Mixing
Real symmetric matrix and real eigenvalues
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As I understood all eigenvalues of real symmetric matrix are real. But is it true that any real matrix with all real eigenvalues is symmetric?
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
asked Jan 30 at 4:58
ChaosPredictorChaosPredictor
1295
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add a comment |
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As I understood all eigenvalues of real symmetric matrix are real. But is it true that any real matrix with all real eigenvalues is symmetric?
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
asked Jan 30 at 4:58
ChaosPredictorChaosPredictor
1295
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Definitely not. You can probably think of a $2$ by $2$ case very easily.
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– MathIsLife12
Jan 30 at 5:01
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Thanks, [1 1; 1 0]. I think prove that it's wrong
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– ChaosPredictor
Jan 30 at 5:06
add a comment |
$begingroup$
As I understood all eigenvalues of real symmetric matrix are real. But is it true that any real matrix with all real eigenvalues is symmetric?
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
asked Jan 30 at 4:58
ChaosPredictorChaosPredictor
1295
$endgroup$
As I understood all eigenvalues of real symmetric matrix are real. But is it true that any real matrix with all real eigenvalues is symmetric?
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices
asked Jan 30 at 4:58
ChaosPredictorChaosPredictor
1295
asked Jan 30 at 4:58
ChaosPredictorChaosPredictor
1295
asked Jan 30 at 4:58
ChaosPredictorChaosPredictor
1295
asked Jan 30 at 4:58
ChaosPredictorChaosPredictor
1295
asked Jan 30 at 4:58
ChaosPredictorChaosPredictor
1295
1295
$begingroup$
Definitely not. You can probably think of a $2$ by $2$ case very easily.
$endgroup$
– MathIsLife12
Jan 30 at 5:01
$begingroup$
Thanks, [1 1; 1 0]. I think prove that it's wrong
$endgroup$
– ChaosPredictor
Jan 30 at 5:06
add a comment |
$begingroup$
Definitely not. You can probably think of a $2$ by $2$ case very easily.
$endgroup$
– MathIsLife12
Jan 30 at 5:01
$begingroup$
Thanks, [1 1; 1 0]. I think prove that it's wrong
$endgroup$
– ChaosPredictor
Jan 30 at 5:06
$begingroup$
Definitely not. You can probably think of a $2$ by $2$ case very easily.
$endgroup$
– MathIsLife12
Jan 30 at 5:01
$begingroup$
Definitely not. You can probably think of a $2$ by $2$ case very easily.
$endgroup$
– MathIsLife12
Jan 30 at 5:01
$begingroup$
Thanks, [1 1; 1 0]. I think prove that it's wrong
$endgroup$
– ChaosPredictor
Jan 30 at 5:06
$begingroup$
Thanks, [1 1; 1 0]. I think prove that it's wrong
$endgroup$
– ChaosPredictor
Jan 30 at 5:06
add a comment |
1 Answer
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No! Take any non zero nilpotent matrix with real entries!
For example, $$begin{pmatrix} 0&1\0&0end{pmatrix}$$
If any real matrix with all real eigenvalues is symmetric, then we have a conclusion: $$text{any real matrix with all real eigenvalues is diagonalizable!}$$ which is in general false!
answered Jan 30 at 5:07
Chinnapparaj RChinnapparaj R
5,8682928
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1 Answer
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1 Answer
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$begingroup$
No! Take any non zero nilpotent matrix with real entries!
For example, $$begin{pmatrix} 0&1\0&0end{pmatrix}$$
If any real matrix with all real eigenvalues is symmetric, then we have a conclusion: $$text{any real matrix with all real eigenvalues is diagonalizable!}$$ which is in general false!
answered Jan 30 at 5:07
Chinnapparaj RChinnapparaj R
5,8682928
$endgroup$
add a comment |
$begingroup$
No! Take any non zero nilpotent matrix with real entries!
For example, $$begin{pmatrix} 0&1\0&0end{pmatrix}$$
If any real matrix with all real eigenvalues is symmetric, then we have a conclusion: $$text{any real matrix with all real eigenvalues is diagonalizable!}$$ which is in general false!
answered Jan 30 at 5:07
Chinnapparaj RChinnapparaj R
5,8682928
$endgroup$
add a comment |
$begingroup$
No! Take any non zero nilpotent matrix with real entries!
For example, $$begin{pmatrix} 0&1\0&0end{pmatrix}$$
If any real matrix with all real eigenvalues is symmetric, then we have a conclusion: $$text{any real matrix with all real eigenvalues is diagonalizable!}$$ which is in general false!
answered Jan 30 at 5:07
Chinnapparaj RChinnapparaj R
5,8682928
$endgroup$
No! Take any non zero nilpotent matrix with real entries!
For example, $$begin{pmatrix} 0&1\0&0end{pmatrix}$$
If any real matrix with all real eigenvalues is symmetric, then we have a conclusion: $$text{any real matrix with all real eigenvalues is diagonalizable!}$$ which is in general false!
answered Jan 30 at 5:07
Chinnapparaj RChinnapparaj R
5,8682928
edited Jan 30 at 5:13
answered Jan 30 at 5:07
Chinnapparaj RChinnapparaj R
5,8682928
answered Jan 30 at 5:07
Chinnapparaj RChinnapparaj R
5,8682928
answered Jan 30 at 5:07
Chinnapparaj RChinnapparaj R
5,8682928
5,8682928
add a comment |
add a comment |
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$begingroup$
Definitely not. You can probably think of a $2$ by $2$ case very easily.
$endgroup$
– MathIsLife12
Jan 30 at 5:01
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Thanks, [1 1; 1 0]. I think prove that it's wrong
$endgroup$
– ChaosPredictor
Jan 30 at 5:06