MATH 240

Wed. September 18th, 2019


Linear Transformations (cont.)

eie_i is defined as the column vector where the iith element is 1 and all others are 0.
Ex. e2e_2 in R3\R^3 is (010)\begin{pmatrix} 0\\1\\0 \end{pmatrix}.


Theorem.

x=(x1x2...xn)=(x10...0)+(0x2...0)+...+(00...xn)=x1e1+x2e2+...+xnenT(x)=T(x1e1+x2e2+...+xnen)=T(x1e1)+T(x2e2)+...+T(xnen)=x1T(e1)+x2T(e2)+...+xnT(en)=[T(e1)T(e2)...T(en)](x1x2...xn)=Ax.\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 AA in this case is called the standard matrix for TT.

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 eie_i vectors.


Ex. For some linear transformation TT, the following are true:
T(100)=(001)T(010)=(100)T(001)=(010)\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 AA of this transformation TT is [T(e1)T(e2)T(e3)]\begin{bmatrix} T(e_1)&T(e_2)&T(e_3) \end{bmatrix}, which is just
(010001100)\begin{pmatrix} 0&1&0\\0&0&1\\1&0&0 \end{pmatrix}
based on the assertions above.


Onto Functions

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

Ex. T:R3R2, T(x)=(100010)x.T: \R^3 \rightarrow \R^2, \ T(x)=\begin{pmatrix} 1&0&0\\0&1&0 \end{pmatrix}x. Is TT onto?
Yes, because this is equivalent to asking whether Ax=bAx=b will always have a solution. AA has a pivot position in every row, so its columns span R2\R^2, so its associated function TT is onto.


One-to-One Functions

A linear transformation T:RnRmT: \R^n \rightarrow \R^m is one-to-one if T(v1)=T(v2)T(v_1)=T(v_2) implies that v1=v2v_1=v_2 (i.e., its inverse will also be a function).

Using the linear transformation from the previous example, T(123)=(12)T\begin{pmatrix} 1\\2\\3 \end{pmatrix}=\begin{pmatrix} 1\\2 \end{pmatrix} and T(124)=(12),T\begin{pmatrix} 1\\2\\4 \end{pmatrix}=\begin{pmatrix} 1\\2 \end{pmatrix}, but (123)(124);\begin{pmatrix} 1\\2\\3 \end{pmatrix}\neq \begin{pmatrix} 1\\2\\4 \end{pmatrix}; therefore, TT is not one-to-one.

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