Singular and eigen values properties…
$begingroup$
Let $Ainmathcal {M}_n(mathbb{R})$, we will denote $lambda_{max}(A)$ the biggest eigenvalue of $A$ in absolute value, as for $Binmathcal M_{m,n}(mathbb{R})$ we will denote $sigma_{max}(B)$ the highest singular value of $B$.
I have $3$ things to show from which I have shown 1 and for the last one I only have an idea.
$1$. with $A$ a symetric matrix show that $lambda_{max}(A)= max_{|v|=1}v^tAv$
$$A=A^timplies max_{|v|=1}(v^tAv)
=max_{|v|=1}(v^tA^tv)
=max_{|v|=1}((Av)^tv)
=max_{|v|=1}((lambda v)^tv)
=max_{|v|=1}(lambda v^tv)
=max_{|v|=1}(lambda |v|^2)
=max_{|v|=1}(lambda)
=max(lambda).$$
$2$. Show that $sigma_{max}(B) = sqrt{lambda_{max}(B^tB)}$.
$3$. with $Cinmathcal M_n(mathbb{R})implieslambda_{max}(frac{C+C^t}{2})leqsigma_{max}(C)$.
Here I think is something relating with AM-GM Mean right?
linear-algebra eigenvalues-eigenvectors singularvalues
edited Dec 6 '18 at 18:04
zxmkn
340213
asked Dec 6 '18 at 17:33
C. CristiC. Cristi
1,636218
$endgroup$
add a comment |
$begingroup$
Let $Ainmathcal {M}_n(mathbb{R})$, we will denote $lambda_{max}(A)$ the biggest eigenvalue of $A$ in absolute value, as for $Binmathcal M_{m,n}(mathbb{R})$ we will denote $sigma_{max}(B)$ the highest singular value of $B$.
I have $3$ things to show from which I have shown 1 and for the last one I only have an idea.
$1$. with $A$ a symetric matrix show that $lambda_{max}(A)= max_{|v|=1}v^tAv$
$$A=A^timplies max_{|v|=1}(v^tAv)
=max_{|v|=1}(v^tA^tv)
=max_{|v|=1}((Av)^tv)
=max_{|v|=1}((lambda v)^tv)
=max_{|v|=1}(lambda v^tv)
=max_{|v|=1}(lambda |v|^2)
=max_{|v|=1}(lambda)
=max(lambda).$$
$2$. Show that $sigma_{max}(B) = sqrt{lambda_{max}(B^tB)}$.
$3$. with $Cinmathcal M_n(mathbb{R})implieslambda_{max}(frac{C+C^t}{2})leqsigma_{max}(C)$.
Here I think is something relating with AM-GM Mean right?
linear-algebra eigenvalues-eigenvectors singularvalues
edited Dec 6 '18 at 18:04
zxmkn
340213
asked Dec 6 '18 at 17:33
C. CristiC. Cristi
1,636218
$endgroup$
$begingroup$
For part 1 to make sense, the definition of $lambda_{max}(A)$ should be the biggest eigenvalue of $A$ not in absolute value.
$endgroup$
– angryavian
Dec 6 '18 at 17:55
add a comment |
$begingroup$
Let $Ainmathcal {M}_n(mathbb{R})$, we will denote $lambda_{max}(A)$ the biggest eigenvalue of $A$ in absolute value, as for $Binmathcal M_{m,n}(mathbb{R})$ we will denote $sigma_{max}(B)$ the highest singular value of $B$.
I have $3$ things to show from which I have shown 1 and for the last one I only have an idea.
$1$. with $A$ a symetric matrix show that $lambda_{max}(A)= max_{|v|=1}v^tAv$
$$A=A^timplies max_{|v|=1}(v^tAv)
=max_{|v|=1}(v^tA^tv)
=max_{|v|=1}((Av)^tv)
=max_{|v|=1}((lambda v)^tv)
=max_{|v|=1}(lambda v^tv)
=max_{|v|=1}(lambda |v|^2)
=max_{|v|=1}(lambda)
=max(lambda).$$
$2$. Show that $sigma_{max}(B) = sqrt{lambda_{max}(B^tB)}$.
$3$. with $Cinmathcal M_n(mathbb{R})implieslambda_{max}(frac{C+C^t}{2})leqsigma_{max}(C)$.
Here I think is something relating with AM-GM Mean right?
linear-algebra eigenvalues-eigenvectors singularvalues
edited Dec 6 '18 at 18:04
zxmkn
340213
asked Dec 6 '18 at 17:33
C. CristiC. Cristi
1,636218
$endgroup$
Let $Ainmathcal {M}_n(mathbb{R})$, we will denote $lambda_{max}(A)$ the biggest eigenvalue of $A$ in absolute value, as for $Binmathcal M_{m,n}(mathbb{R})$ we will denote $sigma_{max}(B)$ the highest singular value of $B$.
I have $3$ things to show from which I have shown 1 and for the last one I only have an idea.
$1$. with $A$ a symetric matrix show that $lambda_{max}(A)= max_{|v|=1}v^tAv$
$$A=A^timplies max_{|v|=1}(v^tAv)
=max_{|v|=1}(v^tA^tv)
=max_{|v|=1}((Av)^tv)
=max_{|v|=1}((lambda v)^tv)
=max_{|v|=1}(lambda v^tv)
=max_{|v|=1}(lambda |v|^2)
=max_{|v|=1}(lambda)
=max(lambda).$$
$2$. Show that $sigma_{max}(B) = sqrt{lambda_{max}(B^tB)}$.
$3$. with $Cinmathcal M_n(mathbb{R})implieslambda_{max}(frac{C+C^t}{2})leqsigma_{max}(C)$.
Here I think is something relating with AM-GM Mean right?
linear-algebra eigenvalues-eigenvectors singularvalues
linear-algebra eigenvalues-eigenvectors singularvalues
edited Dec 6 '18 at 18:04
zxmkn
340213
asked Dec 6 '18 at 17:33
C. CristiC. Cristi
1,636218
edited Dec 6 '18 at 18:04
zxmkn
340213
asked Dec 6 '18 at 17:33
C. CristiC. Cristi
1,636218
edited Dec 6 '18 at 18:04
zxmkn
340213
edited Dec 6 '18 at 18:04
zxmkn
340213
edited Dec 6 '18 at 18:04
zxmkn
340213
340213
asked Dec 6 '18 at 17:33
C. CristiC. Cristi
1,636218
asked Dec 6 '18 at 17:33
C. CristiC. Cristi
1,636218
asked Dec 6 '18 at 17:33
C. CristiC. Cristi
1,636218
1,636218
$begingroup$
For part 1 to make sense, the definition of $lambda_{max}(A)$ should be the biggest eigenvalue of $A$ not in absolute value.
$endgroup$
– angryavian
Dec 6 '18 at 17:55
add a comment |
$begingroup$
For part 1 to make sense, the definition of $lambda_{max}(A)$ should be the biggest eigenvalue of $A$ not in absolute value.
$endgroup$
– angryavian
Dec 6 '18 at 17:55
$begingroup$
For part 1 to make sense, the definition of $lambda_{max}(A)$ should be the biggest eigenvalue of $A$ not in absolute value.
$endgroup$
– angryavian
Dec 6 '18 at 17:55
$begingroup$
For part 1 to make sense, the definition of $lambda_{max}(A)$ should be the biggest eigenvalue of $A$ not in absolute value.
$endgroup$
– angryavian
Dec 6 '18 at 17:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are a few errors in your attempt. When you use $Av = lambda v$ you are assuming $v$ is an eigenvector of $A$, even though the maximum is over all unit norm $v$ which may include non-eigenvectors. Also your attempt does not really use symmetry; note that you could have proceeded as $v^t A v = v^t (lambda v) = lambda |v|^2$ directly (disregarding the eigenvalue issue mentioned above). Symmetry is important here because of the spectral theorem. This may help you fix your proof.
I suppose you are defining singular values from the SVD? Then write $B=USigma V^t$ and note $B^t B = V Sigma^t Sigma V^t$. Since $Sigma^t Sigma$ is a diagonal matrix, this is a diagonalization of $B^t B$, so you can write down the eigenvalues of $B^t B$ in terms of the singular values of $B$.
Let $USigma V^t$ be the SVD of $C$. Note $(C+C^t)/2$ is symmetric. Using the first part, $$lambda_{max}(frac{C+C^t}{2}) = max_{|v|=1} v^t C v = max_{|v|=1} v^t U Sigma V^t v le max_{|x|=|y|=1} x^t Sigma y = sigma_{max}(C),$$
where the inequality comes from the change of variables $x=U^t v$ and $y = V^t v$ and noting that orthogonal matrices are norm-preserving (i.e. $|x|=|v|$).
Edit:
As I mentioned in my comment, I think $lambda_{max}$ should be the largest eigenvalue, not the largest in absolute value.
Response to your comment:
Let $UDU^t$ be the eigendecomposition of $A$. Then $max_{|v|=1} v^t A v = max_{|v|=1} v^t UDU^t v = max_{|w| = 1} w^t D w = lambda_{max}(A)$, where we have made the change of variables $w = U^t v$ and noted that orthogonal matrices are norm-preserving.
I already told you that the eigenvalues of $B^t B$ are the diagonal entries of the diagonal matrix $Sigma^t Sigma$.
answered Dec 6 '18 at 17:45
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
1. Yeah I saw, okay to use the spectral theorem we have that: if $A$ is symetric then there is $Q$ and $D$ such that $A=QDQ^t$ then $A$ is orto-diagonalizable... but how do I use that here, since if I plug in $A$ in there I don't solve anything, can you be more explicit, also, for 2, yes, indeed it's about the SVD, how do I write down the eigenvalues of $B^tB$ in terms of singular values of $B$?
$endgroup$
– C. Cristi
Dec 6 '18 at 17:51
$begingroup$
@C.Cristi See my edit.
$endgroup$
– angryavian
Dec 6 '18 at 18:05
$begingroup$
How did you make the note that the norm of $U^tv=1$ is the same?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:08
$begingroup$
Hey, why $lambda_{max}(frac {C+C^t}{2})=max_{|v|=1}v^tCv$ and not $=max_{|v|=1}v^tfrac{C+C^t}{2}v$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:16
$begingroup$
Orthogonal matrix are norm-preserving because $|Uv|=sqrt{langle Uv,Uvrangle}=sqrt{langle U^tv,Uvrangle}=sqrt{UU^tlangle v, vrangle}= sqrt{langle v, vrangle}$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:22
|
show 2 more comments
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$begingroup$
There are a few errors in your attempt. When you use $Av = lambda v$ you are assuming $v$ is an eigenvector of $A$, even though the maximum is over all unit norm $v$ which may include non-eigenvectors. Also your attempt does not really use symmetry; note that you could have proceeded as $v^t A v = v^t (lambda v) = lambda |v|^2$ directly (disregarding the eigenvalue issue mentioned above). Symmetry is important here because of the spectral theorem. This may help you fix your proof.
I suppose you are defining singular values from the SVD? Then write $B=USigma V^t$ and note $B^t B = V Sigma^t Sigma V^t$. Since $Sigma^t Sigma$ is a diagonal matrix, this is a diagonalization of $B^t B$, so you can write down the eigenvalues of $B^t B$ in terms of the singular values of $B$.
Let $USigma V^t$ be the SVD of $C$. Note $(C+C^t)/2$ is symmetric. Using the first part, $$lambda_{max}(frac{C+C^t}{2}) = max_{|v|=1} v^t C v = max_{|v|=1} v^t U Sigma V^t v le max_{|x|=|y|=1} x^t Sigma y = sigma_{max}(C),$$
where the inequality comes from the change of variables $x=U^t v$ and $y = V^t v$ and noting that orthogonal matrices are norm-preserving (i.e. $|x|=|v|$).
Edit:
As I mentioned in my comment, I think $lambda_{max}$ should be the largest eigenvalue, not the largest in absolute value.
Response to your comment:
Let $UDU^t$ be the eigendecomposition of $A$. Then $max_{|v|=1} v^t A v = max_{|v|=1} v^t UDU^t v = max_{|w| = 1} w^t D w = lambda_{max}(A)$, where we have made the change of variables $w = U^t v$ and noted that orthogonal matrices are norm-preserving.
I already told you that the eigenvalues of $B^t B$ are the diagonal entries of the diagonal matrix $Sigma^t Sigma$.
answered Dec 6 '18 at 17:45
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
1. Yeah I saw, okay to use the spectral theorem we have that: if $A$ is symetric then there is $Q$ and $D$ such that $A=QDQ^t$ then $A$ is orto-diagonalizable... but how do I use that here, since if I plug in $A$ in there I don't solve anything, can you be more explicit, also, for 2, yes, indeed it's about the SVD, how do I write down the eigenvalues of $B^tB$ in terms of singular values of $B$?
$endgroup$
– C. Cristi
Dec 6 '18 at 17:51
$begingroup$
@C.Cristi See my edit.
$endgroup$
– angryavian
Dec 6 '18 at 18:05
$begingroup$
How did you make the note that the norm of $U^tv=1$ is the same?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:08
$begingroup$
Hey, why $lambda_{max}(frac {C+C^t}{2})=max_{|v|=1}v^tCv$ and not $=max_{|v|=1}v^tfrac{C+C^t}{2}v$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:16
$begingroup$
Orthogonal matrix are norm-preserving because $|Uv|=sqrt{langle Uv,Uvrangle}=sqrt{langle U^tv,Uvrangle}=sqrt{UU^tlangle v, vrangle}= sqrt{langle v, vrangle}$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:22
|
show 2 more comments
$begingroup$
There are a few errors in your attempt. When you use $Av = lambda v$ you are assuming $v$ is an eigenvector of $A$, even though the maximum is over all unit norm $v$ which may include non-eigenvectors. Also your attempt does not really use symmetry; note that you could have proceeded as $v^t A v = v^t (lambda v) = lambda |v|^2$ directly (disregarding the eigenvalue issue mentioned above). Symmetry is important here because of the spectral theorem. This may help you fix your proof.
I suppose you are defining singular values from the SVD? Then write $B=USigma V^t$ and note $B^t B = V Sigma^t Sigma V^t$. Since $Sigma^t Sigma$ is a diagonal matrix, this is a diagonalization of $B^t B$, so you can write down the eigenvalues of $B^t B$ in terms of the singular values of $B$.
Let $USigma V^t$ be the SVD of $C$. Note $(C+C^t)/2$ is symmetric. Using the first part, $$lambda_{max}(frac{C+C^t}{2}) = max_{|v|=1} v^t C v = max_{|v|=1} v^t U Sigma V^t v le max_{|x|=|y|=1} x^t Sigma y = sigma_{max}(C),$$
where the inequality comes from the change of variables $x=U^t v$ and $y = V^t v$ and noting that orthogonal matrices are norm-preserving (i.e. $|x|=|v|$).
Edit:
As I mentioned in my comment, I think $lambda_{max}$ should be the largest eigenvalue, not the largest in absolute value.
Response to your comment:
Let $UDU^t$ be the eigendecomposition of $A$. Then $max_{|v|=1} v^t A v = max_{|v|=1} v^t UDU^t v = max_{|w| = 1} w^t D w = lambda_{max}(A)$, where we have made the change of variables $w = U^t v$ and noted that orthogonal matrices are norm-preserving.
I already told you that the eigenvalues of $B^t B$ are the diagonal entries of the diagonal matrix $Sigma^t Sigma$.
answered Dec 6 '18 at 17:45
angryavianangryavian
42.1k23381
$endgroup$
$begingroup$
1. Yeah I saw, okay to use the spectral theorem we have that: if $A$ is symetric then there is $Q$ and $D$ such that $A=QDQ^t$ then $A$ is orto-diagonalizable... but how do I use that here, since if I plug in $A$ in there I don't solve anything, can you be more explicit, also, for 2, yes, indeed it's about the SVD, how do I write down the eigenvalues of $B^tB$ in terms of singular values of $B$?
$endgroup$
– C. Cristi
Dec 6 '18 at 17:51
$begingroup$
@C.Cristi See my edit.
$endgroup$
– angryavian
Dec 6 '18 at 18:05
$begingroup$
How did you make the note that the norm of $U^tv=1$ is the same?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:08
$begingroup$
Hey, why $lambda_{max}(frac {C+C^t}{2})=max_{|v|=1}v^tCv$ and not $=max_{|v|=1}v^tfrac{C+C^t}{2}v$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:16
$begingroup$
Orthogonal matrix are norm-preserving because $|Uv|=sqrt{langle Uv,Uvrangle}=sqrt{langle U^tv,Uvrangle}=sqrt{UU^tlangle v, vrangle}= sqrt{langle v, vrangle}$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:22
|
show 2 more comments
$begingroup$
There are a few errors in your attempt. When you use $Av = lambda v$ you are assuming $v$ is an eigenvector of $A$, even though the maximum is over all unit norm $v$ which may include non-eigenvectors. Also your attempt does not really use symmetry; note that you could have proceeded as $v^t A v = v^t (lambda v) = lambda |v|^2$ directly (disregarding the eigenvalue issue mentioned above). Symmetry is important here because of the spectral theorem. This may help you fix your proof.
I suppose you are defining singular values from the SVD? Then write $B=USigma V^t$ and note $B^t B = V Sigma^t Sigma V^t$. Since $Sigma^t Sigma$ is a diagonal matrix, this is a diagonalization of $B^t B$, so you can write down the eigenvalues of $B^t B$ in terms of the singular values of $B$.
Let $USigma V^t$ be the SVD of $C$. Note $(C+C^t)/2$ is symmetric. Using the first part, $$lambda_{max}(frac{C+C^t}{2}) = max_{|v|=1} v^t C v = max_{|v|=1} v^t U Sigma V^t v le max_{|x|=|y|=1} x^t Sigma y = sigma_{max}(C),$$
where the inequality comes from the change of variables $x=U^t v$ and $y = V^t v$ and noting that orthogonal matrices are norm-preserving (i.e. $|x|=|v|$).
Edit:
As I mentioned in my comment, I think $lambda_{max}$ should be the largest eigenvalue, not the largest in absolute value.
Response to your comment:
Let $UDU^t$ be the eigendecomposition of $A$. Then $max_{|v|=1} v^t A v = max_{|v|=1} v^t UDU^t v = max_{|w| = 1} w^t D w = lambda_{max}(A)$, where we have made the change of variables $w = U^t v$ and noted that orthogonal matrices are norm-preserving.
I already told you that the eigenvalues of $B^t B$ are the diagonal entries of the diagonal matrix $Sigma^t Sigma$.
answered Dec 6 '18 at 17:45
angryavianangryavian
42.1k23381
$endgroup$
There are a few errors in your attempt. When you use $Av = lambda v$ you are assuming $v$ is an eigenvector of $A$, even though the maximum is over all unit norm $v$ which may include non-eigenvectors. Also your attempt does not really use symmetry; note that you could have proceeded as $v^t A v = v^t (lambda v) = lambda |v|^2$ directly (disregarding the eigenvalue issue mentioned above). Symmetry is important here because of the spectral theorem. This may help you fix your proof.
I suppose you are defining singular values from the SVD? Then write $B=USigma V^t$ and note $B^t B = V Sigma^t Sigma V^t$. Since $Sigma^t Sigma$ is a diagonal matrix, this is a diagonalization of $B^t B$, so you can write down the eigenvalues of $B^t B$ in terms of the singular values of $B$.
Let $USigma V^t$ be the SVD of $C$. Note $(C+C^t)/2$ is symmetric. Using the first part, $$lambda_{max}(frac{C+C^t}{2}) = max_{|v|=1} v^t C v = max_{|v|=1} v^t U Sigma V^t v le max_{|x|=|y|=1} x^t Sigma y = sigma_{max}(C),$$
where the inequality comes from the change of variables $x=U^t v$ and $y = V^t v$ and noting that orthogonal matrices are norm-preserving (i.e. $|x|=|v|$).
Edit:
As I mentioned in my comment, I think $lambda_{max}$ should be the largest eigenvalue, not the largest in absolute value.
Response to your comment:
Let $UDU^t$ be the eigendecomposition of $A$. Then $max_{|v|=1} v^t A v = max_{|v|=1} v^t UDU^t v = max_{|w| = 1} w^t D w = lambda_{max}(A)$, where we have made the change of variables $w = U^t v$ and noted that orthogonal matrices are norm-preserving.
I already told you that the eigenvalues of $B^t B$ are the diagonal entries of the diagonal matrix $Sigma^t Sigma$.
answered Dec 6 '18 at 17:45
angryavianangryavian
42.1k23381
edited Dec 6 '18 at 18:01
answered Dec 6 '18 at 17:45
angryavianangryavian
42.1k23381
answered Dec 6 '18 at 17:45
angryavianangryavian
42.1k23381
answered Dec 6 '18 at 17:45
angryavianangryavian
42.1k23381
42.1k23381
$begingroup$
1. Yeah I saw, okay to use the spectral theorem we have that: if $A$ is symetric then there is $Q$ and $D$ such that $A=QDQ^t$ then $A$ is orto-diagonalizable... but how do I use that here, since if I plug in $A$ in there I don't solve anything, can you be more explicit, also, for 2, yes, indeed it's about the SVD, how do I write down the eigenvalues of $B^tB$ in terms of singular values of $B$?
$endgroup$
– C. Cristi
Dec 6 '18 at 17:51
$begingroup$
@C.Cristi See my edit.
$endgroup$
– angryavian
Dec 6 '18 at 18:05
$begingroup$
How did you make the note that the norm of $U^tv=1$ is the same?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:08
$begingroup$
Hey, why $lambda_{max}(frac {C+C^t}{2})=max_{|v|=1}v^tCv$ and not $=max_{|v|=1}v^tfrac{C+C^t}{2}v$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:16
$begingroup$
Orthogonal matrix are norm-preserving because $|Uv|=sqrt{langle Uv,Uvrangle}=sqrt{langle U^tv,Uvrangle}=sqrt{UU^tlangle v, vrangle}= sqrt{langle v, vrangle}$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:22
|
show 2 more comments
$begingroup$
1. Yeah I saw, okay to use the spectral theorem we have that: if $A$ is symetric then there is $Q$ and $D$ such that $A=QDQ^t$ then $A$ is orto-diagonalizable... but how do I use that here, since if I plug in $A$ in there I don't solve anything, can you be more explicit, also, for 2, yes, indeed it's about the SVD, how do I write down the eigenvalues of $B^tB$ in terms of singular values of $B$?
$endgroup$
– C. Cristi
Dec 6 '18 at 17:51
$begingroup$
@C.Cristi See my edit.
$endgroup$
– angryavian
Dec 6 '18 at 18:05
$begingroup$
How did you make the note that the norm of $U^tv=1$ is the same?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:08
$begingroup$
Hey, why $lambda_{max}(frac {C+C^t}{2})=max_{|v|=1}v^tCv$ and not $=max_{|v|=1}v^tfrac{C+C^t}{2}v$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:16
$begingroup$
Orthogonal matrix are norm-preserving because $|Uv|=sqrt{langle Uv,Uvrangle}=sqrt{langle U^tv,Uvrangle}=sqrt{UU^tlangle v, vrangle}= sqrt{langle v, vrangle}$?
$endgroup$
– C. Cristi
Dec 6 '18 at 18:22
$begingroup$
1. Yeah I saw, okay to use the spectral theorem we have that: if $A$ is symetric then there is $Q$ and $D$ such that $A=QDQ^t$ then $A$ is orto-diagonalizable... but how do I use that here, since if I plug in $A$ in there I don't solve anything, can you be more explicit, also, for 2, yes, indeed it's about the SVD, how do I write down the eigenvalues of $B^tB$ in terms of singular values of $B$?
$endgroup$
– C. Cristi
Dec 6 '18 at 17:51
$begingroup$
1. Yeah I saw, okay to use the spectral theorem we have that: if $A$ is symetric then there is $Q$ and $D$ such that $A=QDQ^t$ then $A$ is orto-diagonalizable... but how do I use that here, since if I plug in $A$ in there I don't solve anything, can you be more explicit, also, for 2, yes, indeed it's about the SVD, how do I write down the eigenvalues of $B^tB$ in terms of singular values of $B$?
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– C. Cristi
Dec 6 '18 at 17:51
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@C.Cristi See my edit.
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– angryavian
Dec 6 '18 at 18:05
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@C.Cristi See my edit.
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– angryavian
Dec 6 '18 at 18:05
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How did you make the note that the norm of $U^tv=1$ is the same?
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– C. Cristi
Dec 6 '18 at 18:08
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How did you make the note that the norm of $U^tv=1$ is the same?
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– C. Cristi
Dec 6 '18 at 18:08
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Hey, why $lambda_{max}(frac {C+C^t}{2})=max_{|v|=1}v^tCv$ and not $=max_{|v|=1}v^tfrac{C+C^t}{2}v$?
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– C. Cristi
Dec 6 '18 at 18:16
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Hey, why $lambda_{max}(frac {C+C^t}{2})=max_{|v|=1}v^tCv$ and not $=max_{|v|=1}v^tfrac{C+C^t}{2}v$?
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– C. Cristi
Dec 6 '18 at 18:16
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Orthogonal matrix are norm-preserving because $|Uv|=sqrt{langle Uv,Uvrangle}=sqrt{langle U^tv,Uvrangle}=sqrt{UU^tlangle v, vrangle}= sqrt{langle v, vrangle}$?
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– C. Cristi
Dec 6 '18 at 18:22
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Orthogonal matrix are norm-preserving because $|Uv|=sqrt{langle Uv,Uvrangle}=sqrt{langle U^tv,Uvrangle}=sqrt{UU^tlangle v, vrangle}= sqrt{langle v, vrangle}$?
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– C. Cristi
Dec 6 '18 at 18:22
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For part 1 to make sense, the definition of $lambda_{max}(A)$ should be the biggest eigenvalue of $A$ not in absolute value.
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– angryavian
Dec 6 '18 at 17:55