Invertible Matrix – Is it Always a Change of Basis Matrix?

Question

In the course that I am having, we are treating change of basis matrices as the matrices of the identity operation from one basis S to another basis say B.

So, our instructor introduced a theorem :

If A and B are similar, i.e., S−1AS = B for an invertible matrix S, then they have the same characteristic polynomial. In particular, they have the same eigenvalues, det(A) = det(B) and Trace(A) = Trace(B).

And hence came the question here… Is every invertible matrix a change of basis matrix?
Every change of basis matrix is certainly invertible, is the converse true?

Answer 1

Yes. If $S$ is an invertible matrix, then its columns will be the other basis, as for the standard basis $e_1,\dots,e_n$, we have $\ Se_k=$ the $k$th column of $S$.

Say $b_k:=Se_k$. Then for a linear transformation $A$, the value $ASe_k$ is the image of $b_k$ under $A$, and for a vector (given in standard coordinates), $S^{-1}v$ will give its coordinates in the basis $(b_1,\dots,b_n)$

because if $v=\lambda_1 b_1+\dots+\lambda_n b_n$, then, as the columns of $S$ are just the $b_k$ 's, this equation becomes $\ v=S\cdot\pmatrix{\lambda_1\\ \vdots\\ \lambda_n}$.