# Is every invertible matrix a change of basis matrix?

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?

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 $\ v=S\cdot\pmatrix{\lambda_1\\ \vdots\\ \lambda_n}$.

• And the reverse direction also holds? So can we say that: a matrix is a change of basis matrix $\iff$ it is invertible. Commented Feb 8 at 10:11

Yes it is. Any invertible matrix $\mathbf{A}$ is the change of basis of the basis formed by the columns of $\mathbf{A}$ (which is a basis because $\mathbf{A}$ is invertible) to the canonical basis.

Yes. In fact, the following is true: If $$P$$ is an invertible $$n\times n$$ matrix with coefficients in some field $$\mathbb{K}$$ and $$\beta=(b_1,...,b_n)$$ is a basis for an $$n$$ dimensional vector space $$V$$ over $$\mathbb{K}$$, then

1) There is a basis $$\alpha=(a_1, ... , a_n)$$ for $$V$$ such that $$P$$ is the change of basis matrix from $$\beta$$ to $$\alpha$$.

2) There is a basis $$\alpha'=(a_1', ... , a_n')$$ for $$V$$ such that $$P$$ is the change of basis matrix from $$\alpha'$$ to $$\beta$$.

Let's prove this, assuming that $$V= \mathbb{K}^n$$ (the results will easily carry over to the general case by the isomorphism of V and $$\mathbb{K}^n$$).

Assuming $$V= \mathbb{K}^n$$, we can let $$B$$ be the (invertible) matrix with columns $$b_1, ... ,b_n$$. As $$P$$ is also invertible, we can define a matrix $$A$$ by $$A=B P^{-1}$$. As $$A$$ is invertible, the columns $$a_1, ..., a_n$$ of $$A$$ constitute a basis $$\alpha$$ for $$V$$. We now claim that this is the required basis, i.e. that $$P$$ is the change of basis matrix from $$\beta$$ to $$\alpha$$. To show this, we must show that the columns of matrix $$P$$ are the basis vectors $$b_j$$ expressed in the basis $$\alpha$$, i.e. that $$b_j=\sum_{i=1}^{n} P_{ij}a_i$$ for $$j=1,...,n.$$ But this is indeed the case, as
$$\sum_{i=1}^{n} P_{ij}a_i=A \begin{bmatrix}P_{1j}\\.\\.\\.\\P_{nj}\end{bmatrix}=BP^{-1}\begin{bmatrix}P_{1j}\\.\\.\\.\\P_{nj}\end{bmatrix}=B e_j=b_j,$$ where $$e_j$$ is the j'th element in the standard basis for $$\mathbb{K}^n$$.
For $$j=1,...,n$$, let $$a_j'=\sum_{i=1}^{n} P_{ij}b_i$$. As P is invertible, $$\alpha'=(a_1', ... , a_n')$$ constitutes a basis that is easily seen to meet the requirements.
• I am not sure you need to assume $V = \mathbb{K}^n$, as the role of $e_j$ is quite dummy here..? We have this equality for any $V$ in fact, $BP^{-1}\begin{bmatrix}P_{1j}\\.\\.\\.\\P_{nj}\end{bmatrix}=b_j$ only comes by invertibility of $P$ I think !. If i am wrong, could you please exhibit the solution with the isomorphism, it is important to me to understand technically how it would help to generalise :) Commented Jan 5, 2022 at 19:47