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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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    Thanks everyone. From all the help, if I'm using determinant to proof, I could say $det(K)=det(A^{T})det(A)$. And since all the eigenvalues are positive, det(K)>0. Based on the fact that $det(A^{T})=det(A)$, I could say det(A)=\sqrt{det(K)}>0 and therefore, $A$ is invertible. Using the nullspace to prove is also interesting with the one Mark and Jonas had given below. Thanks everyone for the help.2011-08-06

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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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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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    @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 $Ax\neq 0$.2013-10-19
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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.