How do I find this eigenvector for a symmetric Matrix?
0
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I have a symmetric matrix A, whose eigenvalues are $lambda_1 = 6,~ lambda_2 = 3,~ lambda_3 = 2$ and eigenvectors are $vec{v_1} = (1, 1, 1),~vec{v_2} = (1,1,-1)$. How do I find the third eigenvector $vec{v_3}$?
matrices eigenvalues-eigenvectors symmetric-matrices
asked Dec 5 '18 at 10:45
Henri SödergårdHenri Södergård
1
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add a comment |
0
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I have a symmetric matrix A, whose eigenvalues are $lambda_1 = 6,~ lambda_2 = 3,~ lambda_3 = 2$ and eigenvectors are $vec{v_1} = (1, 1, 1),~vec{v_2} = (1,1,-1)$. How do I find the third eigenvector $vec{v_3}$?
matrices eigenvalues-eigenvectors symmetric-matrices
asked Dec 5 '18 at 10:45
Henri SödergårdHenri Södergård
1
$endgroup$
$begingroup$
Eigenvectors of distinct eigenvalues are pairwise orthogonal.
$endgroup$
– xbh
Dec 5 '18 at 10:46
add a comment |
0
0
0
$begingroup$
I have a symmetric matrix A, whose eigenvalues are $lambda_1 = 6,~ lambda_2 = 3,~ lambda_3 = 2$ and eigenvectors are $vec{v_1} = (1, 1, 1),~vec{v_2} = (1,1,-1)$. How do I find the third eigenvector $vec{v_3}$?
matrices eigenvalues-eigenvectors symmetric-matrices
asked Dec 5 '18 at 10:45
Henri SödergårdHenri Södergård
1
$endgroup$
I have a symmetric matrix A, whose eigenvalues are $lambda_1 = 6,~ lambda_2 = 3,~ lambda_3 = 2$ and eigenvectors are $vec{v_1} = (1, 1, 1),~vec{v_2} = (1,1,-1)$. How do I find the third eigenvector $vec{v_3}$?
matrices eigenvalues-eigenvectors symmetric-matrices
matrices eigenvalues-eigenvectors symmetric-matrices
asked Dec 5 '18 at 10:45
Henri SödergårdHenri Södergård
1
asked Dec 5 '18 at 10:45
Henri SödergårdHenri Södergård
1
asked Dec 5 '18 at 10:45
Henri SödergårdHenri Södergård
1
asked Dec 5 '18 at 10:45
Henri SödergårdHenri Södergård
1
asked Dec 5 '18 at 10:45
Henri SödergårdHenri Södergård
1
1
$begingroup$
Eigenvectors of distinct eigenvalues are pairwise orthogonal.
$endgroup$
– xbh
Dec 5 '18 at 10:46
add a comment |
$begingroup$
Eigenvectors of distinct eigenvalues are pairwise orthogonal.
$endgroup$
– xbh
Dec 5 '18 at 10:46
$begingroup$
Eigenvectors of distinct eigenvalues are pairwise orthogonal.
$endgroup$
– xbh
Dec 5 '18 at 10:46
$begingroup$
Eigenvectors of distinct eigenvalues are pairwise orthogonal.
$endgroup$
– xbh
Dec 5 '18 at 10:46
add a comment |
1 Answer
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Hint: Since the matrix is symmetric, (and real, I presume) it has an orthogonal basis of eigenvectors.
answered Dec 5 '18 at 10:46
José Carlos SantosJosé Carlos Santos
164k22132235
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$begingroup$
But how do I use this in practice?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:08
$begingroup$
By searching for a vector orthogonal to both $vec{v_1}$ and $vec{v_2}$.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:15
$begingroup$
So, it would be okay to take the cross-product of both vectors, as in $vec{v_3} = vec{v_1} times vec{v_2}$ and the resulting $vec{v_3} = (-3,3,0)$ would be what I'm looking for?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:23
$begingroup$
Yes, indeed it is.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:24
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Hint: Since the matrix is symmetric, (and real, I presume) it has an orthogonal basis of eigenvectors.
answered Dec 5 '18 at 10:46
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
$begingroup$
But how do I use this in practice?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:08
$begingroup$
By searching for a vector orthogonal to both $vec{v_1}$ and $vec{v_2}$.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:15
$begingroup$
So, it would be okay to take the cross-product of both vectors, as in $vec{v_3} = vec{v_1} times vec{v_2}$ and the resulting $vec{v_3} = (-3,3,0)$ would be what I'm looking for?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:23
$begingroup$
Yes, indeed it is.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:24
add a comment |
0
$begingroup$
Hint: Since the matrix is symmetric, (and real, I presume) it has an orthogonal basis of eigenvectors.
answered Dec 5 '18 at 10:46
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
$begingroup$
But how do I use this in practice?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:08
$begingroup$
By searching for a vector orthogonal to both $vec{v_1}$ and $vec{v_2}$.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:15
$begingroup$
So, it would be okay to take the cross-product of both vectors, as in $vec{v_3} = vec{v_1} times vec{v_2}$ and the resulting $vec{v_3} = (-3,3,0)$ would be what I'm looking for?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:23
$begingroup$
Yes, indeed it is.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:24
add a comment |
0
0
0
$begingroup$
Hint: Since the matrix is symmetric, (and real, I presume) it has an orthogonal basis of eigenvectors.
answered Dec 5 '18 at 10:46
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
Hint: Since the matrix is symmetric, (and real, I presume) it has an orthogonal basis of eigenvectors.
answered Dec 5 '18 at 10:46
José Carlos SantosJosé Carlos Santos
164k22132235
answered Dec 5 '18 at 10:46
José Carlos SantosJosé Carlos Santos
164k22132235
answered Dec 5 '18 at 10:46
José Carlos SantosJosé Carlos Santos
164k22132235
answered Dec 5 '18 at 10:46
José Carlos SantosJosé Carlos Santos
164k22132235
164k22132235
$begingroup$
But how do I use this in practice?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:08
$begingroup$
By searching for a vector orthogonal to both $vec{v_1}$ and $vec{v_2}$.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:15
$begingroup$
So, it would be okay to take the cross-product of both vectors, as in $vec{v_3} = vec{v_1} times vec{v_2}$ and the resulting $vec{v_3} = (-3,3,0)$ would be what I'm looking for?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:23
$begingroup$
Yes, indeed it is.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:24
add a comment |
$begingroup$
But how do I use this in practice?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:08
$begingroup$
By searching for a vector orthogonal to both $vec{v_1}$ and $vec{v_2}$.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:15
$begingroup$
So, it would be okay to take the cross-product of both vectors, as in $vec{v_3} = vec{v_1} times vec{v_2}$ and the resulting $vec{v_3} = (-3,3,0)$ would be what I'm looking for?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:23
$begingroup$
Yes, indeed it is.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:24
$begingroup$
But how do I use this in practice?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:08
$begingroup$
But how do I use this in practice?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:08
$begingroup$
By searching for a vector orthogonal to both $vec{v_1}$ and $vec{v_2}$.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:15
$begingroup$
By searching for a vector orthogonal to both $vec{v_1}$ and $vec{v_2}$.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:15
$begingroup$
So, it would be okay to take the cross-product of both vectors, as in $vec{v_3} = vec{v_1} times vec{v_2}$ and the resulting $vec{v_3} = (-3,3,0)$ would be what I'm looking for?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:23
$begingroup$
So, it would be okay to take the cross-product of both vectors, as in $vec{v_3} = vec{v_1} times vec{v_2}$ and the resulting $vec{v_3} = (-3,3,0)$ would be what I'm looking for?
$endgroup$
– Henri Södergård
Dec 5 '18 at 11:23
$begingroup$
Yes, indeed it is.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:24
$begingroup$
Yes, indeed it is.
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:24
add a comment |
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$begingroup$
Eigenvectors of distinct eigenvalues are pairwise orthogonal.
$endgroup$
– xbh
Dec 5 '18 at 10:46