If we have a matrices \[A=\begin{bmatrix}a & b \\c & d \end{bmatri

Clifton Sanchez 2021-11-23 Answered
If we have a matrices
\[A=\begin{bmatrix}a & b \\c & d \end{bmatrix} \text{ and } e_{12}(\lambda)=\begin{bmatrix}1 & \lambda \\0 & 1\end{bmatrix}\]
then by doing product \[Ae_{12}(\lambda)=\begin{bmatrix}a & a\lambda+b \\c & c\lambda+d \end{bmatrix} \text{ and } e_{12}(\lambda)A=\begin{bmatrix}a+c\lambda & b+d\lambda \\c & d\end{bmatrix}\]
we can interpret that right multiplication by \(\displaystyle{e}_{{{12}}}\) to A gives a column-operation: add \(\displaystyle\lambda\)-times first column to the second column. In similar way, left multiplication by \(\displaystyle{e}_{{{12}}}{\left(\lambda\right)}\) to A gives row-operation on A.
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?

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Expert Answer

Froldigh
Answered 2021-11-24 Author has 958 answers
Here conceptual and computational ideas go hand in hand. We can see this by looking at the multiplication of A with \(\displaystyle{e}_{{{12}}}{\left(\lambda\right)}\) in some detail.
We have
\[\begin{bmatrix}1 & \lambda \\0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}+\begin{bmatrix}0 & \lambda \\0 & 0 \end{bmatrix}=I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix}\]
It is the position \(\displaystyle_{\left\lbrace{12}\right\rbrace}\) of the blue marked 1 which determines selected row resp. column of A.
We obtain
\[Ae_{12}(\lambda)=A(I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix})=A+\lambda\begin{bmatrix}0 & a \\0 & c \end{bmatrix}\]
\[e_{12}(\lambda)A=(I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix})A=A+\lambda\begin{bmatrix}c & d \\0 & 0 \end{bmatrix}\]
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Louis Smith
Answered 2021-11-25 Author has 764 answers
Let A,B be two matrices of order n.
We can describe B as B=\(\displaystyle{\left({b}_{{1}},\ldots,{b}_{{n}}\right)}\), where \(\displaystyle{b}_{{i}}\) is its column i.
Notice that AB=\(\displaystyle{\left({A}{b}_{{1}},\ldots,{A}{b}_{{n}}\right)}\)
Now apply some column-operation on AB.
For example, let's say that its column i is multiplied by \(\displaystyle\lambda\)
So we obtain \(\displaystyle{\left({A}{b}_{{1}},\ldots,\lambda{A}{b}_{{i}},\ldots,{A}{b}_{{n}}\right)}\)
Notice that \(\displaystyle{\left({A}{b}_{{1}},\ldots,\lambda{A}{b}_{{i}},\ldots,{A}{b}_{{n}}\right)}={A}{\left({b}_{{1}},\ldots,\lambda{b}_{{i}},\ldots,{b}_{{n}}\right)}\)
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.
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