Orthonormal basis of a symmetric matrix [on hold] The 2019 Stack Overflow Developer Survey Results Are InSymmetric matrix decomposition with orthonormal basis of non-eigenvectorseigendecomposition of symmetric matrixAre the eigenvectors of a real symmetric matrix always an orthonormal basis without change?Symmetric matrix and orthonormal eigenvectorsNecessary conditions for a matrix to have orthonormal eigenvectors?If $A$ is orthogonal and symmetric, there is an orthonormal basis consisting of eigenvectors of $A$A symmetric matrix is diagonalized by a matrix of its orthonormal eigenvectorsIs a set of orthonormal eigenvectors directly an orthonormal basis of $mathbbR^n$?finding an orthonormal basis of R^3 of a matrix using eigenvaluesOrthogonally diagonalizable matrix and orthonormal basis
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Orthonormal basis of a symmetric matrix [on hold]
The 2019 Stack Overflow Developer Survey Results Are InSymmetric matrix decomposition with orthonormal basis of non-eigenvectorseigendecomposition of symmetric matrixAre the eigenvectors of a real symmetric matrix always an orthonormal basis without change?Symmetric matrix and orthonormal eigenvectorsNecessary conditions for a matrix to have orthonormal eigenvectors?If $A$ is orthogonal and symmetric, there is an orthonormal basis consisting of eigenvectors of $A$A symmetric matrix is diagonalized by a matrix of its orthonormal eigenvectorsIs a set of orthonormal eigenvectors directly an orthonormal basis of $mathbbR^n$?finding an orthonormal basis of R^3 of a matrix using eigenvaluesOrthogonally diagonalizable matrix and orthonormal basis
$begingroup$
I should show: If A is a symmetric matrix, then there exists an orthonormal basis of eigenvectors of A.
linear-algebra eigenvalues-eigenvectors orthonormal
asked Apr 6 at 19:35
Katrine NicolaisenKatrine Nicolaisen
1
$endgroup$
put on hold as off-topic by Javi, Xander Henderson, Eric Wofsey, egreg, Paul Frost Apr 6 at 22:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Xander Henderson, Eric Wofsey, egreg, Paul Frost
add a comment |
$begingroup$
I should show: If A is a symmetric matrix, then there exists an orthonormal basis of eigenvectors of A.
linear-algebra eigenvalues-eigenvectors orthonormal
asked Apr 6 at 19:35
Katrine NicolaisenKatrine Nicolaisen
1
$endgroup$
put on hold as off-topic by Javi, Xander Henderson, Eric Wofsey, egreg, Paul Frost Apr 6 at 22:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Xander Henderson, Eric Wofsey, egreg, Paul Frost
2
$begingroup$
A lot of textbooks prove this.
$endgroup$
– Lord Shark the Unknown
Apr 6 at 19:42
$begingroup$
You can do manually the case $nle 2$ and then proceed inductively by observing that: 1) if $A=A^t$ and $V$ is a $A$-invariant subspace, then $V^perp$ is $A$-invariant as well; 2) every real matrix $A$ has a $A$-invariant subspace $V$ of dimension either $1$ or $2$.
$endgroup$
– Saucy O'Path
Apr 6 at 20:05
add a comment |
$begingroup$
I should show: If A is a symmetric matrix, then there exists an orthonormal basis of eigenvectors of A.
linear-algebra eigenvalues-eigenvectors orthonormal
asked Apr 6 at 19:35
Katrine NicolaisenKatrine Nicolaisen
1
$endgroup$
I should show: If A is a symmetric matrix, then there exists an orthonormal basis of eigenvectors of A.
linear-algebra eigenvalues-eigenvectors orthonormal
linear-algebra eigenvalues-eigenvectors orthonormal
asked Apr 6 at 19:35
Katrine NicolaisenKatrine Nicolaisen
1
asked Apr 6 at 19:35
Katrine NicolaisenKatrine Nicolaisen
1
asked Apr 6 at 19:35
Katrine NicolaisenKatrine Nicolaisen
1
asked Apr 6 at 19:35
Katrine NicolaisenKatrine Nicolaisen
1
asked Apr 6 at 19:35
Katrine NicolaisenKatrine Nicolaisen
1
1
put on hold as off-topic by Javi, Xander Henderson, Eric Wofsey, egreg, Paul Frost Apr 6 at 22:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Xander Henderson, Eric Wofsey, egreg, Paul Frost
put on hold as off-topic by Javi, Xander Henderson, Eric Wofsey, egreg, Paul Frost Apr 6 at 22:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Xander Henderson, Eric Wofsey, egreg, Paul Frost
2
$begingroup$
A lot of textbooks prove this.
$endgroup$
– Lord Shark the Unknown
Apr 6 at 19:42
$begingroup$
You can do manually the case $nle 2$ and then proceed inductively by observing that: 1) if $A=A^t$ and $V$ is a $A$-invariant subspace, then $V^perp$ is $A$-invariant as well; 2) every real matrix $A$ has a $A$-invariant subspace $V$ of dimension either $1$ or $2$.
$endgroup$
– Saucy O'Path
Apr 6 at 20:05
add a comment |
2
$begingroup$
A lot of textbooks prove this.
$endgroup$
– Lord Shark the Unknown
Apr 6 at 19:42
$begingroup$
You can do manually the case $nle 2$ and then proceed inductively by observing that: 1) if $A=A^t$ and $V$ is a $A$-invariant subspace, then $V^perp$ is $A$-invariant as well; 2) every real matrix $A$ has a $A$-invariant subspace $V$ of dimension either $1$ or $2$.
$endgroup$
– Saucy O'Path
Apr 6 at 20:05
2
2
$begingroup$
A lot of textbooks prove this.
$endgroup$
– Lord Shark the Unknown
Apr 6 at 19:42
$begingroup$
A lot of textbooks prove this.
$endgroup$
– Lord Shark the Unknown
Apr 6 at 19:42
$begingroup$
You can do manually the case $nle 2$ and then proceed inductively by observing that: 1) if $A=A^t$ and $V$ is a $A$-invariant subspace, then $V^perp$ is $A$-invariant as well; 2) every real matrix $A$ has a $A$-invariant subspace $V$ of dimension either $1$ or $2$.
$endgroup$
– Saucy O'Path
Apr 6 at 20:05
$begingroup$
You can do manually the case $nle 2$ and then proceed inductively by observing that: 1) if $A=A^t$ and $V$ is a $A$-invariant subspace, then $V^perp$ is $A$-invariant as well; 2) every real matrix $A$ has a $A$-invariant subspace $V$ of dimension either $1$ or $2$.
$endgroup$
– Saucy O'Path
Apr 6 at 20:05
add a comment |
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2
$begingroup$
A lot of textbooks prove this.
$endgroup$
– Lord Shark the Unknown
Apr 6 at 19:42
$begingroup$
You can do manually the case $nle 2$ and then proceed inductively by observing that: 1) if $A=A^t$ and $V$ is a $A$-invariant subspace, then $V^perp$ is $A$-invariant as well; 2) every real matrix $A$ has a $A$-invariant subspace $V$ of dimension either $1$ or $2$.
$endgroup$
– Saucy O'Path
Apr 6 at 20:05