# Let $M$ and $A$ be matrices for which the product matrix $MA$ is defined.

Let $M$ and $A$ be matrices for which the product matrix $MA$ is defined. If the jth column of $A$ is a linear combination of a set of columns of $A$, prove that the jth column of $MA$ is a linear combination of the corresponding columns of $MA$ with the same corresponding coefficients.

I am unsure of where to start or, quite honestly, what the question is asking. I have spent some time wrestling with this in my head and I am not developing any intuition. Thanks for your help!

Assume we have matrices $A$ and $M$ where the product $MA$ is defined and that the jth column of $A$ is a linear combination of some set of columns of $A$. Consider the jth column of $A$ $A_j= \begin{pmatrix} a_1 \\ a_2\\ \vdots\\ a_n \end{pmatrix} =a_iA_i+a_kA_k+ \dots +a_lA_l = a_i\begin{pmatrix} i_1 \\ i_2\\ \vdots\\ i_n \end{pmatrix} + a_k\begin{pmatrix} k_1 \\ k_2\\ \vdots\\ k_n \end{pmatrix} + \dots + a_l\begin{pmatrix} l_1 \\ l_2\\ \vdots\\ l_n \end{pmatrix}$

If we consider the jth row of $M$ since the jth column of $MA$ will be the product of the $jth$ row of $M$ and the $jth$ column of $A$ we have $M_j= \begin{pmatrix} b_1 & b_2 & \dots & b_n \end{pmatrix}$ We have
$(MA)_j= \begin{pmatrix} b_1a_1 \\ b_2a_2\\ \vdots\\ b_na_n \end{pmatrix} = a_i\begin{pmatrix} b_1i_1 \\ b_2i_2\\ \vdots\\ b_ni_n \end{pmatrix} + a_k\begin{pmatrix} b_1k_1 \\ b_2k_2\\ \vdots\\ b_nk_n \end{pmatrix} + \dots + a_l\begin{pmatrix} b_1l_1 \\ b_2l_2\\ \vdots\\ b_nl_n \end{pmatrix}$

and so $MA_j$ is a linear combination of the corresponding columns of $MA$.

• Perhaps it'd help to represent $A$ as $\pmatrix{a_1 & a_2 & \cdots & a_n}$ where $a_i$ is the $i$th column of $A$. Then $MA = \pmatrix{Ma_1 & Ma_2 & \cdots & Ma_n}$. Can you see where to go from here? – user137731 Oct 25 '15 at 1:17
• I might see where that might help me. Where I am really baffled is in saying the jth column of $A$ is a linear combination of the columns of $A$....? – Aaron Zolotor Oct 25 '15 at 1:19
• Suppose the fifth column of $A$ is $A_5$ and $A_5 = 3A_2 + 4 A_3$. What's the fifth column of $MA$? It's $M A_5$. And what's that equal to? – littleO Oct 25 '15 at 1:20
• Just set $a_k$ (an arbitrary $k$th column) equal to $\alpha a_i + \cdots + \omega a_j$ (a linear combination of arbitrary columns $\{a_i, \dots, a_j\}$). Then you just need to prove that $Ma_k$ is a linear combination of $\{Ma_i, \dots, Ma_j\}$. – user137731 Oct 25 '15 at 1:23
• The question only tells us that one particular column of $A$ , the $j$ th column, is a linear combination of other columns of $A$. – littleO Oct 25 '15 at 1:27

The kicker is that if we write $A$ in terms of its columns as $A=(A_1\:A_2\cdots A_n),$ then you can prove that $$MA=(MA_1\:MA_2\cdots MA_n).$$ So, in general, the $k$th column of $MA$ is $M$ times the $k$th column of $A$. Multiplying on the left by $M$ in the linear combination equation will get us the rest of the way.