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

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.

• Can you show that $A^{T}A$ is positive semidefinite and then show that it is nonsingular? Commented Dec 5, 2020 at 20:51
• No, because I can not understand the connection between the rank of the matrix and the definiteness of the matrix. Commented Dec 5, 2020 at 21:01

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.
• 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 $$A$$.