Mixing
How to show that the ith row $AB$ is the matrix product of the ith row of $A$ with the entire matrix $B$?
$begingroup$
Consider two matrices $A$ and $B$ of dimensions such that the product $AB$
exists. Show that the ith row of $AB$ is the matrix product of the ith row of
$A$ with the entire matrix $B$.
I am trying to use this definition $AB = sum_{l=1}^{n} A_{il} B_{lk} = .$
Unfortunately I am stuck and need some help to proceed further.
linear-algebra matrices
asked Jan 24 at 7:50
OGCOGC
1,43821229
$endgroup$
add a comment |
$begingroup$
Consider two matrices $A$ and $B$ of dimensions such that the product $AB$
exists. Show that the ith row of $AB$ is the matrix product of the ith row of
$A$ with the entire matrix $B$.
I am trying to use this definition $AB = sum_{l=1}^{n} A_{il} B_{lk} = .$
Unfortunately I am stuck and need some help to proceed further.
linear-algebra matrices
asked Jan 24 at 7:50
OGCOGC
1,43821229
$endgroup$
1
$begingroup$
$i$th row and $k$th column of $AB$ is $sum_{l=1}^nA_{il}B_{lk}$ as you rightly said. Varying over $k$ just gives the whole row of $AB$ and we see that every entry of $B$ is used, while only the $i$th row of $A$ is used.
$endgroup$
– Alec B-G
Jan 24 at 12:48
1
$begingroup$
Maybe it helps to define a new row vector $a$ where $a_l = A_{i,l}$. That is, $a$ is the $i$-th row of $A$. Now write the product $aB$ and compare it to the $i$-th row of $AB$.
$endgroup$
– tch
Jan 24 at 15:22
add a comment |
$begingroup$
Consider two matrices $A$ and $B$ of dimensions such that the product $AB$
exists. Show that the ith row of $AB$ is the matrix product of the ith row of
$A$ with the entire matrix $B$.
I am trying to use this definition $AB = sum_{l=1}^{n} A_{il} B_{lk} = .$
Unfortunately I am stuck and need some help to proceed further.
linear-algebra matrices
asked Jan 24 at 7:50
OGCOGC
1,43821229
$endgroup$
Consider two matrices $A$ and $B$ of dimensions such that the product $AB$
exists. Show that the ith row of $AB$ is the matrix product of the ith row of
$A$ with the entire matrix $B$.
I am trying to use this definition $AB = sum_{l=1}^{n} A_{il} B_{lk} = .$
Unfortunately I am stuck and need some help to proceed further.
linear-algebra matrices
linear-algebra matrices
asked Jan 24 at 7:50
OGCOGC
1,43821229
asked Jan 24 at 7:50
OGCOGC
1,43821229
edited Jan 24 at 18:44
OGC
asked Jan 24 at 7:50
OGCOGC
1,43821229
asked Jan 24 at 7:50
OGCOGC
1,43821229
asked Jan 24 at 7:50
OGCOGC
1,43821229
1,43821229
1
$begingroup$
$i$th row and $k$th column of $AB$ is $sum_{l=1}^nA_{il}B_{lk}$ as you rightly said. Varying over $k$ just gives the whole row of $AB$ and we see that every entry of $B$ is used, while only the $i$th row of $A$ is used.
$endgroup$
– Alec B-G
Jan 24 at 12:48
1
$begingroup$
Maybe it helps to define a new row vector $a$ where $a_l = A_{i,l}$. That is, $a$ is the $i$-th row of $A$. Now write the product $aB$ and compare it to the $i$-th row of $AB$.
$endgroup$
– tch
Jan 24 at 15:22
add a comment |
1
$begingroup$
$i$th row and $k$th column of $AB$ is $sum_{l=1}^nA_{il}B_{lk}$ as you rightly said. Varying over $k$ just gives the whole row of $AB$ and we see that every entry of $B$ is used, while only the $i$th row of $A$ is used.
$endgroup$
– Alec B-G
Jan 24 at 12:48
1
$begingroup$
Maybe it helps to define a new row vector $a$ where $a_l = A_{i,l}$. That is, $a$ is the $i$-th row of $A$. Now write the product $aB$ and compare it to the $i$-th row of $AB$.
$endgroup$
– tch
Jan 24 at 15:22
1
1
$begingroup$
$i$th row and $k$th column of $AB$ is $sum_{l=1}^nA_{il}B_{lk}$ as you rightly said. Varying over $k$ just gives the whole row of $AB$ and we see that every entry of $B$ is used, while only the $i$th row of $A$ is used.
$endgroup$
– Alec B-G
Jan 24 at 12:48
$begingroup$
$i$th row and $k$th column of $AB$ is $sum_{l=1}^nA_{il}B_{lk}$ as you rightly said. Varying over $k$ just gives the whole row of $AB$ and we see that every entry of $B$ is used, while only the $i$th row of $A$ is used.
$endgroup$
– Alec B-G
Jan 24 at 12:48
1
1
$begingroup$
Maybe it helps to define a new row vector $a$ where $a_l = A_{i,l}$. That is, $a$ is the $i$-th row of $A$. Now write the product $aB$ and compare it to the $i$-th row of $AB$.
$endgroup$
– tch
Jan 24 at 15:22
$begingroup$
Maybe it helps to define a new row vector $a$ where $a_l = A_{i,l}$. That is, $a$ is the $i$-th row of $A$. Now write the product $aB$ and compare it to the $i$-th row of $AB$.
$endgroup$
– tch
Jan 24 at 15:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For a matrix $M$, define $M_{ij}$ to be its $(i,j)^{th}$ entry, and $M_i$ to be its $i^{th}$ row.
What you want to show is
$$(AB)_i = A_iB$$
And by definition of the regular matrix multiplication,
$$(AB)_{ij} = sum A_{iell}B_{ell j} tag 1$$
$$(A_iB)_{1j} = sum (A_i)_{1ell}B_{ell j} tag 2$$
Since $A_iB$ has only one row, you'll just have to show that $(1)$ and $(2)$ are equal for each $j$.
answered Jan 25 at 3:33
MetricMetric
1,25159
$endgroup$
$begingroup$
@I don't understand one thing. In (2), why does $A_{i}B$ has only one row?
$endgroup$
– OGC
Jan 25 at 21:23
$begingroup$
So for a matrix $M$ with $m$ rows and $p$ columns and a matrix $N$ with $p$ rows and $n$ columns, their product $MN$ is defined and has $m$ rows and $n$ columns. Or, more briefly, if $M in mathbb{R}^{m times p}$ and $N in mathbb{R}^{p times n}$, we have $MN in mathbb{R}^{m times n}$.
$endgroup$
– Metric
Jan 26 at 0:43
1
$begingroup$
Since $A_i in mathbb{R}^{1 times p}$ and $B in mathbb{R}^{p times n}$ for some $p$ and $n$, we have $A_iB in mathbb{R}^{1 times n}$, so $A_iB$ has one row.
$endgroup$
– Metric
Jan 26 at 0:45
add a comment |
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1 Answer
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$begingroup$
For a matrix $M$, define $M_{ij}$ to be its $(i,j)^{th}$ entry, and $M_i$ to be its $i^{th}$ row.
What you want to show is
$$(AB)_i = A_iB$$
And by definition of the regular matrix multiplication,
$$(AB)_{ij} = sum A_{iell}B_{ell j} tag 1$$
$$(A_iB)_{1j} = sum (A_i)_{1ell}B_{ell j} tag 2$$
Since $A_iB$ has only one row, you'll just have to show that $(1)$ and $(2)$ are equal for each $j$.
answered Jan 25 at 3:33
MetricMetric
1,25159
$endgroup$
$begingroup$
@I don't understand one thing. In (2), why does $A_{i}B$ has only one row?
$endgroup$
– OGC
Jan 25 at 21:23
$begingroup$
So for a matrix $M$ with $m$ rows and $p$ columns and a matrix $N$ with $p$ rows and $n$ columns, their product $MN$ is defined and has $m$ rows and $n$ columns. Or, more briefly, if $M in mathbb{R}^{m times p}$ and $N in mathbb{R}^{p times n}$, we have $MN in mathbb{R}^{m times n}$.
$endgroup$
– Metric
Jan 26 at 0:43
1
$begingroup$
Since $A_i in mathbb{R}^{1 times p}$ and $B in mathbb{R}^{p times n}$ for some $p$ and $n$, we have $A_iB in mathbb{R}^{1 times n}$, so $A_iB$ has one row.
$endgroup$
– Metric
Jan 26 at 0:45
add a comment |
$begingroup$
For a matrix $M$, define $M_{ij}$ to be its $(i,j)^{th}$ entry, and $M_i$ to be its $i^{th}$ row.
What you want to show is
$$(AB)_i = A_iB$$
And by definition of the regular matrix multiplication,
$$(AB)_{ij} = sum A_{iell}B_{ell j} tag 1$$
$$(A_iB)_{1j} = sum (A_i)_{1ell}B_{ell j} tag 2$$
Since $A_iB$ has only one row, you'll just have to show that $(1)$ and $(2)$ are equal for each $j$.
answered Jan 25 at 3:33
MetricMetric
1,25159
$endgroup$
$begingroup$
@I don't understand one thing. In (2), why does $A_{i}B$ has only one row?
$endgroup$
– OGC
Jan 25 at 21:23
$begingroup$
So for a matrix $M$ with $m$ rows and $p$ columns and a matrix $N$ with $p$ rows and $n$ columns, their product $MN$ is defined and has $m$ rows and $n$ columns. Or, more briefly, if $M in mathbb{R}^{m times p}$ and $N in mathbb{R}^{p times n}$, we have $MN in mathbb{R}^{m times n}$.
$endgroup$
– Metric
Jan 26 at 0:43
1
$begingroup$
Since $A_i in mathbb{R}^{1 times p}$ and $B in mathbb{R}^{p times n}$ for some $p$ and $n$, we have $A_iB in mathbb{R}^{1 times n}$, so $A_iB$ has one row.
$endgroup$
– Metric
Jan 26 at 0:45
add a comment |
$begingroup$
For a matrix $M$, define $M_{ij}$ to be its $(i,j)^{th}$ entry, and $M_i$ to be its $i^{th}$ row.
What you want to show is
$$(AB)_i = A_iB$$
And by definition of the regular matrix multiplication,
$$(AB)_{ij} = sum A_{iell}B_{ell j} tag 1$$
$$(A_iB)_{1j} = sum (A_i)_{1ell}B_{ell j} tag 2$$
Since $A_iB$ has only one row, you'll just have to show that $(1)$ and $(2)$ are equal for each $j$.
answered Jan 25 at 3:33
MetricMetric
1,25159
$endgroup$
For a matrix $M$, define $M_{ij}$ to be its $(i,j)^{th}$ entry, and $M_i$ to be its $i^{th}$ row.
What you want to show is
$$(AB)_i = A_iB$$
And by definition of the regular matrix multiplication,
$$(AB)_{ij} = sum A_{iell}B_{ell j} tag 1$$
$$(A_iB)_{1j} = sum (A_i)_{1ell}B_{ell j} tag 2$$
Since $A_iB$ has only one row, you'll just have to show that $(1)$ and $(2)$ are equal for each $j$.
answered Jan 25 at 3:33
MetricMetric
1,25159
edited Jan 25 at 3:38
answered Jan 25 at 3:33
MetricMetric
1,25159
answered Jan 25 at 3:33
MetricMetric
1,25159
answered Jan 25 at 3:33
MetricMetric
1,25159
1,25159
$begingroup$
@I don't understand one thing. In (2), why does $A_{i}B$ has only one row?
$endgroup$
– OGC
Jan 25 at 21:23
$begingroup$
So for a matrix $M$ with $m$ rows and $p$ columns and a matrix $N$ with $p$ rows and $n$ columns, their product $MN$ is defined and has $m$ rows and $n$ columns. Or, more briefly, if $M in mathbb{R}^{m times p}$ and $N in mathbb{R}^{p times n}$, we have $MN in mathbb{R}^{m times n}$.
$endgroup$
– Metric
Jan 26 at 0:43
1
$begingroup$
Since $A_i in mathbb{R}^{1 times p}$ and $B in mathbb{R}^{p times n}$ for some $p$ and $n$, we have $A_iB in mathbb{R}^{1 times n}$, so $A_iB$ has one row.
$endgroup$
– Metric
Jan 26 at 0:45
add a comment |
$begingroup$
@I don't understand one thing. In (2), why does $A_{i}B$ has only one row?
$endgroup$
– OGC
Jan 25 at 21:23
$begingroup$
So for a matrix $M$ with $m$ rows and $p$ columns and a matrix $N$ with $p$ rows and $n$ columns, their product $MN$ is defined and has $m$ rows and $n$ columns. Or, more briefly, if $M in mathbb{R}^{m times p}$ and $N in mathbb{R}^{p times n}$, we have $MN in mathbb{R}^{m times n}$.
$endgroup$
– Metric
Jan 26 at 0:43
1
$begingroup$
Since $A_i in mathbb{R}^{1 times p}$ and $B in mathbb{R}^{p times n}$ for some $p$ and $n$, we have $A_iB in mathbb{R}^{1 times n}$, so $A_iB$ has one row.
$endgroup$
– Metric
Jan 26 at 0:45
$begingroup$
@I don't understand one thing. In (2), why does $A_{i}B$ has only one row?
$endgroup$
– OGC
Jan 25 at 21:23
$begingroup$
@I don't understand one thing. In (2), why does $A_{i}B$ has only one row?
$endgroup$
– OGC
Jan 25 at 21:23
$begingroup$
So for a matrix $M$ with $m$ rows and $p$ columns and a matrix $N$ with $p$ rows and $n$ columns, their product $MN$ is defined and has $m$ rows and $n$ columns. Or, more briefly, if $M in mathbb{R}^{m times p}$ and $N in mathbb{R}^{p times n}$, we have $MN in mathbb{R}^{m times n}$.
$endgroup$
– Metric
Jan 26 at 0:43
$begingroup$
So for a matrix $M$ with $m$ rows and $p$ columns and a matrix $N$ with $p$ rows and $n$ columns, their product $MN$ is defined and has $m$ rows and $n$ columns. Or, more briefly, if $M in mathbb{R}^{m times p}$ and $N in mathbb{R}^{p times n}$, we have $MN in mathbb{R}^{m times n}$.
$endgroup$
– Metric
Jan 26 at 0:43
1
1
$begingroup$
Since $A_i in mathbb{R}^{1 times p}$ and $B in mathbb{R}^{p times n}$ for some $p$ and $n$, we have $A_iB in mathbb{R}^{1 times n}$, so $A_iB$ has one row.
$endgroup$
– Metric
Jan 26 at 0:45
$begingroup$
Since $A_i in mathbb{R}^{1 times p}$ and $B in mathbb{R}^{p times n}$ for some $p$ and $n$, we have $A_iB in mathbb{R}^{1 times n}$, so $A_iB$ has one row.
$endgroup$
– Metric
Jan 26 at 0:45
add a comment |
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1
$begingroup$
$i$th row and $k$th column of $AB$ is $sum_{l=1}^nA_{il}B_{lk}$ as you rightly said. Varying over $k$ just gives the whole row of $AB$ and we see that every entry of $B$ is used, while only the $i$th row of $A$ is used.
$endgroup$
– Alec B-G
Jan 24 at 12:48
1
$begingroup$
Maybe it helps to define a new row vector $a$ where $a_l = A_{i,l}$. That is, $a$ is the $i$-th row of $A$. Now write the product $aB$ and compare it to the $i$-th row of $AB$.
$endgroup$
– tch
Jan 24 at 15:22