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# If a matrix A is symmetric then the eigenvectors from different eigens.docx ×

# If a matrix A is symmetric then the eigenvectors from different eigens.docx

If a matrix A is symmetric then the eigenvectors from different eigenspaces are orthogonal.
Is the reverse true?
If yes, can you provide proof.
If no, can you provide an example of a non-symmetric matrix whose eigenvectors from different eigenspaces are arthogonal?
Solution
2 down vote A definition of \"symmetric\" which should prove useful here is: if < , > is a canonically chosen inner product, we say T is symmetric if and only if for all vectors $u,v \\in V$ we have < u, Tv> = < Tu, v>. Assume that the dimension of V is equal to the sum of the dimensions of the eigenspaces $E_1,...,E_k$ associated to the eigenvalues $\\lambda_1,...,\\lambda_k$. Then we can write any vectors $u,v \\in V$ as $\\sum_{i=1}^k u_i$ and $\\sum_{i=1}^k v_i$ respectively. < u, Tv> can then be rewritten <$\\sum u_i$, T $\\sum v_i$> = < $\\sum u_i$, $\\sum T v_i$> = < $\\sum u_i$, $\\sum \\lambda_i v_i$>. Now we use orthogonality. If the eigenspaces are orthogonal to eachother, then this expression becomes $\\sum$ < $u_i$, $\\lambda_i v_i$> = $\\sum \\lambda_i$< $u_i$, $v_i$> = $\\sum$ < $\\lambda_i u_i$, $v_i$> = < $\\sum \\lambda_i u_i$, $\\sum v_i$> = < Tu, v>. The reverse implication isn\'t hard, simply consider the equality < u, Tv> = < Tu, v> when u, v are two eigenvectors corresponding to two different eigenvalues.
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If a matrix A is symmetric then the eigenvectors from different eigenspaces are orthogonal.
Is the reverse true?
If yes, can you provide proof.
If no, can you provide an example of a non-symmetric matrix whose eigenvectors from different eigenspaces are arthogonal?
Solution
2 down vote A definition of \"symmetric\" which should prove useful here is: if < , > is a canonically chosen inner product, we say T is symmetric if and only if for all vectors $u,v \\in V$ we have < u, Tv> = < Tu, v>. Assume that the dimension of V is equal to the sum of the dimensions of the eigenspaces $E_1,...,E_k$ associated to the eigenvalues $\\lambda_1,...,\\lambda_k$. Then we can write any vectors $u,v \\in V$ as $\\sum_{i=1}^k u_i$ and $\\sum_{i=1}^k v_i$ respectively. < u, Tv> can then be rewritten <$\\sum u_i$, T $\\sum v_i$> = < $\\sum u_i$, $\\sum T v_i$> = < $\\sum u_i$, $\\sum \\lambda_i v_i$>. Now we use orthogonality. If the eigenspaces are orthogonal to eachother, then this expression becomes $\\sum$ < $u_i$, $\\lambda_i v_i$> = $\\sum \\lambda_i$< $u_i$, $v_i$> = $\\sum$ < $\\lambda_i u_i$, $v_i$> = < $\\sum \\lambda_i u_i$, $\\sum v_i$> = < Tu, v>. The reverse implication isn\'t hard, simply consider the equality < u, Tv> = < Tu, v> when u, v are two eigenvectors corresponding to two different eigenvalues.
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1. 1. If a matrix A is symmetric then the eigenvectors from different eigenspaces are orthogonal. Is the reverse true? If yes, can you provide proof. If no, can you provide an example of a non-symmetric matrix whose eigenvectors from different eigenspaces are arthogonal? Solution 2 down vote A definition of "symmetric" which should prove useful here is: if < , > is a canonically chosen inner product, we say T is symmetric if and only if for all vectors $u,v in V$ we have < u, Tv> = < Tu, v>. Assume that the dimension of V is equal to the sum of the dimensions of the eigenspaces $E_1,...,E_k$ associated to the eigenvalues $lambda_1,...,lambda_k$. Then we can write any vectors $u,v in V$ as $sum_{i=1}^k u_i$ and $sum_{i=1}^k v_i$ respectively. < u, Tv> can then be rewritten <$sum u_i$, T $sum v_i$> = < $sum u_i$, $sum T v_i$> = < $sum u_i$, $sum lambda_i v_i$>. Now we use orthogonality. If the eigenspaces are orthogonal to eachother, then this expression becomes $sum$ < $u_i$, $lambda_i v_i$> = $sum lambda_i$< $u_i$, $v_i$> = $sum$ < $lambda_i u_i$, $v_i$> = < $sum lambda_i u_i$, $sum v_i$> = < Tu, v>. The reverse implication isn't hard, simply consider the equality < u, Tv> = < Tu, v> when u, v are two eigenvectors corresponding to two different eigenvalues.