{matrix}
$$



$X_2$:



$$
begin{matrix}
0 & 2 & 3 & 1\
2 & 1 & 3 & -1 \
end{matrix}
$$



The product of these matrices being:



begin{matrix}
4 & 6 & 8 & -1\
4 & 6 & 12 & 0 \
0 & 2 & 3 & 1 \
0 & 4 & 6 & 2 \
end{matrix}



$$
$$






linear-algebra







share|cite|improve this question









$endgroup$




I can compute the eigenvalues of the product of these matrices $X_1 X_2$ by multiplying out the matrices, but I'm certain I am missing a trick that will make this computation less tedious. Any hints?



$X_1$:
$$
begin{matrix}
1 & 2 \
0 & 2 \
1 & 0 \
2 & 0 \
end{matrix}
$$



$X_2$:



$$
begin{matrix}
0 & 2 & 3 & 1\
2 & 1 & 3 & -1 \
end{matrix}
$$



The product of these matrices being:



begin{matrix}
4 & 6 & 8 & -1\
4 & 6 & 12 & 0 \
0 & 2 & 3 & 1 \
0 & 4 & 6 & 2 \
end{matrix}



$$
$$






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 Dec 16 '18 at 1:20









user3424575user3424575

124




124












  • $begingroup$
    "I'm certain I am missing a trick that will make this computation less tedious" Why?
    $endgroup$
    – user587192
    Dec 16 '18 at 1:22










  • $begingroup$
    Mainly because this was a question on a previous exam - I doubt the instructor would expect us to find the eigenvalues for a 4x4 matrix.
    $endgroup$
    – user3424575
    Dec 16 '18 at 1:25










  • $begingroup$
    What would a rank of the resulting matrix be?
    $endgroup$
    – user58697
    Dec 16 '18 at 1:36






  • 1




    $begingroup$
    The first two rows of the product matrix are incorrect. I'm not sure if this helps at all with the question though.
    $endgroup$
    – MAX
    Dec 16 '18 at 1:43












  • $begingroup$
    The only thing special I can see is that row 4 is twice row 3, so the matrix is not invertible, making $lambda = 0$ one of the eigenvalues. Also, if you calculate $det(A-lambda I) = 0$, using the zeros in the first column to your advantage should make the determinant a lot less tedious.
    $endgroup$
    – Matthias
    Dec 16 '18 at 1:45




















  • $begingroup$
    "I'm certain I am missing a trick that will make this computation less tedious" Why?
    $endgroup$
    – user587192
    Dec 16 '18 at 1:22










  • $begingroup$
    Mainly because this was a question on a previous exam - I doubt the instructor would expect us to find the eigenvalues for a 4x4 matrix.
    $endgroup$
    – user3424575
    Dec 16 '18 at 1:25










  • $begingroup$
    What would a rank of the resulting matrix be?
    $endgroup$
    – user58697
    Dec 16 '18 at 1:36






  • 1




    $begingroup$
    The first two rows of the product matrix are incorrect. I'm not sure if this helps at all with the question though.
    $endgroup$
    – MAX
    Dec 16 '18 at 1:43












  • $begingroup$
    The only thing special I can see is that row 4 is twice row 3, so the matrix is not invertible, making $lambda = 0$ one of the eigenvalues. Also, if you calculate $det(A-lambda I) = 0$, using the zeros in the first column to your advantage should make the determinant a lot less tedious.
    $endgroup$
    – Matthias
    Dec 16 '18 at 1:45


















$begingroup$
"I'm certain I am missing a trick that will make this computation less tedious" Why?
$endgroup$
– user587192
Dec 16 '18 at 1:22




$begingroup$
"I'm certain I am missing a trick that will make this computation less tedious" Why?
$endgroup$
– user587192
Dec 16 '18 at 1:22












$begingroup$
Mainly because this was a question on a previous exam - I doubt the instructor would expect us to find the eigenvalues for a 4x4 matrix.
$endgroup$
– user3424575
Dec 16 '18 at 1:25




$begingroup$
Mainly because this was a question on a previous exam - I doubt the instructor would expect us to find the eigenvalues for a 4x4 matrix.
$endgroup$
– user3424575
Dec 16 '18 at 1:25












$begingroup$
What would a rank of the resulting matrix be?
$endgroup$
– user58697
Dec 16 '18 at 1:36




$begingroup$
What would a rank of the resulting matrix be?
$endgroup$
– user58697
Dec 16 '18 at 1:36




1




1




$begingroup$
The first two rows of the product matrix are incorrect. I'm not sure if this helps at all with the question though.
$endgroup$
– MAX
Dec 16 '18 at 1:43






$begingroup$
The first two rows of the product matrix are incorrect. I'm not sure if this helps at all with the question though.
$endgroup$
– MAX
Dec 16 '18 at 1:43














$begingroup$
The only thing special I can see is that row 4 is twice row 3, so the matrix is not invertible, making $lambda = 0$ one of the eigenvalues. Also, if you calculate $det(A-lambda I) = 0$, using the zeros in the first column to your advantage should make the determinant a lot less tedious.
$endgroup$
– Matthias
Dec 16 '18 at 1:45






$begingroup$
The only thing special I can see is that row 4 is twice row 3, so the matrix is not invertible, making $lambda = 0$ one of the eigenvalues. Also, if you calculate $det(A-lambda I) = 0$, using the zeros in the first column to your advantage should make the determinant a lot less tedious.
$endgroup$
– Matthias
Dec 16 '18 at 1:45












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The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.






share|cite|improve this answer









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    $begingroup$

    The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.






    share|cite|improve this answer









    $endgroup$


















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      $begingroup$

      The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.






      share|cite|improve this answer









      $endgroup$
















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        2





        $begingroup$

        The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.






        share|cite|improve this answer









        $endgroup$



        The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 3:20









        Stephen Montgomery-SmithStephen Montgomery-Smith

        17.8k12247




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