# $A^{T}A$ positive definite then A is invertible?

Say if $A$ is an $n \times n$ matrix, why is it that if $A^{T}A$ is positive definite, the matrix $A$ is then invertible? All I know is $A^{T}A$ gives a symmetric matrix but what does $A^{T}A$ is positive definite tell or imply or hint about the matrix $A$ itself that leads to the fact that it will be invertible?

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Positive definite implies positive eigenvalues. – Qiaochu Yuan Aug 6 '11 at 16:48
determinants is product of eigenvalues – user9413 Aug 6 '11 at 16:49
But if $K=A^{T}A$ is a positive definite, that is only the matrix $K$ that has positive eigenvalues and does not imply that $A$ will also have positive eigenvalues, am I right? Wouldn't some singular matrices give a positive eigenvalues too? – xenon Aug 6 '11 at 16:52
$A^tA$ will be diagonalizable with positive entries which you can scale to get something like $SPA^tAP^{-1}=I$ to get $P^{-1}SPA^tA=I$ an explicit inverse for $A$ – yoyo Aug 6 '11 at 16:55
${\rm det}(A^{T}A) = {\rm det}(A)^{2}$ when $A$ is square. – Geoff Robinson Aug 6 '11 at 17:03

Actually $\ker(A^T A) = ker(A)$ so one is invertible if and only if the other is invertible. Proof:

If $v \in \ker(A)$ then $Av=0$ so $A^T A v = 0$, hence $v \in \ker(A^T A)$. On the other hand, if $v \in \ker (A^T A)$ then $\left\langle w,A^{T}Av\right\rangle =0$ for all $w \in \mathbb{R}^n$, so $\left\langle Aw,Av\right\rangle =0$ for all $w$, and in particular for $w=v$ you get $Av=0$.

A positive definite matrix is invertible (for instance, because it has positive eigenvalues) so you're done.

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$A^TA$ is always non-negative and I think you don't have any problem with it. (otherwise see that $x^TA^TAx$ is nothing but $\|Ax\|^2_2 \geq 0$)

Suppose A is square and not invertible hence rank deficient. Due to the homework tag, I will just add that when two matrices multiplied they cannot form a matrix that has a higher rank than the individual matrices, for example if you multiply a vector $x x^T$ the resulting matrix cannot be of rank 2.

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Hint: Show that $A$ is injective as a linear transformation on $\mathbb R^n$. This implies invertibility by the rank-nullity theorem.

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But how to prove $A$ is injective? – loganecolss Oct 18 '13 at 9:40
@logancolss: A linear transformation $A$ is injective if and only if for all $x\neq 0$, $Ax\neq 0$. If $A^TA$ is positive definite, then for all $x\neq 0$, $x^TA^TAx>0$, which implies $Ax\neq 0$. – Jonas Meyer Oct 19 '13 at 23:41