# Elementary operations on a matrix and multiplication by elementary matrices

If we have a matrices $$A=\begin{bmatrix} a & b\\ c & d\end{bmatrix} \hskip5mm \mbox{ and } \hskip5mm e_{12}(\lambda)= \begin{bmatrix} 1 & \lambda \\ 0 & 1 \end{bmatrix}$$ then by doing product $$Ae_{12}(\lambda) = \begin{bmatrix} a & a\lambda + b\\ c & c\lambda + d \end{bmatrix} \hskip5mm \mbox{ and } \hskip5mm e_{12}(\lambda)A = \begin{bmatrix} a +c\lambda & b+d\lambda \\ c & d \end{bmatrix}$$ we can interpret that right multiplication by $$e_{12}$$ to $$A$$ gives a column-operation: add $$\lambda$$-times first column to the second column.

In similar way, left multiplication by $$e_{12}(\lambda)$$ to $$A$$ gives row-operation on $$A$$.

Question: Is there any conceptual (not computational, if any) way to see that elementary row and column operations on a matrix can be expressed as multiplication by elementary matrices on left or right, accordingly?

• It seems to me that it should be like this: $$e_{12}(\lambda) A = \begin{bmatrix} a +c\lambda & b+d\lambda \\ c & d \end{bmatrix} \hskip5mm \mbox{ and } \hskip5mm Ae_{12}(\lambda) = \begin{bmatrix} a & a\lambda + b\\ c & c\lambda + d \end{bmatrix}$$ Commented Oct 4, 2021 at 14:56
• Thanks; I edited it; sorry for the mistake. Commented Oct 4, 2021 at 15:20
• You can search for 3Blue1Brown they have provided intuition behind matrix operations.
– user960916
Commented Oct 4, 2021 at 15:50

Let $$A,B$$ be two matrices of order $$n$$.

We can describe $$B$$ as $$B=(b_1,\ldots,b_n)$$, where $$b_i$$ is its column $$i$$.

Notice that $$AB=(Ab_1,\ldots,Ab_n)$$.

Now apply some column-operation on $$AB$$.

For example, let's say that its column $$i$$ is multiplied by $$\lambda$$.

So we obtain $$(Ab_1,\ldots,\lambda Ab_i,\ldots,Ab_n)$$.

Notice that $$(Ab_1,\ldots,\lambda Ab_i,\ldots,Ab_n)=A(b_1,\ldots,\lambda b_i,\ldots,b_n)$$.

So in order to apply some column-operation on $$AB$$, we can first apply it on $$B$$ and then multiply the resulting matrix with $$A$$.

Can you see what happens if $$B=Id$$?

The same reasoning can be used for row-opperations.

Here conceptual and computational ideas go hand in hand. We can see this by looking at the multiplication of $$A$$ with $$e_{12}(\lambda)$$ in some detail.

We have \begin{align*} \begin{bmatrix}1&\lambda\\ 0&1 \end{bmatrix} &=\begin{bmatrix}1&0\\ 0&1 \end{bmatrix} + \begin{bmatrix}0&\lambda\\ 0&0 \end{bmatrix} =I+\lambda\begin{bmatrix}0&\color{blue}{1}\\ 0&0 \end{bmatrix} \end{align*}

It is the position $$_{12}$$ of the blue marked $$1$$ which determines selected row resp. column of $$A$$.

We obtain \begin{align*} Ae_{12}(\lambda)&=A\left(I+\lambda \begin{bmatrix}0&\color{blue}{1}\\ 0&0 \end{bmatrix}\right) =A+\lambda\begin{bmatrix} 0&\color{blue}{a}\\ 0&\color{blue}{c} \end{bmatrix}\\ e_{12}(\lambda)A&=\left(I+\lambda \begin{bmatrix}0&\color{blue}{1}\\ 0&0 \end{bmatrix}\right)A =A+\lambda\begin{bmatrix} \color{blue}{c}&\color{blue}{d}\\ 0&0 \end{bmatrix} \end{align*}