# 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.

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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 $x_2 = \begin{bmatrix} x_2^1 & \cdots & x_2^N \end{bmatrix}^T$.)