# Some simple matrix identities

I've recently been learning some linear algebra and I've isolated what seem to be some important matrix relations (often used tacitly). I would be most grateful if someone could just check that I have isolated them correctly. I would also welcome any important generalisations :)

Many thanks!

NB: $F$ denotes a field.

1) Let $\lambda_1,\ldots,\lambda_n \in F$ and $c_1,\ldots,c_n \in F^{m,T}$. Then

$$\begin{pmatrix} \uparrow & & \uparrow\\ c_1 & \ldots & c_n\\ \downarrow & & \downarrow \end{pmatrix} \begin{pmatrix} \lambda_1\\ \vdots\\ \lambda_n \end{pmatrix}=\lambda_1c_1+\ldots+\lambda_nc_n.$$

$\,$

2) Let $\mu_1,\ldots,\mu_m \in F$ and $r_1,\ldots,r_m \in F^n$. Then

$$\begin{pmatrix} \mu_1 & \ldots & \mu_m \end{pmatrix} \begin{pmatrix} \leftarrow & r_1 & \rightarrow \\ & \vdots & \\ \leftarrow & r_m & \rightarrow \\ \end{pmatrix} =\mu_1r_1+\ldots+\mu_mr_m.$$

$\,$

3) Let $A \in \text{Mat}_{k\times m}(F)$ and $c_1,\ldots,c_n \in F^{m,T}$. Then

$$\begin{pmatrix} & & \\ \; & A & \; \\ & & \end{pmatrix} \begin{pmatrix} \uparrow & & \uparrow\\ c_1 & \ldots & c_n\\ \downarrow & & \downarrow \end{pmatrix}= \begin{pmatrix} \uparrow & & \uparrow\\ Ac_1 & \ldots & Ac_n \\ \downarrow & & \downarrow \end{pmatrix}$$

$\,$

4) Let $B \in \text{Mat}_{n\times k}(F)$ and $r_1,\ldots,r_m \in F^n$. Then

$$\begin{pmatrix} \leftarrow & r_1 & \rightarrow \\ & \vdots & \\ \leftarrow & r_m & \rightarrow \\ \end{pmatrix} \begin{pmatrix} & & \\ \; & B & \; \\ & & \end{pmatrix} =\begin{pmatrix} \leftarrow & r_1B & \rightarrow \\ & \vdots & \\ \leftarrow & r_mB & \rightarrow \\ \end{pmatrix}$$

• Your dimensions in the last two don't look right. If $A$ is $m \times n$ and $C = (c_1, c_2, \ldots c_n)$ is also $m \times n$, then you can't form the product $AC$. Similarly, if $R = (r_1; r_2; \ldots r_m)$ is $m \times n$ then $RA$ doesn't make sense. – Bungo Aug 26 '14 at 22:26
• Yes, I noticed that :) I think it's fixed now. Thank you! – user171862 Aug 26 '14 at 22:33
• The rows in your final matrix should be $r_j B$, assuming $r_j$ are $1 \times n$ vectors. Aside from that, everything looks correct. – Bungo Aug 26 '14 at 22:39
• Oops! So they should. Thank you again :) – user171862 Aug 26 '14 at 22:40

These are all correct. And in fact, if $A$ has rows $r_i$ and $B$ has columns $c_j$, then the $ij$ entry of $AB$ has value $r_i c_j$ (where that's a matrix multiply of a $1 \times n$ matrix by an $n \times 1$ matrix, resulting in a $1 \times 1$ matrix that I'm treating as a number).