# Is $XX^T$ invertible?

In one of the lectures today, the professor said that if $X \in \mathbb{R}^{m \times n}$ matrix, and the columns of $X$ span $\mathbb{R}^m$, then the matrix $XX^T$ is invertible. I am not sure why this is the case unless $m=n$?

If $XX^\top v = 0$, then $\|X^\top v\|^2=v^\top XX^\top v=0$, and thus $X^\top v = 0$. Since the rows of $X^\top$ span $\mathbb{R}^m$, we must have $v = 0$, and thus the kernel of $XX^\top$ is $\{0\}$.

• Can I use the same method on $X^TX$?. If $X^TXv=0$ then $||Xv||^2=v^TX^TXv = 0$ and hence $Xv=0$ but since columns of $X$ span then $v=0$ and so the kernel of $X^TX$ is $\{0\}$?? Jan 21, 2018 at 21:52
• I’m not adept in linear algebra, but it fascinates me, so I follow the tag. Is $\left\| X^{\sf T}\right\|$ the magnitude of the determinant of the transpose of matrix $X$? Jan 21, 2018 at 21:53
• @ChaseRyanTaylor $X^T$ is the transpose of $X$, and $||X^Tv||$ is a norm of the vector $X^Tv$ (without other indication, it's usually the euclidian norm, that is $||u||^2=\sum_i u_i^2$). Jan 21, 2018 at 21:54
• @AspiringMat The error in your comment is that "columns of $X$ are spanning" does not allow you to deduce $v = 0$ from $Xv = 0$. Jan 21, 2018 at 21:56
• @angryavian Isn't $Xv$ a linear combination of the columns on $X$? But these columns are linearly independent, so the only way they can add to 0 is if $v$ is 0? I am a bit confused? Jan 21, 2018 at 21:59

The rows of $X^T$ span, so $X^Tv=0$ just when $v=0$, so $XX^Tv=0$ just when $v=0$, so $XX^T$ has kernel $0$, and is square, so is invertible.

• I see. So if $X$ is full rank, then we always have that $XX^T$ and $X^TX$ are both invertible using the same logic you did here, correct? Jan 21, 2018 at 21:39
• No, not both. Try a simple example, like $X=(1 \ 0)$. Jan 21, 2018 at 21:41
• @AspiringMat The columns of $X^T$ may not span Jan 21, 2018 at 21:42
• @angryavian $XX^Tv=0$ implies $v^TXX^Tv=0$ implies $X^Tv=0$ Jan 21, 2018 at 21:43