# Eigenvalues of $A^TA$

Suppose $A$ be a $n\times n$ matrix in $M(\mathbb{R})$. I'd like to know the eigenvalues of $A^TA$. I believe it's false to assume $\lambda^2$ be eigenvalue of $A^TA$, given $\lambda$ eigenvalue of $A$. Can somebody find a counterexample of it?

Also I guess following claim to be true regarding the eigenvalues of $A^TA$:

If $A$ is normal, then $\lambda^2$ is the eigenvalue of $A^TA$, given $\lambda$ eigenvalue of $A$.

Can somebody prove above claim?

• Does $A$ have real (as opposed to complex) entries? – Omnomnomnom May 10 '15 at 0:53
• Let $A = \begin{pmatrix}0 & 1 \\ -1& 0\end{pmatrix}$. It is normal, with eigenvalues $\pm i$, so squares are $-1$, when $A^TA$ has eigenvalues $1$ – uranix May 10 '15 at 0:58
• $A$ is real matrix. – Math Wizard May 10 '15 at 0:59
• It is, real matrices can have complex eigenvalues – uranix May 10 '15 at 1:00

Much of this is made much easier to see if you know about the spectral theorem. My answers below rely heavily on this.

For your first statement, any non-normal matrix provides a counterexample. In fact, we have the following theorem:

Let $A$ be a square matrix with eigenvalues $\lambda_k$. Let $\sigma_1,\dots,\sigma_n$ denote the eigenvalues of $A^TA$ (which are all positive). Then $$\sum_{k=1}^n |\lambda_k|^2 \leq \sum_{k=1}^n \sigma_k$$ and $A$ is normal if and only if $\sigma_k = |\lambda_k|^2$ for each $k$.

The proof of your second statement (by the spectral theorem) is as follows:

Because $A$ is normal, there exists a unitary matrix $U$ and diagonal matrix $D$ (each with complex entries) such that $A = UDU^*$ where $M^* = \overline{M^T}$ denotes the conjugate-transpose, AKA the adjoint of a complex matrix. Note that $$D = \pmatrix{\lambda_1\\&\ddots \\&& \lambda_n}$$ where $\lambda_k$ are the eigenvalues of $A$. We then have $$A^TA = A^*A = (UDU^*)^*UDU^* = UD^*DU^* = U \pmatrix{|\lambda_1|^2\\ & \ddots \\ && |\lambda_n|^2}U^*$$ Thus, the eigenvalues of $A^TA$ are $|\lambda_k|^2$.

• I thought of your proof, but how to explain if $A$ is the skew-symmetrical matrix as another post in this question suggests that eigenvalue of is $-\lambda^2$? – Math Wizard May 10 '15 at 2:34
• You need to square the absolute value. That what those "$|\cdot|$"s are there for. – Omnomnomnom May 10 '15 at 2:40
• Note that $U^*U=I$ doesn't imply $U^TU=I$. – Omnomnomnom May 10 '15 at 2:42
• To your edit: $A$ is real, so $A^T = \overline{A^T} = A^*$. – Omnomnomnom May 10 '15 at 3:22
• Oh, I understand that if $A$ is the skew-symmetrical matrix, its eigenvalue is $i\lambda$. – Math Wizard May 10 '15 at 3:48

Counterexample:

\begin{align*} \operatorname{eig} \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} &= \{0,1\} \\ \operatorname{eig} \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} &= \{0,2\} \\ \end{align*}

• That is not a normal matrix – uranix May 10 '15 at 1:01
• @uranix As best I can tell there are two parts to the question. First, is it true that if $\lambda$ is an eigenvalue of $A$ then $\lambda^2$ is an eigenvalue of $A^T A$? Second, is the previous statement true if we assume that $A$ is normal? Arkamis' answer addresses the first question. Hence my +1. – Ian May 10 '15 at 1:02
• It is ok to give a example of non normal matrix. – Math Wizard May 10 '15 at 1:03

A real normal matrix is the matrix that satisfies $AA^T = A^T A$. That's what wiki says on normal matrices

Among complex matrices, all unitary, Hermitian, and skew-Hermitian matrices are normal.

Skew-Hermitan matrices are promising for counterexample, since their eigenvalues are purely imaginary. Real skew-Hermitan matrix is just a skew-symmetrical one.

Let $A$ be the skew-symmetrical matrix. Consider eigenvector $x$ of $A$. He have $$A^TA x = -A^2 x = -\lambda^2 x$$