Let B={b1,...,bn} and C={c1,...cn} be bases of a vector space V. Then there is a unique n×n matrix PC←B such that [x]C=PC←B[x]B.
The columns of PC←B are the C-coordinate vectors of the vectors in the basis B, i.e. PC←B=[[b1]C,...,[bn]C].
Ex. Let b1=(−91), b2=(−51), c1=(1−4), c2=(3−5).
B={b1,b2}, C={c1,c2} form bases of R2. Find the change of basis matrix PC←B.
[c1c2][b1]C[c1c2][b2]C=b1=b2 The operations to reduce [c1c2] to the identity will be the same for both equations. Therefore to simplify computations we can place [b1b2] to the right of it, reduce it to the identity, and then whatever ends up on the right will be PC←B (similar to the technique we used before to find the inverse of a matrix).
Eigenvalues and Eigenvectors
Given an n×n matrix A, what points will remained fixed under the transformation x→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 A be an n×n matrix. A scalar λ is called an eigenvalue of A if there is a non-trivial (non-zero) solution x of Ax=λx. Such a vector x is called an eigenvector of λ.
Ex.
If x is an eigenvector of A with an eigenvalue of 2, then Ax=2x.
A2(x)=A(Ax)=A(2x)=2A(x)=4x.
A3(x)=A(A(Ax))=A(4x)=4A(x)=8x.
An(x)=2nx.
This point moves away from the origin in a straight line under repeated multiplication with A.
If x is an eigenvector of A with an eigenvalue of 21, then Ax=21x.
A2(x)=A(Ax)=A(21x)=21A(x)=41x.
A3(x)=A(A(Ax))=A(41x)=41A(x)=81x.
An(x)=2−nx.
This point moves towards the origin in a straight line under repeated multiplication with A.
Ex.
Let A=(1562), u=(6−5), and v=(3−2). Are either u or v eigenvectors of A?
AuAv=(−2420)=−4(6−5)=−4u=λu=(−911)=λv
u is an eigenvector of A with an eigenvalue −4 because Au=−4u.
v is not an eigenvector of A because there is no scalar λ such that Av=λv.
Ex.
Show that 7 is an eigenvalue of (1562).
AxAx−7x(A−7I)x(−656−5)x=7x=0=0=0