Change of basis changes rank of matrix

Question

I have $3\times 3$ matrix $A$ with full rank in basis $\mathcal{B}_1$. If I change the basis to $\mathcal{B}_2$, I get matrix $A'$. But does the rank of $A'$ changes or stays $3$. I would say it stays the same, but I don't know how to prove that.

Answer 1

You are right, the rank stays the same. $A$ and $A'$ represent the same endomorphism but in two different bases. They are said similar. See here for the proof.

Answer 2

If $A$ is the transformation matrix of the endomorphism $f: V \to V$ with respect the basis $B_1$ of the vectorspace $V$ and $A'$ the transformation matrix of $f$ with respect to the basis $B_2$ of the vectorspace V you can find an invertible matrix $P$ such that $A'=P^{-1}AP$ which is the same as saying that $A$ and $A'$ are similar. It immediately follows of the definition of similar martices that $A$ and $A'$ have the same determinant.

Here is a proof that is very intuitive.

Answer 3

Well, a base change can be viewed as a left-multiplication with a full-rank (invertible) matrix $B$, i..e., $A' = BA$. By using determinants $\det (A') = \det(BA) = \det(B)\cdot \det(A)$. Since $A$ has full rank by hypothesis and $B$ has full rank as well, $\det(B)\cdot \det(A)\ne 0$ and so $A'$ has full rank as well.