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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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    Counter-example in $\mathbb{Z}_p$ (see Ted's comment): \begin{bmatrix}1 & \cdots & 1\end{bmatrix} with $p$ columns.2012-10-27

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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.)2012-10-27