Linear algebra: Proving matrix multiplication
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I have an issue with the following homework assignment:
Let A be an m×n matrix and B be an n×n matrix where $b_{kl} = 1$ for fixed $1≤k$, $l≤n$ and $b_{ij} = 0$ for $i neq k$ or $j neq l$. What is the result of AB? Prove your claims!
I understand what the linear map $B$ does, but I don't know how to prove it formally. As far as I understand, matrix (AB) has the $k$th column of A in its $l$th column, and zeros elsewhere.
I would appreciate if you could give me some hints on how to prove this problem.
The problem is the following:
linear-algebra
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add a comment |
$begingroup$
I have an issue with the following homework assignment:
Let A be an m×n matrix and B be an n×n matrix where $b_{kl} = 1$ for fixed $1≤k$, $l≤n$ and $b_{ij} = 0$ for $i neq k$ or $j neq l$. What is the result of AB? Prove your claims!
I understand what the linear map $B$ does, but I don't know how to prove it formally. As far as I understand, matrix (AB) has the $k$th column of A in its $l$th column, and zeros elsewhere.
I would appreciate if you could give me some hints on how to prove this problem.
The problem is the following:
linear-algebra
$endgroup$
1
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HInt: If $B=[b_1|b_1|cdots|b_n]$ where each $b_i$ is a column vector then $AB=[Ab_1|Ab_|cdots|Ab_n]$.
$endgroup$
– user9077
Dec 10 '18 at 13:52
add a comment |
$begingroup$
I have an issue with the following homework assignment:
Let A be an m×n matrix and B be an n×n matrix where $b_{kl} = 1$ for fixed $1≤k$, $l≤n$ and $b_{ij} = 0$ for $i neq k$ or $j neq l$. What is the result of AB? Prove your claims!
I understand what the linear map $B$ does, but I don't know how to prove it formally. As far as I understand, matrix (AB) has the $k$th column of A in its $l$th column, and zeros elsewhere.
I would appreciate if you could give me some hints on how to prove this problem.
The problem is the following:
linear-algebra
$endgroup$
I have an issue with the following homework assignment:
Let A be an m×n matrix and B be an n×n matrix where $b_{kl} = 1$ for fixed $1≤k$, $l≤n$ and $b_{ij} = 0$ for $i neq k$ or $j neq l$. What is the result of AB? Prove your claims!
I understand what the linear map $B$ does, but I don't know how to prove it formally. As far as I understand, matrix (AB) has the $k$th column of A in its $l$th column, and zeros elsewhere.
I would appreciate if you could give me some hints on how to prove this problem.
The problem is the following:
linear-algebra
linear-algebra
asked Dec 10 '18 at 13:46
Ulrich Paul WohakUlrich Paul Wohak
52
52
1
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HInt: If $B=[b_1|b_1|cdots|b_n]$ where each $b_i$ is a column vector then $AB=[Ab_1|Ab_|cdots|Ab_n]$.
$endgroup$
– user9077
Dec 10 '18 at 13:52
add a comment |
1
$begingroup$
HInt: If $B=[b_1|b_1|cdots|b_n]$ where each $b_i$ is a column vector then $AB=[Ab_1|Ab_|cdots|Ab_n]$.
$endgroup$
– user9077
Dec 10 '18 at 13:52
1
1
$begingroup$
HInt: If $B=[b_1|b_1|cdots|b_n]$ where each $b_i$ is a column vector then $AB=[Ab_1|Ab_|cdots|Ab_n]$.
$endgroup$
– user9077
Dec 10 '18 at 13:52
$begingroup$
HInt: If $B=[b_1|b_1|cdots|b_n]$ where each $b_i$ is a column vector then $AB=[Ab_1|Ab_|cdots|Ab_n]$.
$endgroup$
– user9077
Dec 10 '18 at 13:52
add a comment |
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1
$begingroup$
HInt: If $B=[b_1|b_1|cdots|b_n]$ where each $b_i$ is a column vector then $AB=[Ab_1|Ab_|cdots|Ab_n]$.
$endgroup$
– user9077
Dec 10 '18 at 13:52