Proving two simple matrix product properties

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TheSodesa
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Homework Statement


Let ##A## be an n × p matrix and ##B## be an p × m matrix with the following column vector representation,

[tex] B = \begin{bmatrix}<br /> b_1 , & b_2, & ... & ,b_m<br /> \end{bmatrix}[/tex]

Prove that
[tex]AB =<br /> \begin{bmatrix}<br /> Ab_1 , & Ab_2, & ... & , Ab_m<br /> \end{bmatrix}[/tex]

If ##A## is represented with help of its row vectors, prove that

[tex] AB =<br /> \begin{bmatrix}<br /> a^{T}_{1}\\<br /> \vdots\\<br /> a_{n}^{T}<br /> \end{bmatrix} B<br /> =<br /> \begin{bmatrix}<br /> a^{T}_{1} B\\<br /> \vdots\\<br /> a^{T}_{n} B<br /> \end{bmatrix}[/tex]

Homework Equations



The matrix product:
If ##A## is an ##m\times p## matrix and ##B## is a ##p\times n## matrix, then
\begin{equation}
AB = C = (c_{ij})_{m \times n} = (\sum_{k=1}^{p} a_{ik}b_{kj})_{m \times n}
\end{equation}

The Attempt at a Solution



For starters what does proving in this context mean? Should I simply write out the matrix

[tex] C =<br /> \begin{bmatrix}<br /> \sum_{k=1}^{p} a_{1k} b_{k1} & \cdots & \sum_{k=1}^{p} a_{1k} b_{km}\\<br /> \vdots & \ddots & \vdots\\<br /> \sum_{k=1}^{p} a_{mk} b_{k1} & \cdots &\sum_{k=1}^{p} a_{mk} b_{km}<br /> \end{bmatrix}[/tex]
and conclude that each column is essentially ##Ab_{l}## where ##1 < l < m##, since each element of the matrix is the dot (inner) product of a row in ##A## and a column in ##B##?
 
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Oops. This

[tex] C =<br /> \begin{bmatrix}<br /> \sum_{k=1}^{p} a_{1k} b_{k1} & \cdots & \sum_{k=1}^{p} a_{1k} b_{km}\\<br /> \vdots & \ddots & \vdots\\<br /> \sum_{k=1}^{p} a_{mk} b_{k1} & \cdots &\sum_{k=1}^{p} a_{mk} b_{km}<br /> \end{bmatrix}[/tex]

should be this
[tex] C =<br /> \begin{bmatrix}<br /> \sum_{k=1}^{p} a_{1k} b_{k1} & \cdots & \sum_{k=1}^{p} a_{1k} b_{km}\\<br /> \vdots & \ddots & \vdots\\<br /> \sum_{k=1}^{p} a_{nk} b_{k1} & \cdots &\sum_{k=1}^{p} a_{mk} b_{km}<br /> \end{bmatrix}[/tex]
since it's an ##n \times m## matrix, not an ##m \times m## matrix.