MATH 240

Fri. November 1st, 2019


Change of Basis Review

Let B={b1,...,bn}B=\{b_1,...,b_n\} and C={c1,...cn}C=\{c_1,...c_n\} be bases of a vector space VV. Then there is a unique n×nn\times n matrix PCBP_{C\leftarrow B} such that [x]C=PCB[x]B[x]_C=P_{C\leftarrow B}[x]_B.

The columns of PCBP_{C\leftarrow B} are the CC-coordinate vectors of the vectors in the basis BB, i.e. PCB=[[b1]C,...,[bn]C]P_{C\leftarrow B}=\begin{bmatrix} [b_1]_C,...,[b_n]_C \end{bmatrix}.


Ex. Let b1=(91), b2=(51), c1=(14), c2=(35)b_1=\begin{pmatrix} -9\\1 \end{pmatrix},\ b_2=\begin{pmatrix} -5\\1 \end{pmatrix},\ c_1=\begin{pmatrix} 1\\-4 \end{pmatrix},\ c_2=\begin{pmatrix} 3\\-5 \end{pmatrix}.

B={b1,b2}, C={c1,c2}B=\{b_1,b_2\},\ C=\{c_1,c_2\} form bases of R2\R^2. Find the change of basis matrix PCBP_{C\leftarrow B}.

[c1c2][b1]C=b1[c1c2][b2]C=b2\begin{aligned} \begin{bmatrix} c_1&c_2 \end{bmatrix}[b_1]_C&=b_1\\ \begin{bmatrix} c_1&c_2 \end{bmatrix}[b_2]_C&=b_2\\ \end{aligned} The operations to reduce [c1c2]\begin{bmatrix} c_1&c_2 \end{bmatrix} to the identity will be the same for both equations. Therefore to simplify computations we can place [b1b2]\begin{bmatrix} b_1&b_2 \end{bmatrix} to the right of it, reduce it to the identity, and then whatever ends up on the right will be PCBP_{C\leftarrow B} (similar to the technique we used before to find the inverse of a matrix).


Eigenvalues and Eigenvectors

Given an n×nn\times n matrix AA, what points will remained fixed under the transformation xAxx\rightarrow Ax?

This question is significant because the result of applying the transformation to that point any number of times will be easily computable.

Definition. Let AA be an n×nn\times n matrix. A scalar λ\lambda is called an eigenvalue of AA if there is a non-trivial (non-zero) solution xx of Ax=λxAx=\lambda x. Such a vector xx is called an eigenvector of λ\lambda.


Ex.
If xx is an eigenvector of AA with an eigenvalue of 22, then Ax=2xAx=2x.
A2(x)=A(Ax)=A(2x)=2A(x)=4xA^2(x)=A(Ax)=A(2x)=2A(x)=4x.
A3(x)=A(A(Ax))=A(4x)=4A(x)=8x.A^3(x)=A(A(Ax))=A(4x)=4A(x)=8x.
An(x)=2nxA^n(x)=2^nx.
This point moves away from the origin in a straight line under repeated multiplication with AA.

If xx is an eigenvector of AA with an eigenvalue of 12\frac{1}{2}, then Ax=12xAx=\frac{1}{2}x.
A2(x)=A(Ax)=A(12x)=12A(x)=14xA^2(x)=A(Ax)=A(\frac{1}{2}x)=\frac{1}{2}A(x)=\frac{1}{4}x.
A3(x)=A(A(Ax))=A(14x)=14A(x)=18x.A^3(x)=A(A(Ax))=A(\frac{1}{4}x)=\frac{1}{4}A(x)=\frac{1}{8}x.
An(x)=2nxA^n(x)=2^{-n}x.
This point moves towards the origin in a straight line under repeated multiplication with AA.


Ex.
Let A=(1652)A=\begin{pmatrix} 1&6\\5&2 \end{pmatrix}, u=(65)u=\begin{pmatrix} 6\\-5 \end{pmatrix}, and v=(32)v=\begin{pmatrix} 3\\-2 \end{pmatrix}. Are either uu or vv eigenvectors of AA?

Au=(2420)=4(65)=4u=λuAv=(911)λv\begin{aligned} Au&=\begin{pmatrix} -24\\20 \end{pmatrix}\\ &=-4\begin{pmatrix} 6\\-5 \end{pmatrix}\\ &=-4u\\ &=\lambda u\\ \\ Av&=\begin{pmatrix} -9\\11 \end{pmatrix}\\ &\neq \lambda v \end{aligned}
uu is an eigenvector of AA with an eigenvalue 4-4 because Au=4uAu=-4u.
vv is not an eigenvector of AA because there is no scalar λ\lambda such that Av=λvAv=\lambda v.


Ex.
Show that 77 is an eigenvalue of (1652)\begin{pmatrix} 1&6\\5&2 \end{pmatrix}.

Ax=7xAx7x=0(A7I)x=0(6655)x=0\begin{aligned} Ax&=7x\\ Ax-7x&=0\\ (A-7I)x&=0\\ \begin{pmatrix} -6&6\\5&-5 \end{pmatrix}x&=0 \end{aligned}