Raising the bar

Geometry of linear equations: matrix as a mapping

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

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

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

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

See how this works for the function $latex f(x)=0,$ $latex x\leq 0,$ $latex f(x)=x,$ $latex x>0.$ 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 $latex A$ be a matrix of size $latex n\times k.$ It generates a mapping $latex f:R^k\rightarrow R^n$ according to $latex f(x)=Ax$ ($latex x$ is written as a $latex k\times 1$ column). Following the common practice, we identify the mapping $latex f$ with the matrix $latex A.$

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

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

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

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

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

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

Proof. By (1)

$latex A(ax+by)=(ax+by)_1A^1+...+(ax+by)_kA^k$

$latex =(ax_1A^1+...+ax_kA^k)+(by_1A^1+...+by_kA^k)$

$latex =a(x_1A^1+...+x_kA^k)+b(y_1A^1+...+y_kA^k)=aAx+bAy.$

Remark. In (1) we silently used the multiplication rule for partitioned matrices. Here is the statement of the rule in a simple situation. Let $latex A,B$ be two matrices compatible for multiplication. Let us partition them into smaller matrices

$latex A=\left(\begin{array}{cc}A_{11}&A_{12}\\A_{21}&A_{22}\end{array}\right),$  $latex B=\left(\begin{array}{cc}B_{11}&B_{12}\\B_{21}&B_{22}\end{array}\right)$.

Then the product $latex AB$ can be found as if those blocks were numbers:

$latex AB=\left(\begin{array}{cc}A_{11}&A_{12}\\A_{21}&A_{22}\end{array}\right)\left(\begin{array}{cc}B_{11}&B_{12}\\B_{21}&B_{22}\end{array}\right)=\left(\begin{array}{cc}A_{11}B_{11}+A_{12}B_{21}&A_{11}B_{12}+A_{12}B_{22}\\A_{21}B_{11}+A_{22}B_{21}&A_{21}B_{12}+A_{22}B_{22}\end{array}\right).$

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

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