$X \in \mathbb{R}^{N \times n}$ and $X$ is full rank, meaning that rank($X$) = n. $K \in \mathbb{R}^{N \times N}$ and invertible. In addition, $n < N$.

Is this enough to prove that $X^{\top} K X$ is invertible?

Here's what I'm thinking..............

Let $col(\cdot)$ and $nul(\cdot)$ denote the column space and null space of some matrix

We know that $X$ is full rank, this means that $nul(X) = \{0\}$. If we can prove that $col(X^{\top} K X) = col(X^{\top})$, then we can conclude that for $X^{\top} K X y = 0$ only if $y = 0$, thus $X^{\top} K X$ has linearly independent columns and also a square matrix, hence invertible.

I was able to prove $col(X^{\top} K X) \subseteq col(X^{\top})$, but how do I prove the equality or is $col(X^{\top} K X) = col(X^{\top})$? Or is there another way to go about this? Or $X^{\top} K X$ is not actually invertible?

  • $\begingroup$ will be true if $K$ is positive definite $\endgroup$ – lion Dec 6 '17 at 22:30

No. Consider $X=[1\quad 0]^T,\,K=\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.