# How to prove that det($A^{T}A$) is nonnegative?

Why is the determinant of the product of a matrix and its transpose nonnegative?

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What properties of the determinant do you know? – EuYu Mar 5 '13 at 3:56
Is $A$ real? Is it square? – 1015 Mar 5 '13 at 4:03
Bella: You need to clarify if $A$ is a square $n\times n$ matrix, for some $n$ – amWhy Mar 5 '13 at 4:11
As you can see, I removed that portion of my comment. We still need Bella to respond and clarify. Thanks! You're correct, another approach may be needed. But I'm just sensing from the level of question that the question concerns square matrices with real entries. – amWhy Mar 5 '13 at 4:15
Given the answer of Alex Jordan, there is no more need now. – awllower Mar 5 '13 at 4:18

I assume that $A$ is real.

$A^TA$ is symmetric, so it is (orthogonally) diagonalizable. So its determinant is the product of its eigenvalues. Let $\lambda$ be an eigenvalue for $\vec{v}$, where $\vec{v}$ is an eigenvector of $A^TA$.

Using inner product notation: $$0\le\langle A\vec{v},A\vec{v}\rangle=(A\vec{v})^T(A\vec{v})=\vec{v}^TA^TA\vec{v}=\vec{v}^T\lambda\vec{v}=\lambda\langle\vec{v},\vec{v}\rangle$$

This implies that $\lambda$ is nonnegative, since $\langle\vec{v},\vec{v}\rangle>0$. So the determinant is a product of nonnegative real numbers, and therefore a nonnegative real number. Note that this shows something much more specific about $A^TA$ than merely having positive determinant.

(If you prefer dot product notation: $$0\le( A\vec{v})\cdot(A\vec{v})=(A\vec{v})^T(A\vec{v})=\vec{v}^TA^TA\vec{v}=\vec{v}^T\lambda\vec{v}=\lambda\vec{v}\cdot\vec{v}$$)

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+1 btw $\quad \ddot\smile$ – amWhy Mar 5 '13 at 4:21
(+1) nice answer. – Mhenni Benghorbal Mar 5 '13 at 4:31

Assuming $A$ is square, (hence $\det (A)$ is defined):

Recall $$\det(A) = \det(A^T)$$ $$\det(A^TA) = \det(A^T)\det(A)$$

What does this imply about $\det(A^TA)$ if

• If $\det A > 0?$
• If $\det A < 0$?
• If $\det A = 0$?
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What if A is not square? – bonsoon Mar 5 '13 at 4:03
It's not clear from OP's question that A is square; $\det(A)$ might not be defined. – alex.jordan Mar 5 '13 at 4:05
$A^TA$ is always square regardless of what $A$ is, and always can have its determinant taken. – bonsoon Mar 5 '13 at 4:11
You're all correct. I tried to state my assumptions clearly, with no clarification from the OP. I think this is a good question for two levels at which the OP may be asking it: first encounter with determinants of square matrices/properties of determinants, or at a place where learning about eigenvalues... – amWhy Mar 5 '13 at 4:19