How to show that $A′A$ is a positive definite matrix given rank$(A) = k$?

Question

I have difficulty in trying to prove the following question.Let $A = n \times k, k ≤ n$. Show that, if rank$(A) = k$, then $A′A$ is a positive definite matrix ($A'$ denotes the transpose of $A$). Do you have any idea how it can be solved? Thanks a lot.

Answer 1

The linear mapping $T : \mathbb{R}^{k} \to \mathbb{R}^{n}$ with $k \le n$ induced by the matrix $A$ is injective. This is because the dimension of the image of $T$ equals $k$ (because rank $A = k$) so the dimension of the kernel of $T$ equals $0$. So for all $x \in \mathbb{R}^{k}$ with $x \ne 0$ we have $Ax \ne 0$. This implies

$x^{T}(A^{T}A)x = (Ax)^{T}(Ax) = \|Ax\|^{2} > 0$

therefore the quadratic form $A^{T}A$ is positive definite by definition.

Answer 2

• Show that $A'A$ is positive semidefinite, i.e. its eigenvalues are nonnegative. (Hint: if $v$ is a $\lambda$-eigenvector of $A'A$, then $\lambda \|v\|^2 = v^\top A^\top A v = \|Av\|^2 \ge 0$.)
• Thus it suffices to show that the eigenvalues of $A'A$ are nonzero, i.e. that $A'A$ is invertible. It may help to note that $A'A$ is a $k \times k$ matrix with the same rank as $A$.