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Assuming $F$ is a matrix with full rank but more columns than rows - why is $F\cdot F^T$ invertible?

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What just happened?! – Lord_Farin Oct 27 '12 at 21:23
Invertible..."things"? You mean like a dog doing a rolling trick or a capsizing ship? Really... – DonAntonio Oct 27 '12 at 21:25
Counter-example in $\mathbb{Z}_p$ (see Ted's comment): $$\begin{bmatrix}1 & \cdots & 1\end{bmatrix}$$ with $p$ columns. – wj32 Oct 27 '12 at 21:59
up vote 5 down vote accepted

It suffices to show the null space of $F F^T$ is $\{0\}$.

Suppose $F F^T x = 0$. Then \begin{align*} & x^T F F^T x = 0 \\ \implies & (F^T x)^T (F^T x) = 0\\ \implies & \|F^T x \|^2 = 0 \\ \implies & F^T x = 0 \\ \implies & x = 0. \end{align*}

(Because $F^T$ is a skinny matrix with full rank, its null space is $\{0\}$.)

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Assuming the matrix is over the real numbers, of course. (Or other field where sum of nonzero squares cannot be zero.) – Ted Oct 27 '12 at 0:33

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