Matrix inequality with its norm
$begingroup$
Let M $in mathbb{R}^{ntimes n}$ with its norm given by $|{M}|$.
I am trying to find a suitable upper bound of this matrix.
Is this inequality correct?
$M leq |M|mathbb{I}$, where $mathbb{I} in mathbb{R}^{ntimes n}$ is a matrix of all 1s and the inequality is implied entry wise.
linear-algebra matrices
asked Dec 25 '18 at 3:16
jjgarrisonjjgarrison
316
$endgroup$
add a comment |
$begingroup$
Let M $in mathbb{R}^{ntimes n}$ with its norm given by $|{M}|$.
I am trying to find a suitable upper bound of this matrix.
Is this inequality correct?
$M leq |M|mathbb{I}$, where $mathbb{I} in mathbb{R}^{ntimes n}$ is a matrix of all 1s and the inequality is implied entry wise.
linear-algebra matrices
asked Dec 25 '18 at 3:16
jjgarrisonjjgarrison
316
$endgroup$
add a comment |
$begingroup$
Let M $in mathbb{R}^{ntimes n}$ with its norm given by $|{M}|$.
I am trying to find a suitable upper bound of this matrix.
Is this inequality correct?
$M leq |M|mathbb{I}$, where $mathbb{I} in mathbb{R}^{ntimes n}$ is a matrix of all 1s and the inequality is implied entry wise.
linear-algebra matrices
asked Dec 25 '18 at 3:16
jjgarrisonjjgarrison
316
$endgroup$
Let M $in mathbb{R}^{ntimes n}$ with its norm given by $|{M}|$.
I am trying to find a suitable upper bound of this matrix.
Is this inequality correct?
$M leq |M|mathbb{I}$, where $mathbb{I} in mathbb{R}^{ntimes n}$ is a matrix of all 1s and the inequality is implied entry wise.
linear-algebra matrices
linear-algebra matrices
asked Dec 25 '18 at 3:16
jjgarrisonjjgarrison
316
asked Dec 25 '18 at 3:16
jjgarrisonjjgarrison
316
asked Dec 25 '18 at 3:16
jjgarrisonjjgarrison
316
asked Dec 25 '18 at 3:16
jjgarrisonjjgarrison
316
asked Dec 25 '18 at 3:16
jjgarrisonjjgarrison
316
316
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, that inequality is correct.
Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.
Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.
Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.
Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.
answered Dec 25 '18 at 3:28
JimmyK4542JimmyK4542
41.8k248110
$endgroup$
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$begingroup$
Yes, that inequality is correct.
Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.
Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.
Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.
Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.
answered Dec 25 '18 at 3:28
JimmyK4542JimmyK4542
41.8k248110
$endgroup$
add a comment |
$begingroup$
Yes, that inequality is correct.
Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.
Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.
Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.
Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.
answered Dec 25 '18 at 3:28
JimmyK4542JimmyK4542
41.8k248110
$endgroup$
add a comment |
$begingroup$
Yes, that inequality is correct.
Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.
Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.
Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.
Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.
answered Dec 25 '18 at 3:28
JimmyK4542JimmyK4542
41.8k248110
$endgroup$
Yes, that inequality is correct.
Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.
Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.
Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.
Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.
answered Dec 25 '18 at 3:28
JimmyK4542JimmyK4542
41.8k248110
answered Dec 25 '18 at 3:28
JimmyK4542JimmyK4542
41.8k248110
answered Dec 25 '18 at 3:28
JimmyK4542JimmyK4542
41.8k248110
answered Dec 25 '18 at 3:28
JimmyK4542JimmyK4542
41.8k248110
41.8k248110
add a comment |
add a comment |
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