# Matrix Multiplying Column Vectors of Another Matrix = Matrix?

$$\color{green}{ \begin{bmatrix}1&2\\3&4\end{bmatrix} }\cdot \begin{bmatrix}1&2\\3&4\end{bmatrix} \mathop{=}^{\huge{\bigstar}} \begin{bmatrix} \color{green}{ \begin{bmatrix}1&2\\3&4\end{bmatrix} }\cdot \begin{bmatrix}1\\3\end{bmatrix} & \color{green}{ \begin{bmatrix}1&2\\3&4\end{bmatrix} }\cdot \begin{bmatrix}2\\4\end{bmatrix} \end{bmatrix} = \left[ \begin{matrix} \left[ \begin{matrix} 7\\ 15\\ \end{matrix}\right] & \left[ \begin{matrix} 10\\ 22\\ \end{matrix}\right]\\ \end{matrix}\right] \mathop{=}^{\huge{\blacklozenge}} \left[ \begin{matrix} 7 & 10\\ 15 & 22\\ \end{matrix}\right].$$

Alex P. wrote: "there is a slight notational abuse; the last two matrices aren't exactly the same".
Yet litteO wrote : "I think there is no abuse of notation, and the last two matrices are exactly the same. You're using block notation."

$\Large{{1,2.}}$ Are the two equalities signalised by the big star and black lozenge authentic and true?
Alex P. wrote that "These things are all matrices, so it's standard matrix multiplication."
Nevertheless, I don't perceive how either qualifies as "standard matrix multiplication."
Please elucidate and expound on why or why not? I'm more interested in intuition than proofs.

So my interpretation is that we're simply using "block notation". If $x_1$ and $x_2$ are $N \times 1$ column vectors, then $\begin{bmatrix} x_1 & x_2 \end{bmatrix}$ denotes an $N \times 2$ matrix. In other words, $\begin{bmatrix} x_1 & x_2 \end{bmatrix}$ is just a short way of writing \begin{bmatrix} x_1^1 & x_2^1 \\ x_1^2 & x_2^2 \\ \vdots & \vdots \\ x_1^N & x_2^N \end{bmatrix} (where $x_1 = \begin{bmatrix} x_1^1 & \cdots & x_1^N \end{bmatrix}^T$ and $x_2 = \begin{bmatrix} x_2^1 & \cdots & x_2^N \end{bmatrix}^T$.)