Mixing
If A is a nxn singular matrix, then it has a singular value = 0
$begingroup$
This is a question on a testexam.
But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?
linear-algebra matrices singularvalues
asked Jan 21 at 7:59
TournaTourna
126
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add a comment |
$begingroup$
This is a question on a testexam.
But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?
linear-algebra matrices singularvalues
asked Jan 21 at 7:59
TournaTourna
126
$endgroup$
$begingroup$
How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:07
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I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
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– Tourna
Jan 21 at 8:11
$begingroup$
Yes, that is correct, and your argument is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:19
add a comment |
$begingroup$
This is a question on a testexam.
But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?
linear-algebra matrices singularvalues
asked Jan 21 at 7:59
TournaTourna
126
$endgroup$
This is a question on a testexam.
But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?
linear-algebra matrices singularvalues
linear-algebra matrices singularvalues
asked Jan 21 at 7:59
TournaTourna
126
asked Jan 21 at 7:59
TournaTourna
126
asked Jan 21 at 7:59
TournaTourna
126
asked Jan 21 at 7:59
TournaTourna
126
asked Jan 21 at 7:59
TournaTourna
126
126
$begingroup$
How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:07
$begingroup$
I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
$endgroup$
– Tourna
Jan 21 at 8:11
$begingroup$
Yes, that is correct, and your argument is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:19
add a comment |
$begingroup$
How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:07
$begingroup$
I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
$endgroup$
– Tourna
Jan 21 at 8:11
$begingroup$
Yes, that is correct, and your argument is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:19
$begingroup$
How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:07
$begingroup$
How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:07
$begingroup$
I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
$endgroup$
– Tourna
Jan 21 at 8:11
$begingroup$
I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
$endgroup$
– Tourna
Jan 21 at 8:11
$begingroup$
Yes, that is correct, and your argument is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:19
$begingroup$
Yes, that is correct, and your argument is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:19
add a comment |
2 Answers
2
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oldest
votes
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A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.
answered Jan 21 at 8:22
José Carlos SantosJosé Carlos Santos
165k22132235
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add a comment |
$begingroup$
If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.
If $A$ is singular, then $A^*A$ is singular.... Conclusion ?
answered Jan 21 at 8:13
FredFred
47.6k1849
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.
answered Jan 21 at 8:22
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.
answered Jan 21 at 8:22
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.
answered Jan 21 at 8:22
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
A $ntimes n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.
answered Jan 21 at 8:22
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 21 at 8:22
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 21 at 8:22
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 21 at 8:22
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
add a comment |
add a comment |
$begingroup$
If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.
If $A$ is singular, then $A^*A$ is singular.... Conclusion ?
answered Jan 21 at 8:13
FredFred
47.6k1849
$endgroup$
add a comment |
$begingroup$
If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.
If $A$ is singular, then $A^*A$ is singular.... Conclusion ?
answered Jan 21 at 8:13
FredFred
47.6k1849
$endgroup$
add a comment |
$begingroup$
If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.
If $A$ is singular, then $A^*A$ is singular.... Conclusion ?
answered Jan 21 at 8:13
FredFred
47.6k1849
$endgroup$
If $A$ is a square matrix, then $ lambda $ is a singular value of $A$ , then there is an eigenvalue $mu$ of $A^*A$ such that $lambda= mu^{1/2}$.
If $A$ is singular, then $A^*A$ is singular.... Conclusion ?
answered Jan 21 at 8:13
FredFred
47.6k1849
answered Jan 21 at 8:13
FredFred
47.6k1849
answered Jan 21 at 8:13
FredFred
47.6k1849
answered Jan 21 at 8:13
FredFred
47.6k1849
47.6k1849
add a comment |
add a comment |
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$begingroup$
How does a zero eigenvalue imply a zero singular value? If you explain this, then you are okay.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:07
$begingroup$
I thought that singular values were the squareroot of the A(transpose)A. And I thought that At and A have the same eigenvalues?
$endgroup$
– Tourna
Jan 21 at 8:11
$begingroup$
Yes, that is correct, and your argument is fine.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 21 at 8:19