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Jul 18

## Geometry of linear equations: matrix as a mapping ## Geometry of linear equations: matrix as a mapping

General idea. Let $f(x)$ be a function with the domain $D(f).$ We would like to know for which $y$ the equation $f(x)=y$ has solutions, and if it does, then how many (one or more). For the existence part, let the argument $x$ run over $D(f)$ and see what set the values $f(x)$ belong to. It is the image $\text{Img}(f)=\{ f(x):x\in D(f)\}.$ This definition directly implies

Basic observation 1. The equation $f(x)=y$ has solutions if and only if $y\in\text{Img}(f).$

For the uniqueness part, fix $y\in\text{Img}(f)$ and see for which arguments $x\in D(f)$ the value $f(x)$ equals that $y.$ This is the counter-image $f^{-1}(y)=\{ x\in D(f):f(x)=y\}$ of $y.$

Basic observation 2. If $f^{-1}(y)$ consists of one point, you have uniqueness.

See how this works for the function $f(x)=0,$ $x\leq 0,$ $f(x)=x,$ $x>0.$ This function is not linear. The function generated by a matrix is linear, and a lot more can be said in addition to the above observations.

### Matrix as a mapping

Definition 1. Let $A$ be a matrix of size $n\times k.$ It generates a mapping $f:R^k\rightarrow R^n$ according to $f(x)=Ax$ ( $x$ is written as a $k\times 1$ column). Following the common practice, we identify the mapping $f$ with the matrix $A.$

Exercise 1 (first characterization of matrix image) Show that the image $\text{Img}(A)$ consists of linear combinations of the columns of $A.$

Solution. Partitioning $A$ into columns, for any $x$ we have

(1) $Ax=\left(A^1,...,A^k\right)\left(\begin{array}{c}x_1\\...\\x_k\end{array}\right)=x_1A^1+...+x_kA^k.$

This means that when $x$ runs over $D(A)=R^k,$ the images $Ax$ are linear combinations of the column-vectors $A^1,...,A^k.$

Exercise 2. The mapping from Definition 1 is linear: for any vectors $x,y$ and numbers $a,b$ one has

(2) $A(ax+by)=aAx+bAy.$

Proof. By (1) $A(ax+by)=(ax+by)_1A^1+...+(ax+by)_kA^k$ $=(ax_1A^1+...+ax_kA^k)+(by_1A^1+...+by_kA^k)$ $=a(x_1A^1+...+x_kA^k)+b(y_1A^1+...+y_kA^k)=aAx+bAy.$

Remark. In (1) we silently used the multiplication rule for partitioned matrices. Here is the statement of the rule in a simple situation. Let $A,B$ be two matrices compatible for multiplication. Let us partition them into smaller matrices $A=\left(\begin{array}{cc}A_{11}&A_{12}\\A_{21}&A_{22}\end{array}\right),$ $B=\left(\begin{array}{cc}B_{11}&B_{12}\\B_{21}&B_{22}\end{array}\right)$.

Then the product $AB$ can be found as if those blocks were numbers: $AB=\left(\begin{array}{cc}A_{11}&A_{12}\\A_{21}&A_{22}\end{array}\right)\left(\begin{array}{cc}B_{11}&B_{12}\\B_{21}&B_{22}\end{array}\right)=\left(\begin{array}{cc}A_{11}B_{11}+A_{12}B_{21}&A_{11}B_{12}+A_{12}B_{22}\\A_{21}B_{11}+A_{22}B_{21}&A_{21}B_{12}+A_{22}B_{22}\end{array}\right).$

The only requirement for this to be true is that the blocks $A_{ij},$ $B_{ij}$ be compatible for multiplication. You will not be bored with the proof. In (1) the multiplication is performed as if $A^1,...,A^k$ were numbers.