MATH 240

Wed. September 18th, 2019

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Linear Transformations (cont.)

$e_i$ is defined as the column vector where the $i$th element is 1 and all others are 0.
Ex. $e_2$ in $\R^3$ is $\begin{pmatrix} 0\\1\\0 \end{pmatrix}$.

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Theorem.

$\begin{aligned} x&=\begin{pmatrix} x_1\\x_2\\...\\x_n \end{pmatrix}\\ &=\begin{pmatrix} x_1\\0\\...\\0 \end{pmatrix}+\begin{pmatrix} 0\\x_2\\...\\0 \end{pmatrix}+...+\begin{pmatrix} 0\\0\\...\\x_n \end{pmatrix}\\ &=x_1e_1+x_2e_2+...+x_ne_n\\ \\ T(x)&=T(x_1e_1+x_2e_2+...+x_ne_n)\\ &=T(x_1e_1)+T(x_2e_2)+...+T(x_ne_n)\\ &=x_1T(e_1)+x_2T(e_2)+...+x_nT(e_n)\\ &=\begin{bmatrix} T(e_1)&T(e_2)&...&T(e_n) \end{bmatrix}\begin{pmatrix} x_1\\x_2\\...\\x_n \end{pmatrix}\\ &=Ax. \end{aligned}$
The matrix $A$ in this case is called the standard matrix for $T$.

This theorem tells us that all we don’t have to know what happens to every possible input vector to describe the linear transformation; we just have to know what happens to the $e_i$ vectors.

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Ex. For some linear transformation $T$, the following are true:
$\begin{aligned} T\begin{pmatrix} 1\\0\\0 \end{pmatrix}&=\begin{pmatrix} 0\\0\\1 \end{pmatrix}\\ T\begin{pmatrix} 0\\1\\0 \end{pmatrix}&=\begin{pmatrix} 1\\0\\0 \end{pmatrix}\\ T\begin{pmatrix} 0\\0\\1 \end{pmatrix}&=\begin{pmatrix} 0\\1\\0 \end{pmatrix} \end{aligned}$
The standard matrix $A$ of this transformation $T$ is $\begin{bmatrix} T(e_1)&T(e_2)&T(e_3) \end{bmatrix}$, which is just
$\begin{pmatrix} 0&1&0\\0&0&1\\1&0&0 \end{pmatrix}$
based on the assertions above.

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Onto Functions

A linear transformation $T: \R^n \rightarrow \R^m$ is onto if, for any $b$ in $\R^m$, there is an $x$ in $\R^n$ such that $T(x)=b$.

Ex. $T: \R^3 \rightarrow \R^2, \ T(x)=\begin{pmatrix} 1&0&0\\0&1&0 \end{pmatrix}x.$ Is $T$ onto?
Yes, because this is equivalent to asking whether $Ax=b$ will always have a solution. $A$ has a pivot position in every row, so its columns span $\R^2$, so its associated function $T$ is onto.

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One-to-One Functions

A linear transformation $T: \R^n \rightarrow \R^m$ is one-to-one if $T(v_1)=T(v_2)$ implies that $v_1=v_2$ (i.e., its inverse will also be a function).

Using the linear transformation from the previous example, $T\begin{pmatrix} 1\\2\\3 \end{pmatrix}=\begin{pmatrix} 1\\2 \end{pmatrix}$ and $T\begin{pmatrix} 1\\2\\4 \end{pmatrix}=\begin{pmatrix} 1\\2 \end{pmatrix},$ but $\begin{pmatrix} 1\\2\\3 \end{pmatrix}\neq \begin{pmatrix} 1\\2\\4 \end{pmatrix};$ therefore, $T$ is not one-to-one.

If $T(x)=0$ has only the trivial solution, then $T$ is one-to-one. This is the same definition as linear independence for the column vectors of the standard matrix of T!