This question is predicated upon this answer and the comments thereunder:

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

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

This means the equations in this question are simply true, with no abuse of notation, only a perfectly correct use of block notation.

However, it's possible that my way of looking at this is non-standard, so I'll be curious to hear any other viewpoints.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.