Proving two simple matrix product properties

AI Thread Summary
The discussion focuses on proving two properties of matrix products involving matrices A and B. The first property asserts that the product AB can be expressed as a matrix of the form [Ab_1, Ab_2, ..., Ab_m], where each column is the product of A and the corresponding column vector of B. The second property shows that if A is represented by its row vectors, the product AB can also be expressed as a matrix where each row vector of A multiplies the entire matrix B. Participants clarify the need to correctly represent the dimensions of the resulting matrix and ensure that calculations align with the definitions of matrix multiplication. The thread emphasizes the importance of accurately following matrix multiplication rules to validate the properties.
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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,

<br /> B = \begin{bmatrix}<br /> b_1 , &amp; b_2, &amp; ... &amp; ,b_m<br /> \end{bmatrix}<br />

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

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

<br /> 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}<br />

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

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

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

should be this
<br /> C =<br /> \begin{bmatrix}<br /> \sum_{k=1}^{p} a_{1k} b_{k1} &amp; \cdots &amp; \sum_{k=1}^{p} a_{1k} b_{km}\\<br /> \vdots &amp; \ddots &amp; \vdots\\<br /> \sum_{k=1}^{p} a_{nk} b_{k1} &amp; \cdots &amp;\sum_{k=1}^{p} a_{mk} b_{km}<br /> \end{bmatrix}<br />
since it's an ##n \times m## matrix, not an ##m \times m## matrix.
 
1. Calculate C = AB.
2. Calculate D = [A b1 ... A bm]
3. Show Dij = Cij for any i,j
 
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