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Given a matrix $A$ with dimensions $m \times n$, is $B=AA^T$ invertible if and only if the rows of $A$ are linearly independent?

So far, I've tried writing A as row vectors,

$$A = \begin{bmatrix}v_1\\ v_2\\ \vdots \\ v_m\end{bmatrix}$$

where $B_{i,j} = (v_i v_j) $

The Wikipedia page on the Gram Determinant suggests that the condition I stated above is necessary and sufficient, but does not provide a proof.

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  • $\begingroup$ If the columns/rows of $A$ are orthnormal, that should be enough. $\endgroup$ – IAmNoOne Feb 17 '15 at 2:26
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Let $D=A'$ so that $B=D'D$. The necessary and sufficient condition for the invertibility of $B$ is that $D$ has full column rank (i.e. $A$ has full row rank).

Necessary: suppose that $D'D$ is invertible and suppose that there exists some nonzero $x$ such that $Dx=0$. But then $Dx=0$ implies $D'Dx=0$ violating the nonsingularity of $D'D$.

Sufficient: because $D'D$ is square, it's enough to show that it has full column rank. Suppose not: there is some nonzero $x$ such that $D'Dx=0$. But this implies $0=x'D'Dx=|Dx|^2$, that is $Dx=0$, violating $D$ having full column rank.

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    $\begingroup$ Note that this answer assumes $A$ is a real matrix. The condition is not sufficient for complex matrices (eg $A=\begin{bmatrix}1&i\end{bmatrix}$) or over finite fields. $\endgroup$ – stewbasic Aug 14 '17 at 4:54

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