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$X \in \mathbb{R}^{N \times n}$ 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?

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  • $\begingroup$ will be true if $K$ is positive definite $\endgroup$ – lion Dec 6 '17 at 22:30
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No. Consider $X=[1\quad 0]^T,\,K=\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$.

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