Finding eigenvector only knowing others eigenvectors.
$begingroup$
The matrix $A in M_3(mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$?
Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.
The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.
linear-algebra eigenvalues-eigenvectors
asked 1 hour ago
rodorgasrodorgas
1915
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add a comment |
$begingroup$
The matrix $A in M_3(mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$?
Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.
The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.
linear-algebra eigenvalues-eigenvectors
asked 1 hour ago
rodorgasrodorgas
1915
$endgroup$
add a comment |
$begingroup$
The matrix $A in M_3(mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$?
Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.
The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.
linear-algebra eigenvalues-eigenvectors
asked 1 hour ago
rodorgasrodorgas
1915
$endgroup$
The matrix $A in M_3(mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$?
Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.
The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked 1 hour ago
rodorgasrodorgas
1915
asked 1 hour ago
rodorgasrodorgas
1915
asked 1 hour ago
rodorgasrodorgas
1915
asked 1 hour ago
rodorgasrodorgas
1915
asked 1 hour ago
rodorgasrodorgas
1915
1915
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.
So, find which of the vectors is orthogonal to the first two.
(1,1,-3) is.
answered 1 hour ago
Chris CusterChris Custer
13.5k3827
$endgroup$
$begingroup$
I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
$endgroup$
– Chris Custer
53 mins ago
1
$begingroup$
The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
$endgroup$
– amd
46 mins ago
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@amd true. Good catch.
$endgroup$
– Chris Custer
40 mins ago
$begingroup$
@amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
$endgroup$
– Chris Custer
30 mins ago
$begingroup$
@ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
$endgroup$
– angryavian
9 mins ago
add a comment |
$begingroup$
Hint: the condition $A^t = A$ allows you to use the spectral theorem.
Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.
answered 1 hour ago
angryavianangryavian
41.2k23380
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.
So, find which of the vectors is orthogonal to the first two.
(1,1,-3) is.
answered 1 hour ago
Chris CusterChris Custer
13.5k3827
$endgroup$
$begingroup$
I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
$endgroup$
– Chris Custer
53 mins ago
1
$begingroup$
The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
$endgroup$
– amd
46 mins ago
$begingroup$
@amd true. Good catch.
$endgroup$
– Chris Custer
40 mins ago
$begingroup$
@amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
$endgroup$
– Chris Custer
30 mins ago
$begingroup$
@ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
$endgroup$
– angryavian
9 mins ago
add a comment |
$begingroup$
Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.
So, find which of the vectors is orthogonal to the first two.
(1,1,-3) is.
answered 1 hour ago
Chris CusterChris Custer
13.5k3827
$endgroup$
$begingroup$
I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
$endgroup$
– Chris Custer
53 mins ago
1
$begingroup$
The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
$endgroup$
– amd
46 mins ago
$begingroup$
@amd true. Good catch.
$endgroup$
– Chris Custer
40 mins ago
$begingroup$
@amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
$endgroup$
– Chris Custer
30 mins ago
$begingroup$
@ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
$endgroup$
– angryavian
9 mins ago
add a comment |
$begingroup$
Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.
So, find which of the vectors is orthogonal to the first two.
(1,1,-3) is.
answered 1 hour ago
Chris CusterChris Custer
13.5k3827
$endgroup$
Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.
So, find which of the vectors is orthogonal to the first two.
(1,1,-3) is.
answered 1 hour ago
Chris CusterChris Custer
13.5k3827
edited 51 mins ago
answered 1 hour ago
Chris CusterChris Custer
13.5k3827
answered 1 hour ago
Chris CusterChris Custer
13.5k3827
answered 1 hour ago
Chris CusterChris Custer
13.5k3827
13.5k3827
$begingroup$
I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
$endgroup$
– Chris Custer
53 mins ago
1
$begingroup$
The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
$endgroup$
– amd
46 mins ago
$begingroup$
@amd true. Good catch.
$endgroup$
– Chris Custer
40 mins ago
$begingroup$
@amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
$endgroup$
– Chris Custer
30 mins ago
$begingroup$
@ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
$endgroup$
– angryavian
9 mins ago
add a comment |
$begingroup$
I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
$endgroup$
– Chris Custer
53 mins ago
1
$begingroup$
The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
$endgroup$
– amd
46 mins ago
$begingroup$
@amd true. Good catch.
$endgroup$
– Chris Custer
40 mins ago
$begingroup$
@amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
$endgroup$
– Chris Custer
30 mins ago
$begingroup$
@ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
$endgroup$
– angryavian
9 mins ago
$begingroup$
I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
$endgroup$
– Chris Custer
53 mins ago
$begingroup$
I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
$endgroup$
– Chris Custer
53 mins ago
1
1
$begingroup$
The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
$endgroup$
– amd
46 mins ago
$begingroup$
The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
$endgroup$
– amd
46 mins ago
$begingroup$
@amd true. Good catch.
$endgroup$
– Chris Custer
40 mins ago
$begingroup$
@amd true. Good catch.
$endgroup$
– Chris Custer
40 mins ago
$begingroup$
@amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
$endgroup$
– Chris Custer
30 mins ago
$begingroup$
@amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
$endgroup$
– Chris Custer
30 mins ago
$begingroup$
@ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
$endgroup$
– angryavian
9 mins ago
$begingroup$
@ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
$endgroup$
– angryavian
9 mins ago
add a comment |
$begingroup$
Hint: the condition $A^t = A$ allows you to use the spectral theorem.
Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.
answered 1 hour ago
angryavianangryavian
41.2k23380
$endgroup$
add a comment |
$begingroup$
Hint: the condition $A^t = A$ allows you to use the spectral theorem.
Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.
answered 1 hour ago
angryavianangryavian
41.2k23380
$endgroup$
add a comment |
$begingroup$
Hint: the condition $A^t = A$ allows you to use the spectral theorem.
Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.
answered 1 hour ago
angryavianangryavian
41.2k23380
$endgroup$
Hint: the condition $A^t = A$ allows you to use the spectral theorem.
Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.
answered 1 hour ago
angryavianangryavian
41.2k23380
answered 1 hour ago
angryavianangryavian
41.2k23380
answered 1 hour ago
angryavianangryavian
41.2k23380
answered 1 hour ago
angryavianangryavian
41.2k23380
41.2k23380
add a comment |
add a comment |
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