Mixing
How to show $|A|_2 leq t$ implies $ t^2I-A^TA succeq 0$?
2
$begingroup$
Let $A in mathbb{R}^{n times n}$ and $|A|_2$ be the spectral norm of the matrix defined as the largest singular values of the matrix.
How to show when the maximum singular value is bounded, then $t^2I-A^TA$ is positive semi-definite. Formally,
$|A|_2 leq t$ implies $ t^2I-A^TA succeq 0$?
linear-algebra norm
asked Jan 14 at 4:14
SaeedSaeed
1,036310
$endgroup$
add a comment |
2
$begingroup$
Let $A in mathbb{R}^{n times n}$ and $|A|_2$ be the spectral norm of the matrix defined as the largest singular values of the matrix.
How to show when the maximum singular value is bounded, then $t^2I-A^TA$ is positive semi-definite. Formally,
$|A|_2 leq t$ implies $ t^2I-A^TA succeq 0$?
linear-algebra norm
asked Jan 14 at 4:14
SaeedSaeed
1,036310
$endgroup$
add a comment |
2
2
2
$begingroup$
Let $A in mathbb{R}^{n times n}$ and $|A|_2$ be the spectral norm of the matrix defined as the largest singular values of the matrix.
How to show when the maximum singular value is bounded, then $t^2I-A^TA$ is positive semi-definite. Formally,
$|A|_2 leq t$ implies $ t^2I-A^TA succeq 0$?
linear-algebra norm
asked Jan 14 at 4:14
SaeedSaeed
1,036310
$endgroup$
Let $A in mathbb{R}^{n times n}$ and $|A|_2$ be the spectral norm of the matrix defined as the largest singular values of the matrix.
How to show when the maximum singular value is bounded, then $t^2I-A^TA$ is positive semi-definite. Formally,
$|A|_2 leq t$ implies $ t^2I-A^TA succeq 0$?
linear-algebra norm
linear-algebra norm
asked Jan 14 at 4:14
SaeedSaeed
1,036310
asked Jan 14 at 4:14
SaeedSaeed
1,036310
edited Jan 14 at 4:23
Saeed
asked Jan 14 at 4:14
SaeedSaeed
1,036310
asked Jan 14 at 4:14
SaeedSaeed
1,036310
asked Jan 14 at 4:14
SaeedSaeed
1,036310
1,036310
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
Hint: For each $x in mathbb R^d$ you have
$$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2 geq 0$$
answered Jan 14 at 4:25
N. S.N. S.
104k7112208
$endgroup$
$begingroup$
How can we show the reverse? I mean when $t^2I-A^TA$ is P.S.D. how to show $|A|_2 leq t$.
$endgroup$
– Saeed
Jan 14 at 4:52
$begingroup$
@Saeed Exactly the same way: $$t^2|x |_2^2 - | Ax |_2^2=x^T(t^2I-A^TA)x geq 0$$... The point is that you have $$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2$$ Therefore, $$x^T(t^2I-A^TA)x geq 0 mbox{ iff } t^2|x |_2^2 - | Ax |_2^2 geq 0$$
$endgroup$
– N. S.
Jan 14 at 4:57
$begingroup$
$x^T(t^2I-A^TA)x geq 0 ,,,forall x in mathbb{R}^n$, then $|Ax|_2^2 leq t^2|x|_2^2$. How can I get $|A|_2 leq t$? It is true that $|Ax|_2^2 leq |A|_2^2|x|_2^2$ but Cauchy-Schwarz is not a good idea because one may not be sure the upper bound is less than $t^2|x|_2^2$? You see my point?
$endgroup$
– Saeed
Jan 14 at 5:00
$begingroup$
@Saeed $$|Ax|_2^2 leq t^2|x|_2^2 Rightarrow |Ax|_2 leq t|x|_2$$ Now simply use the definition of $| A|_2$, it is usually defined as the smallest $C$ such that $ |Ax|_2 leq C|x|_2$ for all $x$ or as the supremum of ${ frac{ | A x |_2}{|x|_2}: x neq 0 }$... I think you are confusing $| A |_2$ with the so called Frobenius norm of the matrix $| A |_F$.
$endgroup$
– N. S.
Jan 14 at 5:08
$begingroup$
I know $|A|_2$ is the maximum singular value of $A$, is this what you are talking about?
$endgroup$
– Saeed
Jan 14 at 5:15
|
show 3 more comments
1
$begingroup$
The following facts are useful to know:
$lambda$ is an eigenvalue of $M$ if and only if $lambda + c$ is an eigenvalue of $cI + M$. This is true because $Mx = lambda x iff (M + cI)x = (lambda + c)x$.- The spectral norm of a matrix $A$ with real entries is the square root of the maximum eigenvalue of $A^T A$.
- A real symmetric matrix is positive semidefinite if and only if its eigenvalues are nonnegative.
The eigenvalues of $t^2 I - A^T A$ are given by $t^2-lambda$, where $lambda$ is an eigenvalue of $A^T A$. Because $|A|_2 leq t$, we know that the square root of the maximum eigenvalue of $A^T A$ is less than or equal to $t$. It follows that $t^2 I - A^T A$ has nonnegative eigenvalues. In other words, $t^2 I - A^T A$ is positive semidefinite.
answered Jan 14 at 4:24
littleOlittleO
29.8k646109
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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2
$begingroup$
Hint: For each $x in mathbb R^d$ you have
$$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2 geq 0$$
answered Jan 14 at 4:25
N. S.N. S.
104k7112208
$endgroup$
$begingroup$
How can we show the reverse? I mean when $t^2I-A^TA$ is P.S.D. how to show $|A|_2 leq t$.
$endgroup$
– Saeed
Jan 14 at 4:52
$begingroup$
@Saeed Exactly the same way: $$t^2|x |_2^2 - | Ax |_2^2=x^T(t^2I-A^TA)x geq 0$$... The point is that you have $$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2$$ Therefore, $$x^T(t^2I-A^TA)x geq 0 mbox{ iff } t^2|x |_2^2 - | Ax |_2^2 geq 0$$
$endgroup$
– N. S.
Jan 14 at 4:57
$begingroup$
$x^T(t^2I-A^TA)x geq 0 ,,,forall x in mathbb{R}^n$, then $|Ax|_2^2 leq t^2|x|_2^2$. How can I get $|A|_2 leq t$? It is true that $|Ax|_2^2 leq |A|_2^2|x|_2^2$ but Cauchy-Schwarz is not a good idea because one may not be sure the upper bound is less than $t^2|x|_2^2$? You see my point?
$endgroup$
– Saeed
Jan 14 at 5:00
$begingroup$
@Saeed $$|Ax|_2^2 leq t^2|x|_2^2 Rightarrow |Ax|_2 leq t|x|_2$$ Now simply use the definition of $| A|_2$, it is usually defined as the smallest $C$ such that $ |Ax|_2 leq C|x|_2$ for all $x$ or as the supremum of ${ frac{ | A x |_2}{|x|_2}: x neq 0 }$... I think you are confusing $| A |_2$ with the so called Frobenius norm of the matrix $| A |_F$.
$endgroup$
– N. S.
Jan 14 at 5:08
$begingroup$
I know $|A|_2$ is the maximum singular value of $A$, is this what you are talking about?
$endgroup$
– Saeed
Jan 14 at 5:15
|
show 3 more comments
2
$begingroup$
Hint: For each $x in mathbb R^d$ you have
$$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2 geq 0$$
answered Jan 14 at 4:25
N. S.N. S.
104k7112208
$endgroup$
$begingroup$
How can we show the reverse? I mean when $t^2I-A^TA$ is P.S.D. how to show $|A|_2 leq t$.
$endgroup$
– Saeed
Jan 14 at 4:52
$begingroup$
@Saeed Exactly the same way: $$t^2|x |_2^2 - | Ax |_2^2=x^T(t^2I-A^TA)x geq 0$$... The point is that you have $$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2$$ Therefore, $$x^T(t^2I-A^TA)x geq 0 mbox{ iff } t^2|x |_2^2 - | Ax |_2^2 geq 0$$
$endgroup$
– N. S.
Jan 14 at 4:57
$begingroup$
$x^T(t^2I-A^TA)x geq 0 ,,,forall x in mathbb{R}^n$, then $|Ax|_2^2 leq t^2|x|_2^2$. How can I get $|A|_2 leq t$? It is true that $|Ax|_2^2 leq |A|_2^2|x|_2^2$ but Cauchy-Schwarz is not a good idea because one may not be sure the upper bound is less than $t^2|x|_2^2$? You see my point?
$endgroup$
– Saeed
Jan 14 at 5:00
$begingroup$
@Saeed $$|Ax|_2^2 leq t^2|x|_2^2 Rightarrow |Ax|_2 leq t|x|_2$$ Now simply use the definition of $| A|_2$, it is usually defined as the smallest $C$ such that $ |Ax|_2 leq C|x|_2$ for all $x$ or as the supremum of ${ frac{ | A x |_2}{|x|_2}: x neq 0 }$... I think you are confusing $| A |_2$ with the so called Frobenius norm of the matrix $| A |_F$.
$endgroup$
– N. S.
Jan 14 at 5:08
$begingroup$
I know $|A|_2$ is the maximum singular value of $A$, is this what you are talking about?
$endgroup$
– Saeed
Jan 14 at 5:15
|
show 3 more comments
2
2
2
$begingroup$
Hint: For each $x in mathbb R^d$ you have
$$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2 geq 0$$
answered Jan 14 at 4:25
N. S.N. S.
104k7112208
$endgroup$
Hint: For each $x in mathbb R^d$ you have
$$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2 geq 0$$
answered Jan 14 at 4:25
N. S.N. S.
104k7112208
answered Jan 14 at 4:25
N. S.N. S.
104k7112208
answered Jan 14 at 4:25
N. S.N. S.
104k7112208
answered Jan 14 at 4:25
N. S.N. S.
104k7112208
104k7112208
$begingroup$
How can we show the reverse? I mean when $t^2I-A^TA$ is P.S.D. how to show $|A|_2 leq t$.
$endgroup$
– Saeed
Jan 14 at 4:52
$begingroup$
@Saeed Exactly the same way: $$t^2|x |_2^2 - | Ax |_2^2=x^T(t^2I-A^TA)x geq 0$$... The point is that you have $$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2$$ Therefore, $$x^T(t^2I-A^TA)x geq 0 mbox{ iff } t^2|x |_2^2 - | Ax |_2^2 geq 0$$
$endgroup$
– N. S.
Jan 14 at 4:57
$begingroup$
$x^T(t^2I-A^TA)x geq 0 ,,,forall x in mathbb{R}^n$, then $|Ax|_2^2 leq t^2|x|_2^2$. How can I get $|A|_2 leq t$? It is true that $|Ax|_2^2 leq |A|_2^2|x|_2^2$ but Cauchy-Schwarz is not a good idea because one may not be sure the upper bound is less than $t^2|x|_2^2$? You see my point?
$endgroup$
– Saeed
Jan 14 at 5:00
$begingroup$
@Saeed $$|Ax|_2^2 leq t^2|x|_2^2 Rightarrow |Ax|_2 leq t|x|_2$$ Now simply use the definition of $| A|_2$, it is usually defined as the smallest $C$ such that $ |Ax|_2 leq C|x|_2$ for all $x$ or as the supremum of ${ frac{ | A x |_2}{|x|_2}: x neq 0 }$... I think you are confusing $| A |_2$ with the so called Frobenius norm of the matrix $| A |_F$.
$endgroup$
– N. S.
Jan 14 at 5:08
$begingroup$
I know $|A|_2$ is the maximum singular value of $A$, is this what you are talking about?
$endgroup$
– Saeed
Jan 14 at 5:15
|
show 3 more comments
$begingroup$
How can we show the reverse? I mean when $t^2I-A^TA$ is P.S.D. how to show $|A|_2 leq t$.
$endgroup$
– Saeed
Jan 14 at 4:52
$begingroup$
@Saeed Exactly the same way: $$t^2|x |_2^2 - | Ax |_2^2=x^T(t^2I-A^TA)x geq 0$$... The point is that you have $$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2$$ Therefore, $$x^T(t^2I-A^TA)x geq 0 mbox{ iff } t^2|x |_2^2 - | Ax |_2^2 geq 0$$
$endgroup$
– N. S.
Jan 14 at 4:57
$begingroup$
$x^T(t^2I-A^TA)x geq 0 ,,,forall x in mathbb{R}^n$, then $|Ax|_2^2 leq t^2|x|_2^2$. How can I get $|A|_2 leq t$? It is true that $|Ax|_2^2 leq |A|_2^2|x|_2^2$ but Cauchy-Schwarz is not a good idea because one may not be sure the upper bound is less than $t^2|x|_2^2$? You see my point?
$endgroup$
– Saeed
Jan 14 at 5:00
$begingroup$
@Saeed $$|Ax|_2^2 leq t^2|x|_2^2 Rightarrow |Ax|_2 leq t|x|_2$$ Now simply use the definition of $| A|_2$, it is usually defined as the smallest $C$ such that $ |Ax|_2 leq C|x|_2$ for all $x$ or as the supremum of ${ frac{ | A x |_2}{|x|_2}: x neq 0 }$... I think you are confusing $| A |_2$ with the so called Frobenius norm of the matrix $| A |_F$.
$endgroup$
– N. S.
Jan 14 at 5:08
$begingroup$
I know $|A|_2$ is the maximum singular value of $A$, is this what you are talking about?
$endgroup$
– Saeed
Jan 14 at 5:15
$begingroup$
How can we show the reverse? I mean when $t^2I-A^TA$ is P.S.D. how to show $|A|_2 leq t$.
$endgroup$
– Saeed
Jan 14 at 4:52
$begingroup$
How can we show the reverse? I mean when $t^2I-A^TA$ is P.S.D. how to show $|A|_2 leq t$.
$endgroup$
– Saeed
Jan 14 at 4:52
$begingroup$
@Saeed Exactly the same way: $$t^2|x |_2^2 - | Ax |_2^2=x^T(t^2I-A^TA)x geq 0$$... The point is that you have $$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2$$ Therefore, $$x^T(t^2I-A^TA)x geq 0 mbox{ iff } t^2|x |_2^2 - | Ax |_2^2 geq 0$$
$endgroup$
– N. S.
Jan 14 at 4:57
$begingroup$
@Saeed Exactly the same way: $$t^2|x |_2^2 - | Ax |_2^2=x^T(t^2I-A^TA)x geq 0$$... The point is that you have $$x^T(t^2I-A^TA)x=t^2|x |_2^2 - | Ax |_2^2$$ Therefore, $$x^T(t^2I-A^TA)x geq 0 mbox{ iff } t^2|x |_2^2 - | Ax |_2^2 geq 0$$
$endgroup$
– N. S.
Jan 14 at 4:57
$begingroup$
$x^T(t^2I-A^TA)x geq 0 ,,,forall x in mathbb{R}^n$, then $|Ax|_2^2 leq t^2|x|_2^2$. How can I get $|A|_2 leq t$? It is true that $|Ax|_2^2 leq |A|_2^2|x|_2^2$ but Cauchy-Schwarz is not a good idea because one may not be sure the upper bound is less than $t^2|x|_2^2$? You see my point?
$endgroup$
– Saeed
Jan 14 at 5:00
$begingroup$
$x^T(t^2I-A^TA)x geq 0 ,,,forall x in mathbb{R}^n$, then $|Ax|_2^2 leq t^2|x|_2^2$. How can I get $|A|_2 leq t$? It is true that $|Ax|_2^2 leq |A|_2^2|x|_2^2$ but Cauchy-Schwarz is not a good idea because one may not be sure the upper bound is less than $t^2|x|_2^2$? You see my point?
$endgroup$
– Saeed
Jan 14 at 5:00
$begingroup$
@Saeed $$|Ax|_2^2 leq t^2|x|_2^2 Rightarrow |Ax|_2 leq t|x|_2$$ Now simply use the definition of $| A|_2$, it is usually defined as the smallest $C$ such that $ |Ax|_2 leq C|x|_2$ for all $x$ or as the supremum of ${ frac{ | A x |_2}{|x|_2}: x neq 0 }$... I think you are confusing $| A |_2$ with the so called Frobenius norm of the matrix $| A |_F$.
$endgroup$
– N. S.
Jan 14 at 5:08
$begingroup$
@Saeed $$|Ax|_2^2 leq t^2|x|_2^2 Rightarrow |Ax|_2 leq t|x|_2$$ Now simply use the definition of $| A|_2$, it is usually defined as the smallest $C$ such that $ |Ax|_2 leq C|x|_2$ for all $x$ or as the supremum of ${ frac{ | A x |_2}{|x|_2}: x neq 0 }$... I think you are confusing $| A |_2$ with the so called Frobenius norm of the matrix $| A |_F$.
$endgroup$
– N. S.
Jan 14 at 5:08
$begingroup$
I know $|A|_2$ is the maximum singular value of $A$, is this what you are talking about?
$endgroup$
– Saeed
Jan 14 at 5:15
$begingroup$
I know $|A|_2$ is the maximum singular value of $A$, is this what you are talking about?
$endgroup$
– Saeed
Jan 14 at 5:15
|
show 3 more comments
1
$begingroup$
The following facts are useful to know:
$lambda$ is an eigenvalue of $M$ if and only if $lambda + c$ is an eigenvalue of $cI + M$. This is true because $Mx = lambda x iff (M + cI)x = (lambda + c)x$.- The spectral norm of a matrix $A$ with real entries is the square root of the maximum eigenvalue of $A^T A$.
- A real symmetric matrix is positive semidefinite if and only if its eigenvalues are nonnegative.
The eigenvalues of $t^2 I - A^T A$ are given by $t^2-lambda$, where $lambda$ is an eigenvalue of $A^T A$. Because $|A|_2 leq t$, we know that the square root of the maximum eigenvalue of $A^T A$ is less than or equal to $t$. It follows that $t^2 I - A^T A$ has nonnegative eigenvalues. In other words, $t^2 I - A^T A$ is positive semidefinite.
answered Jan 14 at 4:24
littleOlittleO
29.8k646109
$endgroup$
add a comment |
1
$begingroup$
The following facts are useful to know:
$lambda$ is an eigenvalue of $M$ if and only if $lambda + c$ is an eigenvalue of $cI + M$. This is true because $Mx = lambda x iff (M + cI)x = (lambda + c)x$.- The spectral norm of a matrix $A$ with real entries is the square root of the maximum eigenvalue of $A^T A$.
- A real symmetric matrix is positive semidefinite if and only if its eigenvalues are nonnegative.
The eigenvalues of $t^2 I - A^T A$ are given by $t^2-lambda$, where $lambda$ is an eigenvalue of $A^T A$. Because $|A|_2 leq t$, we know that the square root of the maximum eigenvalue of $A^T A$ is less than or equal to $t$. It follows that $t^2 I - A^T A$ has nonnegative eigenvalues. In other words, $t^2 I - A^T A$ is positive semidefinite.
answered Jan 14 at 4:24
littleOlittleO
29.8k646109
$endgroup$
add a comment |
1
1
1
$begingroup$
The following facts are useful to know:
$lambda$ is an eigenvalue of $M$ if and only if $lambda + c$ is an eigenvalue of $cI + M$. This is true because $Mx = lambda x iff (M + cI)x = (lambda + c)x$.- The spectral norm of a matrix $A$ with real entries is the square root of the maximum eigenvalue of $A^T A$.
- A real symmetric matrix is positive semidefinite if and only if its eigenvalues are nonnegative.
The eigenvalues of $t^2 I - A^T A$ are given by $t^2-lambda$, where $lambda$ is an eigenvalue of $A^T A$. Because $|A|_2 leq t$, we know that the square root of the maximum eigenvalue of $A^T A$ is less than or equal to $t$. It follows that $t^2 I - A^T A$ has nonnegative eigenvalues. In other words, $t^2 I - A^T A$ is positive semidefinite.
answered Jan 14 at 4:24
littleOlittleO
29.8k646109
$endgroup$
The following facts are useful to know:
$lambda$ is an eigenvalue of $M$ if and only if $lambda + c$ is an eigenvalue of $cI + M$. This is true because $Mx = lambda x iff (M + cI)x = (lambda + c)x$.- The spectral norm of a matrix $A$ with real entries is the square root of the maximum eigenvalue of $A^T A$.
- A real symmetric matrix is positive semidefinite if and only if its eigenvalues are nonnegative.
The eigenvalues of $t^2 I - A^T A$ are given by $t^2-lambda$, where $lambda$ is an eigenvalue of $A^T A$. Because $|A|_2 leq t$, we know that the square root of the maximum eigenvalue of $A^T A$ is less than or equal to $t$. It follows that $t^2 I - A^T A$ has nonnegative eigenvalues. In other words, $t^2 I - A^T A$ is positive semidefinite.
answered Jan 14 at 4:24
littleOlittleO
29.8k646109
edited Jan 14 at 5:10
answered Jan 14 at 4:24
littleOlittleO
29.8k646109
answered Jan 14 at 4:24
littleOlittleO
29.8k646109
answered Jan 14 at 4:24
littleOlittleO
29.8k646109
29.8k646109
add a comment |
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
