# invertible subspaces and SVD

Suppose $X$ is a $m\times d$ real matrix. I am trying to prove the following claim:

$XX^T$ is invertible iff $\text{span}\,\{x_1,x_2,...,x_m\}=\mathbb{R}^d$ such that $x_1, x_2, ..., x_m$ are X's columns.

I've been suggested to use SVD, but i dont undertand the realation between $rank(X)$ and $rank(XX^T)$.

They are equal. It is obvious that the image of $XX^T$ is included in the image of $X$. For the converse, you have to note that $\ker X=(\text{ran}\,X^T)^\perp$.