Let $A,B$ be $n\times n$ matrices in $\mathbb{R}^n$ where $AB=I$. Prove that rank$B=n$.
I have reviewed the Systems Rank Theorem. I am thinking I have to prove that B in linearly independent which would result in rank$B=n$ but I am not sure how to do that.
Consider $A=\pmatrix{\mathbf{a_1} \\ \mathbf{a_2} \\ \vdots \\ \mathbf{a_n}}$ and $B=\pmatrix{\mathbf{b_1} & \mathbf{b_2} & \cdots \mathbf{b_n}}$.
That is, lets look at the rows of $A$ and columns of $B$.
Since $AB=I$, $a_1 \cdot b_j = 0$ when $1\ne j$ and $a_1 \cdot b_1 = 1$. So $a_1$ is orthogonal to each row of $B$, other than the first row. We thus see that $b_1$ cannot be written as a linear combination of the other columns. Similarly for every column of $B$. Thus the columns of $B$ are linearly independent. This is a definition of $\textrm{rank }B =n$, the number of linearly independent columns.