Mixing
Matrix multiplication notation
$begingroup$
This is from my textbook:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
I know how to multiply matrices but I don't understand this notation : $c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$
Can someone explain what that represents by giving me an example? And how did we get that formula?
linear-algebra matrices notation
asked Dec 18 '16 at 10:16
lmclmc
1,073723
$endgroup$
add a comment |
$begingroup$
This is from my textbook:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
I know how to multiply matrices but I don't understand this notation : $c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$
Can someone explain what that represents by giving me an example? And how did we get that formula?
linear-algebra matrices notation
asked Dec 18 '16 at 10:16
lmclmc
1,073723
$endgroup$
$begingroup$
Example: To find $c_{11}$, you multiply the first row of matrix $A$ with the first column of matrix $B$.
$endgroup$
– Prahlad Vaidyanathan
Dec 18 '16 at 10:18
$begingroup$
$c_{ij}$ is the element on $i$-th row and $j$-th column of $C$. Similarly, $a_{ik}$, $b_{kj}$ are elements of matrices $B$ and $C$.
$endgroup$
– Peter Franek
Dec 18 '16 at 10:19
$begingroup$
If you know coding, this is how we multiply two matrices.
$endgroup$
– Rohan
Dec 18 '16 at 10:20
add a comment |
$begingroup$
This is from my textbook:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
I know how to multiply matrices but I don't understand this notation : $c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$
Can someone explain what that represents by giving me an example? And how did we get that formula?
linear-algebra matrices notation
asked Dec 18 '16 at 10:16
lmclmc
1,073723
$endgroup$
This is from my textbook:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
I know how to multiply matrices but I don't understand this notation : $c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$
Can someone explain what that represents by giving me an example? And how did we get that formula?
linear-algebra matrices notation
linear-algebra matrices notation
asked Dec 18 '16 at 10:16
lmclmc
1,073723
asked Dec 18 '16 at 10:16
lmclmc
1,073723
asked Dec 18 '16 at 10:16
lmclmc
1,073723
asked Dec 18 '16 at 10:16
lmclmc
1,073723
asked Dec 18 '16 at 10:16
lmclmc
1,073723
1,073723
$begingroup$
Example: To find $c_{11}$, you multiply the first row of matrix $A$ with the first column of matrix $B$.
$endgroup$
– Prahlad Vaidyanathan
Dec 18 '16 at 10:18
$begingroup$
$c_{ij}$ is the element on $i$-th row and $j$-th column of $C$. Similarly, $a_{ik}$, $b_{kj}$ are elements of matrices $B$ and $C$.
$endgroup$
– Peter Franek
Dec 18 '16 at 10:19
$begingroup$
If you know coding, this is how we multiply two matrices.
$endgroup$
– Rohan
Dec 18 '16 at 10:20
add a comment |
$begingroup$
Example: To find $c_{11}$, you multiply the first row of matrix $A$ with the first column of matrix $B$.
$endgroup$
– Prahlad Vaidyanathan
Dec 18 '16 at 10:18
$begingroup$
$c_{ij}$ is the element on $i$-th row and $j$-th column of $C$. Similarly, $a_{ik}$, $b_{kj}$ are elements of matrices $B$ and $C$.
$endgroup$
– Peter Franek
Dec 18 '16 at 10:19
$begingroup$
If you know coding, this is how we multiply two matrices.
$endgroup$
– Rohan
Dec 18 '16 at 10:20
$begingroup$
Example: To find $c_{11}$, you multiply the first row of matrix $A$ with the first column of matrix $B$.
$endgroup$
– Prahlad Vaidyanathan
Dec 18 '16 at 10:18
$begingroup$
Example: To find $c_{11}$, you multiply the first row of matrix $A$ with the first column of matrix $B$.
$endgroup$
– Prahlad Vaidyanathan
Dec 18 '16 at 10:18
$begingroup$
$c_{ij}$ is the element on $i$-th row and $j$-th column of $C$. Similarly, $a_{ik}$, $b_{kj}$ are elements of matrices $B$ and $C$.
$endgroup$
– Peter Franek
Dec 18 '16 at 10:19
$begingroup$
$c_{ij}$ is the element on $i$-th row and $j$-th column of $C$. Similarly, $a_{ik}$, $b_{kj}$ are elements of matrices $B$ and $C$.
$endgroup$
– Peter Franek
Dec 18 '16 at 10:19
$begingroup$
If you know coding, this is how we multiply two matrices.
$endgroup$
– Rohan
Dec 18 '16 at 10:20
$begingroup$
If you know coding, this is how we multiply two matrices.
$endgroup$
– Rohan
Dec 18 '16 at 10:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Visualisation might help. I'll use your notations and dimensions of the given matrices:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
You say you know how to multiply matrices, so take a look at one specific element in the product $C=AB$, namely the element on position $(i,j)$, i.e. in the $i$th row and $j$th column.
To obtain this element, you:
- first multiply all elements of the $i$th row of the matrix $A$ pairwise with all the elements of the $j$th column of the matrix $B$;
- and then you add these $n$ products.
You have to repeat this procedure for every element of $C$, but let's zoom in on that one specific (but arbitrary) element on position $(i,j)$ for now:
$$begin{pmatrix}
a_{11} &ldots &a_{1n}\
vdots& ddots &vdots\
color{blue}{mathbf{a_{i1}}} &color{blue}{rightarrow} &color{blue}{mathbf{a_{in}}}\
vdots& ddots &vdots\
a_{m1} &ldots &a_{mn}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&ldots &color{red}{mathbf{b_{1j}}} &ldots &b_{1p}\
vdots& ddots &color{red}{downarrow} & ddots &vdots\
b_{n1}&ldots &color{red}{mathbf{b_{nj}}}&ldots &b_{np}
end{pmatrix}
=
begin{pmatrix}
c_{11}&ldots& c_{1j} &ldots &c_{1p}\
vdots& ddots & & &vdots\
c_{i1}& & color{purple}{mathbf{c_{ij}}} & &c_{ip}\
vdots& & & ddots &vdots\
c_{m1} &ldots& c_{mj} &ldots &c_{mp}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{ij}}}$ equal to:
$$mathbf{color{purple}{c_{ij}} = color{blue}{a_{i1}} color{red}{b_{1j}} + color{blue}{a_{i2}} color{red}{b_{2j}} + cdots + color{blue}{a_{in}} color{red}{b_{nj}}}$$
Now notice that in the sum above, the left outer index is always $i$ ($i$th row of $A$) and the right outer index is always $j$ ($j$th column of $B$). The inner indices run from $1$ to $n$ so you can introduce a summation index $k$ and write this sum compactly using summation notation:
$$color{purple}{mathbf{c_{ij}}}=sum_{k=1}^{n} color{blue}{mathbf{a_{ik}}}color{red}{mathbf{b_{kj}}}$$
The formule above thus gives you the element on position $(i,j)$ in the product matrix $C=AB$ and therefore completely defines $C$ by letting $i=1,...,m$ and $j=1,...,p$.
Can someone explain what that represents by giving me an example? And how did we get that formula?
The illustration above should give you an idea of the general formula, but here's a concrete example where I took $3 times 3$ matrices for $A$ and $B$ and focus on the element on position $(2,3)$:
$$begin{pmatrix}
a_{11} & a_{12} &a_{13}\
color{blue}{1} &color{blue}{2} &color{blue}{3}\
a_{31} & a_{32} &a_{33}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&b_{12} &color{red}{6}\
b_{21}&b_{22} &color{red}{5}\
b_{31}&b_{32} &color{red}{4}
end{pmatrix}
=
begin{pmatrix}
c_{11}& c_{12} &c_{13}\
c_{21}& c_{22} &color{purple}{mathbf{c_{23}}}\
c_{31}& c_{32} &c_{33}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{23}}}$ equal to:
$$begin{array}{rccccccc}
color{purple}{c_{23}}
& = & color{blue}{a_{21}} color{red}{b_{13}} &+& color{blue}{a_{22}} color{red}{b_{23}} &+& color{blue}{a_{23}} color{red}{b_{33}}
& = & displaystyle sum_{k=1}^{3} color{blue}{a_{2k}}color{red}{b_{k3}} \
& = & color{blue}{1} cdot color{red}{6} &+& color{blue}{2} cdot color{red}{5} &+& color{blue}{3} cdot color{red}{4} \[8pt]
& = & 6&+&10&+&12 & =& 28
end{array}$$
answered Dec 18 '16 at 11:15
StackTDStackTD
22.9k2152
$endgroup$
add a comment |
$begingroup$
As you can see the sum run over $k$ from $1$ to $n$, but after you fix $i,j$. You said that you know how to multiply matrices, so you know that if $C=AB$, then $(i,j)-th$ element $c_{ij}$ is obtained fixing the $i-th$ row of $A$, the $j-th$ column of $B$, and then you multiply the elements $a_{i1}$ with $b_{1j}$, $a_{i2}$ with $b_{2j}$, $dots$, $a_{in}$ with $b_{nj}$, and finally you sum all products. Thus
$$c_{ij}=sum_{k=1}^n a_{ik}b_{kj}$$
answered Dec 18 '16 at 10:27
InsideOutInsideOut
5,12431034
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
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oldest
votes
$begingroup$
Visualisation might help. I'll use your notations and dimensions of the given matrices:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
You say you know how to multiply matrices, so take a look at one specific element in the product $C=AB$, namely the element on position $(i,j)$, i.e. in the $i$th row and $j$th column.
To obtain this element, you:
- first multiply all elements of the $i$th row of the matrix $A$ pairwise with all the elements of the $j$th column of the matrix $B$;
- and then you add these $n$ products.
You have to repeat this procedure for every element of $C$, but let's zoom in on that one specific (but arbitrary) element on position $(i,j)$ for now:
$$begin{pmatrix}
a_{11} &ldots &a_{1n}\
vdots& ddots &vdots\
color{blue}{mathbf{a_{i1}}} &color{blue}{rightarrow} &color{blue}{mathbf{a_{in}}}\
vdots& ddots &vdots\
a_{m1} &ldots &a_{mn}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&ldots &color{red}{mathbf{b_{1j}}} &ldots &b_{1p}\
vdots& ddots &color{red}{downarrow} & ddots &vdots\
b_{n1}&ldots &color{red}{mathbf{b_{nj}}}&ldots &b_{np}
end{pmatrix}
=
begin{pmatrix}
c_{11}&ldots& c_{1j} &ldots &c_{1p}\
vdots& ddots & & &vdots\
c_{i1}& & color{purple}{mathbf{c_{ij}}} & &c_{ip}\
vdots& & & ddots &vdots\
c_{m1} &ldots& c_{mj} &ldots &c_{mp}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{ij}}}$ equal to:
$$mathbf{color{purple}{c_{ij}} = color{blue}{a_{i1}} color{red}{b_{1j}} + color{blue}{a_{i2}} color{red}{b_{2j}} + cdots + color{blue}{a_{in}} color{red}{b_{nj}}}$$
Now notice that in the sum above, the left outer index is always $i$ ($i$th row of $A$) and the right outer index is always $j$ ($j$th column of $B$). The inner indices run from $1$ to $n$ so you can introduce a summation index $k$ and write this sum compactly using summation notation:
$$color{purple}{mathbf{c_{ij}}}=sum_{k=1}^{n} color{blue}{mathbf{a_{ik}}}color{red}{mathbf{b_{kj}}}$$
The formule above thus gives you the element on position $(i,j)$ in the product matrix $C=AB$ and therefore completely defines $C$ by letting $i=1,...,m$ and $j=1,...,p$.
Can someone explain what that represents by giving me an example? And how did we get that formula?
The illustration above should give you an idea of the general formula, but here's a concrete example where I took $3 times 3$ matrices for $A$ and $B$ and focus on the element on position $(2,3)$:
$$begin{pmatrix}
a_{11} & a_{12} &a_{13}\
color{blue}{1} &color{blue}{2} &color{blue}{3}\
a_{31} & a_{32} &a_{33}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&b_{12} &color{red}{6}\
b_{21}&b_{22} &color{red}{5}\
b_{31}&b_{32} &color{red}{4}
end{pmatrix}
=
begin{pmatrix}
c_{11}& c_{12} &c_{13}\
c_{21}& c_{22} &color{purple}{mathbf{c_{23}}}\
c_{31}& c_{32} &c_{33}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{23}}}$ equal to:
$$begin{array}{rccccccc}
color{purple}{c_{23}}
& = & color{blue}{a_{21}} color{red}{b_{13}} &+& color{blue}{a_{22}} color{red}{b_{23}} &+& color{blue}{a_{23}} color{red}{b_{33}}
& = & displaystyle sum_{k=1}^{3} color{blue}{a_{2k}}color{red}{b_{k3}} \
& = & color{blue}{1} cdot color{red}{6} &+& color{blue}{2} cdot color{red}{5} &+& color{blue}{3} cdot color{red}{4} \[8pt]
& = & 6&+&10&+&12 & =& 28
end{array}$$
answered Dec 18 '16 at 11:15
StackTDStackTD
22.9k2152
$endgroup$
add a comment |
$begingroup$
Visualisation might help. I'll use your notations and dimensions of the given matrices:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
You say you know how to multiply matrices, so take a look at one specific element in the product $C=AB$, namely the element on position $(i,j)$, i.e. in the $i$th row and $j$th column.
To obtain this element, you:
- first multiply all elements of the $i$th row of the matrix $A$ pairwise with all the elements of the $j$th column of the matrix $B$;
- and then you add these $n$ products.
You have to repeat this procedure for every element of $C$, but let's zoom in on that one specific (but arbitrary) element on position $(i,j)$ for now:
$$begin{pmatrix}
a_{11} &ldots &a_{1n}\
vdots& ddots &vdots\
color{blue}{mathbf{a_{i1}}} &color{blue}{rightarrow} &color{blue}{mathbf{a_{in}}}\
vdots& ddots &vdots\
a_{m1} &ldots &a_{mn}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&ldots &color{red}{mathbf{b_{1j}}} &ldots &b_{1p}\
vdots& ddots &color{red}{downarrow} & ddots &vdots\
b_{n1}&ldots &color{red}{mathbf{b_{nj}}}&ldots &b_{np}
end{pmatrix}
=
begin{pmatrix}
c_{11}&ldots& c_{1j} &ldots &c_{1p}\
vdots& ddots & & &vdots\
c_{i1}& & color{purple}{mathbf{c_{ij}}} & &c_{ip}\
vdots& & & ddots &vdots\
c_{m1} &ldots& c_{mj} &ldots &c_{mp}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{ij}}}$ equal to:
$$mathbf{color{purple}{c_{ij}} = color{blue}{a_{i1}} color{red}{b_{1j}} + color{blue}{a_{i2}} color{red}{b_{2j}} + cdots + color{blue}{a_{in}} color{red}{b_{nj}}}$$
Now notice that in the sum above, the left outer index is always $i$ ($i$th row of $A$) and the right outer index is always $j$ ($j$th column of $B$). The inner indices run from $1$ to $n$ so you can introduce a summation index $k$ and write this sum compactly using summation notation:
$$color{purple}{mathbf{c_{ij}}}=sum_{k=1}^{n} color{blue}{mathbf{a_{ik}}}color{red}{mathbf{b_{kj}}}$$
The formule above thus gives you the element on position $(i,j)$ in the product matrix $C=AB$ and therefore completely defines $C$ by letting $i=1,...,m$ and $j=1,...,p$.
Can someone explain what that represents by giving me an example? And how did we get that formula?
The illustration above should give you an idea of the general formula, but here's a concrete example where I took $3 times 3$ matrices for $A$ and $B$ and focus on the element on position $(2,3)$:
$$begin{pmatrix}
a_{11} & a_{12} &a_{13}\
color{blue}{1} &color{blue}{2} &color{blue}{3}\
a_{31} & a_{32} &a_{33}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&b_{12} &color{red}{6}\
b_{21}&b_{22} &color{red}{5}\
b_{31}&b_{32} &color{red}{4}
end{pmatrix}
=
begin{pmatrix}
c_{11}& c_{12} &c_{13}\
c_{21}& c_{22} &color{purple}{mathbf{c_{23}}}\
c_{31}& c_{32} &c_{33}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{23}}}$ equal to:
$$begin{array}{rccccccc}
color{purple}{c_{23}}
& = & color{blue}{a_{21}} color{red}{b_{13}} &+& color{blue}{a_{22}} color{red}{b_{23}} &+& color{blue}{a_{23}} color{red}{b_{33}}
& = & displaystyle sum_{k=1}^{3} color{blue}{a_{2k}}color{red}{b_{k3}} \
& = & color{blue}{1} cdot color{red}{6} &+& color{blue}{2} cdot color{red}{5} &+& color{blue}{3} cdot color{red}{4} \[8pt]
& = & 6&+&10&+&12 & =& 28
end{array}$$
answered Dec 18 '16 at 11:15
StackTDStackTD
22.9k2152
$endgroup$
add a comment |
$begingroup$
Visualisation might help. I'll use your notations and dimensions of the given matrices:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
You say you know how to multiply matrices, so take a look at one specific element in the product $C=AB$, namely the element on position $(i,j)$, i.e. in the $i$th row and $j$th column.
To obtain this element, you:
- first multiply all elements of the $i$th row of the matrix $A$ pairwise with all the elements of the $j$th column of the matrix $B$;
- and then you add these $n$ products.
You have to repeat this procedure for every element of $C$, but let's zoom in on that one specific (but arbitrary) element on position $(i,j)$ for now:
$$begin{pmatrix}
a_{11} &ldots &a_{1n}\
vdots& ddots &vdots\
color{blue}{mathbf{a_{i1}}} &color{blue}{rightarrow} &color{blue}{mathbf{a_{in}}}\
vdots& ddots &vdots\
a_{m1} &ldots &a_{mn}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&ldots &color{red}{mathbf{b_{1j}}} &ldots &b_{1p}\
vdots& ddots &color{red}{downarrow} & ddots &vdots\
b_{n1}&ldots &color{red}{mathbf{b_{nj}}}&ldots &b_{np}
end{pmatrix}
=
begin{pmatrix}
c_{11}&ldots& c_{1j} &ldots &c_{1p}\
vdots& ddots & & &vdots\
c_{i1}& & color{purple}{mathbf{c_{ij}}} & &c_{ip}\
vdots& & & ddots &vdots\
c_{m1} &ldots& c_{mj} &ldots &c_{mp}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{ij}}}$ equal to:
$$mathbf{color{purple}{c_{ij}} = color{blue}{a_{i1}} color{red}{b_{1j}} + color{blue}{a_{i2}} color{red}{b_{2j}} + cdots + color{blue}{a_{in}} color{red}{b_{nj}}}$$
Now notice that in the sum above, the left outer index is always $i$ ($i$th row of $A$) and the right outer index is always $j$ ($j$th column of $B$). The inner indices run from $1$ to $n$ so you can introduce a summation index $k$ and write this sum compactly using summation notation:
$$color{purple}{mathbf{c_{ij}}}=sum_{k=1}^{n} color{blue}{mathbf{a_{ik}}}color{red}{mathbf{b_{kj}}}$$
The formule above thus gives you the element on position $(i,j)$ in the product matrix $C=AB$ and therefore completely defines $C$ by letting $i=1,...,m$ and $j=1,...,p$.
Can someone explain what that represents by giving me an example? And how did we get that formula?
The illustration above should give you an idea of the general formula, but here's a concrete example where I took $3 times 3$ matrices for $A$ and $B$ and focus on the element on position $(2,3)$:
$$begin{pmatrix}
a_{11} & a_{12} &a_{13}\
color{blue}{1} &color{blue}{2} &color{blue}{3}\
a_{31} & a_{32} &a_{33}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&b_{12} &color{red}{6}\
b_{21}&b_{22} &color{red}{5}\
b_{31}&b_{32} &color{red}{4}
end{pmatrix}
=
begin{pmatrix}
c_{11}& c_{12} &c_{13}\
c_{21}& c_{22} &color{purple}{mathbf{c_{23}}}\
c_{31}& c_{32} &c_{33}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{23}}}$ equal to:
$$begin{array}{rccccccc}
color{purple}{c_{23}}
& = & color{blue}{a_{21}} color{red}{b_{13}} &+& color{blue}{a_{22}} color{red}{b_{23}} &+& color{blue}{a_{23}} color{red}{b_{33}}
& = & displaystyle sum_{k=1}^{3} color{blue}{a_{2k}}color{red}{b_{k3}} \
& = & color{blue}{1} cdot color{red}{6} &+& color{blue}{2} cdot color{red}{5} &+& color{blue}{3} cdot color{red}{4} \[8pt]
& = & 6&+&10&+&12 & =& 28
end{array}$$
answered Dec 18 '16 at 11:15
StackTDStackTD
22.9k2152
$endgroup$
Visualisation might help. I'll use your notations and dimensions of the given matrices:
If $A=(a_{ij})in M_{mn}(Bbb F), B=(b_{ij})in M_{np}(Bbb F)$ then $C=Atimes B=(c_{ij})in M_{mp}(Bbb F)$.
$c_{ij}=sum_{k=1}^{n} a_{ik}b_{kj}$ where $i=1,...m, j=1,...p$
You say you know how to multiply matrices, so take a look at one specific element in the product $C=AB$, namely the element on position $(i,j)$, i.e. in the $i$th row and $j$th column.
To obtain this element, you:
- first multiply all elements of the $i$th row of the matrix $A$ pairwise with all the elements of the $j$th column of the matrix $B$;
- and then you add these $n$ products.
You have to repeat this procedure for every element of $C$, but let's zoom in on that one specific (but arbitrary) element on position $(i,j)$ for now:
$$begin{pmatrix}
a_{11} &ldots &a_{1n}\
vdots& ddots &vdots\
color{blue}{mathbf{a_{i1}}} &color{blue}{rightarrow} &color{blue}{mathbf{a_{in}}}\
vdots& ddots &vdots\
a_{m1} &ldots &a_{mn}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&ldots &color{red}{mathbf{b_{1j}}} &ldots &b_{1p}\
vdots& ddots &color{red}{downarrow} & ddots &vdots\
b_{n1}&ldots &color{red}{mathbf{b_{nj}}}&ldots &b_{np}
end{pmatrix}
=
begin{pmatrix}
c_{11}&ldots& c_{1j} &ldots &c_{1p}\
vdots& ddots & & &vdots\
c_{i1}& & color{purple}{mathbf{c_{ij}}} & &c_{ip}\
vdots& & & ddots &vdots\
c_{m1} &ldots& c_{mj} &ldots &c_{mp}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{ij}}}$ equal to:
$$mathbf{color{purple}{c_{ij}} = color{blue}{a_{i1}} color{red}{b_{1j}} + color{blue}{a_{i2}} color{red}{b_{2j}} + cdots + color{blue}{a_{in}} color{red}{b_{nj}}}$$
Now notice that in the sum above, the left outer index is always $i$ ($i$th row of $A$) and the right outer index is always $j$ ($j$th column of $B$). The inner indices run from $1$ to $n$ so you can introduce a summation index $k$ and write this sum compactly using summation notation:
$$color{purple}{mathbf{c_{ij}}}=sum_{k=1}^{n} color{blue}{mathbf{a_{ik}}}color{red}{mathbf{b_{kj}}}$$
The formule above thus gives you the element on position $(i,j)$ in the product matrix $C=AB$ and therefore completely defines $C$ by letting $i=1,...,m$ and $j=1,...,p$.
Can someone explain what that represents by giving me an example? And how did we get that formula?
The illustration above should give you an idea of the general formula, but here's a concrete example where I took $3 times 3$ matrices for $A$ and $B$ and focus on the element on position $(2,3)$:
$$begin{pmatrix}
a_{11} & a_{12} &a_{13}\
color{blue}{1} &color{blue}{2} &color{blue}{3}\
a_{31} & a_{32} &a_{33}
end{pmatrix}
cdot
begin{pmatrix}
b_{11}&b_{12} &color{red}{6}\
b_{21}&b_{22} &color{red}{5}\
b_{31}&b_{32} &color{red}{4}
end{pmatrix}
=
begin{pmatrix}
c_{11}& c_{12} &c_{13}\
c_{21}& c_{22} &color{purple}{mathbf{c_{23}}}\
c_{31}& c_{32} &c_{33}
end{pmatrix}$$
with element $color{purple}{mathbf{c_{23}}}$ equal to:
$$begin{array}{rccccccc}
color{purple}{c_{23}}
& = & color{blue}{a_{21}} color{red}{b_{13}} &+& color{blue}{a_{22}} color{red}{b_{23}} &+& color{blue}{a_{23}} color{red}{b_{33}}
& = & displaystyle sum_{k=1}^{3} color{blue}{a_{2k}}color{red}{b_{k3}} \
& = & color{blue}{1} cdot color{red}{6} &+& color{blue}{2} cdot color{red}{5} &+& color{blue}{3} cdot color{red}{4} \[8pt]
& = & 6&+&10&+&12 & =& 28
end{array}$$
answered Dec 18 '16 at 11:15
StackTDStackTD
22.9k2152
edited Feb 20 at 8:54
answered Dec 18 '16 at 11:15
StackTDStackTD
22.9k2152
answered Dec 18 '16 at 11:15
StackTDStackTD
22.9k2152
answered Dec 18 '16 at 11:15
StackTDStackTD
22.9k2152
22.9k2152
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$begingroup$
As you can see the sum run over $k$ from $1$ to $n$, but after you fix $i,j$. You said that you know how to multiply matrices, so you know that if $C=AB$, then $(i,j)-th$ element $c_{ij}$ is obtained fixing the $i-th$ row of $A$, the $j-th$ column of $B$, and then you multiply the elements $a_{i1}$ with $b_{1j}$, $a_{i2}$ with $b_{2j}$, $dots$, $a_{in}$ with $b_{nj}$, and finally you sum all products. Thus
$$c_{ij}=sum_{k=1}^n a_{ik}b_{kj}$$
answered Dec 18 '16 at 10:27
InsideOutInsideOut
5,12431034
$endgroup$
add a comment |
$begingroup$
As you can see the sum run over $k$ from $1$ to $n$, but after you fix $i,j$. You said that you know how to multiply matrices, so you know that if $C=AB$, then $(i,j)-th$ element $c_{ij}$ is obtained fixing the $i-th$ row of $A$, the $j-th$ column of $B$, and then you multiply the elements $a_{i1}$ with $b_{1j}$, $a_{i2}$ with $b_{2j}$, $dots$, $a_{in}$ with $b_{nj}$, and finally you sum all products. Thus
$$c_{ij}=sum_{k=1}^n a_{ik}b_{kj}$$
answered Dec 18 '16 at 10:27
InsideOutInsideOut
5,12431034
$endgroup$
add a comment |
$begingroup$
As you can see the sum run over $k$ from $1$ to $n$, but after you fix $i,j$. You said that you know how to multiply matrices, so you know that if $C=AB$, then $(i,j)-th$ element $c_{ij}$ is obtained fixing the $i-th$ row of $A$, the $j-th$ column of $B$, and then you multiply the elements $a_{i1}$ with $b_{1j}$, $a_{i2}$ with $b_{2j}$, $dots$, $a_{in}$ with $b_{nj}$, and finally you sum all products. Thus
$$c_{ij}=sum_{k=1}^n a_{ik}b_{kj}$$
answered Dec 18 '16 at 10:27
InsideOutInsideOut
5,12431034
$endgroup$
As you can see the sum run over $k$ from $1$ to $n$, but after you fix $i,j$. You said that you know how to multiply matrices, so you know that if $C=AB$, then $(i,j)-th$ element $c_{ij}$ is obtained fixing the $i-th$ row of $A$, the $j-th$ column of $B$, and then you multiply the elements $a_{i1}$ with $b_{1j}$, $a_{i2}$ with $b_{2j}$, $dots$, $a_{in}$ with $b_{nj}$, and finally you sum all products. Thus
$$c_{ij}=sum_{k=1}^n a_{ik}b_{kj}$$
answered Dec 18 '16 at 10:27
InsideOutInsideOut
5,12431034
answered Dec 18 '16 at 10:27
InsideOutInsideOut
5,12431034
answered Dec 18 '16 at 10:27
InsideOutInsideOut
5,12431034
answered Dec 18 '16 at 10:27
InsideOutInsideOut
5,12431034
5,12431034
add a comment |
add a comment |
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$begingroup$
Example: To find $c_{11}$, you multiply the first row of matrix $A$ with the first column of matrix $B$.
$endgroup$
– Prahlad Vaidyanathan
Dec 18 '16 at 10:18
$begingroup$
$c_{ij}$ is the element on $i$-th row and $j$-th column of $C$. Similarly, $a_{ik}$, $b_{kj}$ are elements of matrices $B$ and $C$.
$endgroup$
– Peter Franek
Dec 18 '16 at 10:19
$begingroup$
If you know coding, this is how we multiply two matrices.
$endgroup$
– Rohan
Dec 18 '16 at 10:20