Matrix Equation? Maybe showing $AB=I_n$ works.

Question

Let $A, B$ be two $n*n$ matrices with the property that $ABX = X$, $\forall X$ a matrix $n*1$. Prove that $A$ and $B$ are invertible matrices.

Answer 1

The linear map $AB$ is determined on the basis $(e_1,\ldots ,e_n)$ of the vector space. By assumption $ABe_i=e_i$ so that $AB=I_n$. It follows that $\det(A)\det(B)=\det(AB)=\det(I_n)=1$, so that both $\det(A)$ and $\det(B)$ are nonzero. Hence $A$ and $B$ are invertible.

Answer 2

Hint: for $i=1,\dots,n$, let $X_i$ be a column vector. We then have $$ M\pmatrix{X_1 &X_2 & \cdots & X_n}= \pmatrix{M X_1& MX_2&\cdots & M X_n} $$ Now, show that $ABI_n= I_n$.

Answer 3

First, show that $B$ is invertible. If not, then there exists a non-zero $x$ such that $Bx=0$. But then $A(Bx)=A\cdot 0=0\neq x$... contrary to your assumption. Thus, $B$ is invertible.

Now, show that $A$ is invertible. If not, there exists a non-zero vector $z$ so that $Az=0$. Since $B$ is invertible, you can find $x$ so that $Bx=z$. Since $x$ is not zero, $Bx\neq 0$ (since $B$ has full rank). But, $A(Bx) = Az = 0$ and $x\neq 0$. Thus $A$ is also invertible.