# Does this inequality hold $\operatorname{Trace}(A^TA) \ge \rho(A)$?

Suppose $$A \in M_n(\mathbb R)$$ is an arbitrary square matrix and $$\rho(A)$$ is the spectral radius of $$A$$. Does this inequality hold: $$\text{Trace}(A^{\top}A ) \ge \rho(A)?$$

• If you scale $A$ by a factor $c>0$, the LHS is quadratic in $c$ while the RHS is linear in $c$. Therefore the inequality doesn't hold if the LHS is nonzero and $c$ is small. – user1551 Apr 23 at 8:11

As @user1551 explained, a suitable counterexample could be $$A = \begin{pmatrix} c & 0 \\ c & 0 \end{pmatrix}$$ for some $$c \in (0,\frac{1}{2})$$. Then, we have $$\text{Tr}(A^T A) = \text{Tr} \begin{pmatrix} 2 c^2 & 0 \\ 0 & 0 \end{pmatrix} = 2c^2 < c = \rho(A).$$