Mixing
Let A be skew-symmetric, and denote its singular values by $sigma_1geq sigma_2geq dots sigma_ngeq0$.
$begingroup$
Let A be skew-symmetric, and denote its singular values by $sigma_1geq sigma_2geq dots sigma_ngeq0$. Show that
a) If n is even, then $sigma_{2k}=sigma_{2k-1}geq 0, k= 1,2,dots n/2.$ If n is odd, then the same relationship holds up to $k=(n-1)/2$ and also $sigma_n=0$.
b) The eigenvalues $lambda_j=(-1)^jisigma_j$, $j=1,2,dots,n$.
I know that skew symmetric means $-A=A^T$ and I know that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. I am not able to get this though and I have been trying all week...
Thanks for your time.
linear-algebra matrices eigenvalues-eigenvectors
asked Oct 5 '16 at 18:35
MathIsHardMathIsHard
1,318516
$endgroup$
add a comment |
$begingroup$
Let A be skew-symmetric, and denote its singular values by $sigma_1geq sigma_2geq dots sigma_ngeq0$. Show that
a) If n is even, then $sigma_{2k}=sigma_{2k-1}geq 0, k= 1,2,dots n/2.$ If n is odd, then the same relationship holds up to $k=(n-1)/2$ and also $sigma_n=0$.
b) The eigenvalues $lambda_j=(-1)^jisigma_j$, $j=1,2,dots,n$.
I know that skew symmetric means $-A=A^T$ and I know that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. I am not able to get this though and I have been trying all week...
Thanks for your time.
linear-algebra matrices eigenvalues-eigenvectors
asked Oct 5 '16 at 18:35
MathIsHardMathIsHard
1,318516
$endgroup$
add a comment |
$begingroup$
Let A be skew-symmetric, and denote its singular values by $sigma_1geq sigma_2geq dots sigma_ngeq0$. Show that
a) If n is even, then $sigma_{2k}=sigma_{2k-1}geq 0, k= 1,2,dots n/2.$ If n is odd, then the same relationship holds up to $k=(n-1)/2$ and also $sigma_n=0$.
b) The eigenvalues $lambda_j=(-1)^jisigma_j$, $j=1,2,dots,n$.
I know that skew symmetric means $-A=A^T$ and I know that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. I am not able to get this though and I have been trying all week...
Thanks for your time.
linear-algebra matrices eigenvalues-eigenvectors
asked Oct 5 '16 at 18:35
MathIsHardMathIsHard
1,318516
$endgroup$
Let A be skew-symmetric, and denote its singular values by $sigma_1geq sigma_2geq dots sigma_ngeq0$. Show that
a) If n is even, then $sigma_{2k}=sigma_{2k-1}geq 0, k= 1,2,dots n/2.$ If n is odd, then the same relationship holds up to $k=(n-1)/2$ and also $sigma_n=0$.
b) The eigenvalues $lambda_j=(-1)^jisigma_j$, $j=1,2,dots,n$.
I know that skew symmetric means $-A=A^T$ and I know that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. I am not able to get this though and I have been trying all week...
Thanks for your time.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Oct 5 '16 at 18:35
MathIsHardMathIsHard
1,318516
asked Oct 5 '16 at 18:35
MathIsHardMathIsHard
1,318516
asked Oct 5 '16 at 18:35
MathIsHardMathIsHard
1,318516
asked Oct 5 '16 at 18:35
MathIsHardMathIsHard
1,318516
asked Oct 5 '16 at 18:35
MathIsHardMathIsHard
1,318516
1,318516
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Hint: The singular values of $A$ are the (positive) square roots of the eigenvalues of the matrix $A^TA = -A^2$. So, if $lambda$ is an eigenvalue of $A$, then $sqrt{-lambda^2} = |lambda|$ is a singular value.
Because $A$ is real, its complex (i.e. non-real) eigenvalues must come in complex-conjugate pairs.
answered Oct 5 '16 at 19:22
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Thank you. It was my understanding that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. Is this true? Wouldn't that mean they don't come in conjugate pairs?
$endgroup$
– MathIsHard
Oct 5 '16 at 20:13
$begingroup$
Yes, that's true, but it does not mean that they don't come in conjugate pairs. For every non-zero eigenvalue $ai$, $-ai$ must also be an eigenvalue.
$endgroup$
– Omnomnomnom
Oct 5 '16 at 20:20
$begingroup$
Ok. I am still stuck.. I understand what you have said though... Sorry this problem and I don't get along.
$endgroup$
– MathIsHard
Oct 5 '16 at 20:36
$begingroup$
Isn't $sqrt{-lambda^2} = i|lambda| not= |lambda|$? I feel that I am missing something obvious.
$endgroup$
– hasManyStupidQuestions
Jan 26 at 4:15
1
$begingroup$
@hasManyStupidQuestions Right. Regarding your first comment, note that $lambda$ is actually an imaginary number in this context
$endgroup$
– Omnomnomnom
Jan 28 at 17:50
|
show 2 more comments
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1 Answer
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1 Answer
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$begingroup$
Hint: The singular values of $A$ are the (positive) square roots of the eigenvalues of the matrix $A^TA = -A^2$. So, if $lambda$ is an eigenvalue of $A$, then $sqrt{-lambda^2} = |lambda|$ is a singular value.
Because $A$ is real, its complex (i.e. non-real) eigenvalues must come in complex-conjugate pairs.
answered Oct 5 '16 at 19:22
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Thank you. It was my understanding that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. Is this true? Wouldn't that mean they don't come in conjugate pairs?
$endgroup$
– MathIsHard
Oct 5 '16 at 20:13
$begingroup$
Yes, that's true, but it does not mean that they don't come in conjugate pairs. For every non-zero eigenvalue $ai$, $-ai$ must also be an eigenvalue.
$endgroup$
– Omnomnomnom
Oct 5 '16 at 20:20
$begingroup$
Ok. I am still stuck.. I understand what you have said though... Sorry this problem and I don't get along.
$endgroup$
– MathIsHard
Oct 5 '16 at 20:36
$begingroup$
Isn't $sqrt{-lambda^2} = i|lambda| not= |lambda|$? I feel that I am missing something obvious.
$endgroup$
– hasManyStupidQuestions
Jan 26 at 4:15
1
$begingroup$
@hasManyStupidQuestions Right. Regarding your first comment, note that $lambda$ is actually an imaginary number in this context
$endgroup$
– Omnomnomnom
Jan 28 at 17:50
|
show 2 more comments
$begingroup$
Hint: The singular values of $A$ are the (positive) square roots of the eigenvalues of the matrix $A^TA = -A^2$. So, if $lambda$ is an eigenvalue of $A$, then $sqrt{-lambda^2} = |lambda|$ is a singular value.
Because $A$ is real, its complex (i.e. non-real) eigenvalues must come in complex-conjugate pairs.
answered Oct 5 '16 at 19:22
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
$begingroup$
Thank you. It was my understanding that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. Is this true? Wouldn't that mean they don't come in conjugate pairs?
$endgroup$
– MathIsHard
Oct 5 '16 at 20:13
$begingroup$
Yes, that's true, but it does not mean that they don't come in conjugate pairs. For every non-zero eigenvalue $ai$, $-ai$ must also be an eigenvalue.
$endgroup$
– Omnomnomnom
Oct 5 '16 at 20:20
$begingroup$
Ok. I am still stuck.. I understand what you have said though... Sorry this problem and I don't get along.
$endgroup$
– MathIsHard
Oct 5 '16 at 20:36
$begingroup$
Isn't $sqrt{-lambda^2} = i|lambda| not= |lambda|$? I feel that I am missing something obvious.
$endgroup$
– hasManyStupidQuestions
Jan 26 at 4:15
1
$begingroup$
@hasManyStupidQuestions Right. Regarding your first comment, note that $lambda$ is actually an imaginary number in this context
$endgroup$
– Omnomnomnom
Jan 28 at 17:50
|
show 2 more comments
$begingroup$
Hint: The singular values of $A$ are the (positive) square roots of the eigenvalues of the matrix $A^TA = -A^2$. So, if $lambda$ is an eigenvalue of $A$, then $sqrt{-lambda^2} = |lambda|$ is a singular value.
Because $A$ is real, its complex (i.e. non-real) eigenvalues must come in complex-conjugate pairs.
answered Oct 5 '16 at 19:22
OmnomnomnomOmnomnomnom
129k792186
$endgroup$
Hint: The singular values of $A$ are the (positive) square roots of the eigenvalues of the matrix $A^TA = -A^2$. So, if $lambda$ is an eigenvalue of $A$, then $sqrt{-lambda^2} = |lambda|$ is a singular value.
Because $A$ is real, its complex (i.e. non-real) eigenvalues must come in complex-conjugate pairs.
answered Oct 5 '16 at 19:22
OmnomnomnomOmnomnomnom
129k792186
edited Jan 28 at 17:47
answered Oct 5 '16 at 19:22
OmnomnomnomOmnomnomnom
129k792186
answered Oct 5 '16 at 19:22
OmnomnomnomOmnomnomnom
129k792186
answered Oct 5 '16 at 19:22
OmnomnomnomOmnomnomnom
129k792186
129k792186
$begingroup$
Thank you. It was my understanding that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. Is this true? Wouldn't that mean they don't come in conjugate pairs?
$endgroup$
– MathIsHard
Oct 5 '16 at 20:13
$begingroup$
Yes, that's true, but it does not mean that they don't come in conjugate pairs. For every non-zero eigenvalue $ai$, $-ai$ must also be an eigenvalue.
$endgroup$
– Omnomnomnom
Oct 5 '16 at 20:20
$begingroup$
Ok. I am still stuck.. I understand what you have said though... Sorry this problem and I don't get along.
$endgroup$
– MathIsHard
Oct 5 '16 at 20:36
$begingroup$
Isn't $sqrt{-lambda^2} = i|lambda| not= |lambda|$? I feel that I am missing something obvious.
$endgroup$
– hasManyStupidQuestions
Jan 26 at 4:15
1
$begingroup$
@hasManyStupidQuestions Right. Regarding your first comment, note that $lambda$ is actually an imaginary number in this context
$endgroup$
– Omnomnomnom
Jan 28 at 17:50
|
show 2 more comments
$begingroup$
Thank you. It was my understanding that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. Is this true? Wouldn't that mean they don't come in conjugate pairs?
$endgroup$
– MathIsHard
Oct 5 '16 at 20:13
$begingroup$
Yes, that's true, but it does not mean that they don't come in conjugate pairs. For every non-zero eigenvalue $ai$, $-ai$ must also be an eigenvalue.
$endgroup$
– Omnomnomnom
Oct 5 '16 at 20:20
$begingroup$
Ok. I am still stuck.. I understand what you have said though... Sorry this problem and I don't get along.
$endgroup$
– MathIsHard
Oct 5 '16 at 20:36
$begingroup$
Isn't $sqrt{-lambda^2} = i|lambda| not= |lambda|$? I feel that I am missing something obvious.
$endgroup$
– hasManyStupidQuestions
Jan 26 at 4:15
1
$begingroup$
@hasManyStupidQuestions Right. Regarding your first comment, note that $lambda$ is actually an imaginary number in this context
$endgroup$
– Omnomnomnom
Jan 28 at 17:50
$begingroup$
Thank you. It was my understanding that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. Is this true? Wouldn't that mean they don't come in conjugate pairs?
$endgroup$
– MathIsHard
Oct 5 '16 at 20:13
$begingroup$
Thank you. It was my understanding that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. Is this true? Wouldn't that mean they don't come in conjugate pairs?
$endgroup$
– MathIsHard
Oct 5 '16 at 20:13
$begingroup$
Yes, that's true, but it does not mean that they don't come in conjugate pairs. For every non-zero eigenvalue $ai$, $-ai$ must also be an eigenvalue.
$endgroup$
– Omnomnomnom
Oct 5 '16 at 20:20
$begingroup$
Yes, that's true, but it does not mean that they don't come in conjugate pairs. For every non-zero eigenvalue $ai$, $-ai$ must also be an eigenvalue.
$endgroup$
– Omnomnomnom
Oct 5 '16 at 20:20
$begingroup$
Ok. I am still stuck.. I understand what you have said though... Sorry this problem and I don't get along.
$endgroup$
– MathIsHard
Oct 5 '16 at 20:36
$begingroup$
Ok. I am still stuck.. I understand what you have said though... Sorry this problem and I don't get along.
$endgroup$
– MathIsHard
Oct 5 '16 at 20:36
$begingroup$
Isn't $sqrt{-lambda^2} = i|lambda| not= |lambda|$? I feel that I am missing something obvious.
$endgroup$
– hasManyStupidQuestions
Jan 26 at 4:15
$begingroup$
Isn't $sqrt{-lambda^2} = i|lambda| not= |lambda|$? I feel that I am missing something obvious.
$endgroup$
– hasManyStupidQuestions
Jan 26 at 4:15
1
1
$begingroup$
@hasManyStupidQuestions Right. Regarding your first comment, note that $lambda$ is actually an imaginary number in this context
$endgroup$
– Omnomnomnom
Jan 28 at 17:50
$begingroup$
@hasManyStupidQuestions Right. Regarding your first comment, note that $lambda$ is actually an imaginary number in this context
$endgroup$
– Omnomnomnom
Jan 28 at 17:50
|
show 2 more comments
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