Finding the eigenvalues - missing a trick?
$begingroup$
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
asked Dec 16 '18 at 1:20
user3424575user3424575
124
$endgroup$
|
show 1 more comment
$begingroup$
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
asked Dec 16 '18 at 1:20
user3424575user3424575
124
$endgroup$
$begingroup$
"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
|
show 1 more comment
$begingroup$
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
asked Dec 16 '18 at 1:20
user3424575user3424575
124
$endgroup$
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
asked Dec 16 '18 at 1:20
user3424575user3424575
124
asked Dec 16 '18 at 1:20
user3424575user3424575
124
asked Dec 16 '18 at 1:20
user3424575user3424575
124
asked Dec 16 '18 at 1:20
user3424575user3424575
124
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
|
show 1 more comment
$begingroup$
"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
|
show 1 more comment
1 Answer
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$begingroup$
The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.
answered Dec 16 '18 at 3:20
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
add a comment |
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$begingroup$
The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.
answered Dec 16 '18 at 3:20
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
add a comment |
$begingroup$
The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.
answered Dec 16 '18 at 3:20
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
add a comment |
$begingroup$
The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.
answered Dec 16 '18 at 3:20
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
The non-zero eigenvalues of $X_1 X_2$ are the same as the non-zero eigenvalues of $X_2 X_1$.
answered Dec 16 '18 at 3:20
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered Dec 16 '18 at 3:20
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered Dec 16 '18 at 3:20
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered Dec 16 '18 at 3:20
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
17.8k12247
add a comment |
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$begingroup$
"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