On finding the $2$-norm of a matrix

Question

When finding the 2-norm of a matrix, you are to take the square root of the largest eigenvalue found of the matrix $A^TA$. This is just the largest eigenvalue? I do not take the absolute values of the eigenvalues first and see which one has the largest magnitude?

Answer 1

As pointed out by Bungo in the comments, all eigenvalues of $A^TA$ are non-negative and real. So there is no need to take absolute values before you compare them.

The 2-norm of a linear transformation $A$ is (assuming finite dimensions) the maximal value of $\|Av\|$ among all unit vectors $v$ in the domain of $A$. We have $$ \|Av\| = \sqrt{v^TA^TAv} $$ The spectral theorem ($A^TA$ is a symmetric matrix) says that the domain of $A^TA$ (which is also the domain of $A$) has an orthonormal basis consisting of eigenvectors of $A^TA$. Thus if we decompose $v$ into this basis, we see that the largest possible value of $v^TA^TAv$ is exactly the largest eigenvalue of $A^TA$.