why is matrix scalar multiplication we multiply every entry with scalar? and only multiply one row in determinants

Question

why is matrix scalar multiplication we multiply every entry with scalar and only multiply one row in determinants ? in determinants we only multiply one row so. why do we define multiplication that way ?

Answer 1

Think of matrices as linear transformations of a space. That is to say, if we drew a square grid on say $\mathbb{R}^2$, like you'd find in a copy book, in loose sense, when we multiply this vector space by the matrix, all of the lines change and form a grid of parallelograms. To see this, multiply any $2\times 2$ matrix by the standard basis vectors $(1,0)$ and $(0,1)$ in $\mathbb{R}^2$. These vectors once formed two sides of a square and now form two sides of a parallelogram. The same transformation happens to the whole space.

When we multiply a matrix by a scalar, all we do is multiply the side lengths of this new grid by the scaling factor. For example, if we had $A=\begin{pmatrix}1&1\\2&1\end{pmatrix}$, we would find that $\begin{pmatrix}1&1\\2&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix}$ and $\begin{pmatrix}1&1\\2&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}$.

Plot these vectors here to get a good idea of what's going on.

If we defined $B=\lambda A$ and set $\lambda=2$ we would find that $B=\begin{pmatrix}2&2\\4&2\end{pmatrix}$ and $\begin{pmatrix}2&2\\4&2\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}2\\4\end{pmatrix}$ and $\begin{pmatrix}2&2\\4&2\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}2\\2\end{pmatrix}$ which is the same as just scaling the vectors. That is to say $(\lambda A)v=\lambda(Av)$ for any vector $v$.

We have to multiply every element of $A$ by the scaling factor. This is so that the basis vectors which define the new grid get scaled too, just like scaling a vector normally

Answer 2

Consider it as-if it were a "shorthand" technique that just happened to work beyond its specific purpose.

You have some matrix $\mathbf{M}$ that looks like this: $$ \mathbf{M}=\begin{pmatrix}m_{11}&m_{12}\\m_{21}&m_{22}\end{pmatrix} $$ Now you want to represent some operation that multiplies all the values by some constant scalar value. You could do this by matrix multiplication using something like the following: $$ \begin{pmatrix}c&0\\0&c\end{pmatrix}\begin{pmatrix}m_{11}&m_{12}\\m_{21}&m_{22}\end{pmatrix}=\begin{pmatrix}c\ m_{11}&c\ m_{12}\\c \ m_{21}&c \ m_{22}\end{pmatrix} $$ which follows all the "matrix multiplication" steps you expect. But this feels like so much extra carpel-tunnel-inducing repetitive writing! Can't we write it more compactly? It's just an identity matrix where every entry is multiplied by that same constant! So what if we just say, as a shorthand, that: $$ c\mathbf{I}=c\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}c&0\\0&c\end{pmatrix} $$ That would make things simpler to write! We don't have to write out a whole diagonal matrix of $c$'s to show what's happening, we can just write $c\mathbf{I}$ now, so that: $$ c\mathbf{I}\mathbf{M}=\begin{pmatrix}c&0\\0&c\end{pmatrix}\begin{pmatrix}m_{11}&m_{12}\\m_{21}&m_{22}\end{pmatrix}=\begin{pmatrix}c\ m_{11}&c\ m_{12}\\c\ m_{21}&c\ m_{22}\end{pmatrix} $$ but wait... isn't $\mathbf{I}\mathbf{M}=\mathbf{M}$ one of the most important identities/definitions that we are taught! What now? $$ c\mathbf{M}=c\mathbf{I}\mathbf{M}=\begin{pmatrix}c&0\\0&c\end{pmatrix}\begin{pmatrix}m_{11}&m_{12}\\m_{21}&m_{22}\end{pmatrix}=\begin{pmatrix}c\ m_{11}&c\ m_{12}\\c\ m_{21}&c\ m_{22}\end{pmatrix} $$ But that would mean... $$ c\mathbf{M}=c \ \begin{pmatrix}m_{11}&m_{12}\\m_{21}&m_{22}\end{pmatrix}=\begin{pmatrix}c\ m_{11}&c\ m_{12}\\c\ m_{21}&c\ m_{22}\end{pmatrix} $$ Neat! It looks like the "shorthand" for that operation still works just fine for full matrices too. (Rather than just the simple identity matrix case we wanted). Maybe this shorthand works even better than we'd hoped!

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Of course the above is mostly just an ad hoc intuitive explanation to make "scalar multiplication" in an introductory linear algebra "matrix equation" world feel less like and odd-duck and instead just feel like an natural extension of the other "matrix multiplication" concepts you're learning.

The exact "why" reason ties into the "...you want to represent some operation that multiplies all the values by some constant scalar value..." place I started. As explained by AP's answer, sometimes you need to re-scale or transform the different vector spaces at play in a given problem and need exactly that kind of rescale-by-a-constant operation to do so.