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For $A\in M_{n,m}$ and $B\in M_{n, p}$, the matrix product $AB$ is given by $AB=[c_{ik}]\in M_{m, p}$ where
$$\begin{align*}
c_{ik}=\sum_{j=1}^na_{ij}b_{jk}
\end{align*}$$.
Note that $AB$ is only defined if the dimensions match up ie. $A$ is $m\times n$ and $B$ is $n\times p$.
Note that $AB$ is only defined if the dimensions match up ie. $A$ is $m\times n$ and $B$ is $n\times p$.
Concepts
Matrix multiplication does is not communitive in general.Matrix multiplication is associative.
Consider a composition of transformations $A\circ B:\mathbb{F}^p\to\mathbb{F}^m$, where $A=[a_{ij}]$ and $B=[b_{ij}]$.Then $(A\circ B)(x)=A(B(x))$ for $A=[a_{ij}]$, $B=[b_{ij}]$ and $x\in\mathbb{F}^n$.
Recall that $$\begin{align*} [Ax]_i=\begin{bmatrix}\sum_{j=1}^na_{ij}x_j\end{bmatrix} \end{align*}$$ and $$\begin{align*} [Bx]_i=\begin{bmatrix}\sum_{k=1}^pb_{ik}x_k\end{bmatrx}. \end{align*}$$ Let $y=Bx\in\mathbb{F}^n$. Then $$\begin{align*} [y]_j=\begin{bmatrix}\sum_{k=1}^pb_{ik}x_k\end{bmatrix} \end{align*}$$ and hence $$\begin{align*} [A(Bx)]_j&=[Ay]_i \\ &=\begin{bmatrix}sum_{j=1}^n a_{ij}y_j\end{bmatrix} \\ &=\sum_{j=1}^n a_{ij}(\sum_{k=1}^pb_{jk}x_k) \\ &=\sum_{j=1}^n\sum_{k=1}^pa_{ij}b_{ik}x_k \end{align*}$$.
If we'd liked to define $C\in M_{mp}(\mathbb{F})$ by $$\begin{align*} c_{ik}=\sum_{j=1}^na_{ij}b_{jk} \end{align*}$$ then for $x\in\mathbb{F}^p$, $$\begin{align*} [Cx]_i&=\sum_{k=1}^p c_{ik}x_k \\ &=\sum_{k=1}^p\sum_{j=1}^n a_{ij}b_{jk}x_k \\ &=[(A\circ B)(x)]_i \end{align*}$$
Used In
- Compositions of functions
- Powers of matrices
- Matrix roots
Hypothesis
Coming soonResults
Coming soon
For $A\in M_{n,m}$ and $B\in M_{n, p}$, the matrix product $AB$ is given by $AB=[c_{ik}]\in M_{m, p}$ where
$$\begin{align*}
c_{ik}=\sum_{j=1}^na_{ij}b_{jk}
\end{align*}$$.
Note that $AB$ is only defined if the dimensions match up ie. $A$ is $m\times n$ and $B$ is $n\times p$.
Note that $AB$ is only defined if the dimensions match up ie. $A$ is $m\times n$ and $B$ is $n\times p$.
Concepts
Matrix multiplication does is not communitive in general.Matrix multiplication is associative.
Consider a composition of transformations $A\circ B:\mathbb{F}^p\to\mathbb{F}^m$, where $A=[a_{ij}]$ and $B=[b_{ij}]$.Then $(A\circ B)(x)=A(B(x))$ for $A=[a_{ij}]$, $B=[b_{ij}]$ and $x\in\mathbb{F}^n$.
Recall that $$\begin{align*} [Ax]_i=\begin{bmatrix}\sum_{j=1}^na_{ij}x_j\end{bmatrix} \end{align*}$$ and $$\begin{align*} [Bx]_i=\begin{bmatrix}\sum_{k=1}^pb_{ik}x_k\end{bmatrx}. \end{align*}$$ Let $y=Bx\in\mathbb{F}^n$. Then $$\begin{align*} [y]_j=\begin{bmatrix}\sum_{k=1}^pb_{ik}x_k\end{bmatrix} \end{align*}$$ and hence $$\begin{align*} [A(Bx)]_j&=[Ay]_i \\ &=\begin{bmatrix}sum_{j=1}^n a_{ij}y_j\end{bmatrix} \\ &=\sum_{j=1}^n a_{ij}(\sum_{k=1}^pb_{jk}x_k) \\ &=\sum_{j=1}^n\sum_{k=1}^pa_{ij}b_{ik}x_k \end{align*}$$.
If we'd liked to define $C\in M_{mp}(\mathbb{F})$ by $$\begin{align*} c_{ik}=\sum_{j=1}^na_{ij}b_{jk} \end{align*}$$ then for $x\in\mathbb{F}^p$, $$\begin{align*} [Cx]_i&=\sum_{k=1}^p c_{ik}x_k \\ &=\sum_{k=1}^p\sum_{j=1}^n a_{ij}b_{jk}x_k \\ &=[(A\circ B)(x)]_i \end{align*}$$
Used In
- Compositions of functions
- Powers of matrices
- Matrix roots
Hypothesis
Coming soonResults
Coming soon
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