# l_2 norm matrix eigenvalue equality

Hello how to show the following:

$||A||_2$ = $\sqrt{ \text{largest eigenvalue of } A^{T}A}$

for any $m\times n$ matrix $A$.

Thank you

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Hello, how to ask questions? Thank you. – Did Jan 24 '13 at 8:55

Let $A^TA = U\Lambda U^T$ where $U$ is an orthogonal matrix and $\Lambda$ is a diagonal matrix of eigenvalues of $A^TA$. Then \begin{align*} \|A\|_2^2 & = \sup_{x \ne 0} \frac{\|Ax\|_2^2}{\|x\|_2^2} = \sup_{x \ne 0} \frac{x^TA^TAx}{x^Tx} = \sup_{x \ne 0} \frac{x^TU\Lambda U^Tx}{x^TUU^Tx} = \sup_{y \ne 0} \frac{y^T\Lambda y}{y^Ty} = \text{largest eigenvalue of $A^TA$}. \end{align*}
By the change of variable $y = U^Tx \Leftrightarrow x = Uy$. – Tunococ Jan 24 '13 at 21:28
Because $\Lambda$ is diagonal. The optimal choice is to pick $y = e_i$, the $i$-th basis vector, when the $i$-th element of $\Lambda$ is the largest eigenvalue. – Tunococ Jan 25 '13 at 3:00
We can easily generalize Tunococ's proof considering $A\in \mathbb{C}^{m\times n}$ and $P\in \mathbb{C}^{n\times n}$ unitary matrix such that $P^hA^hAP=\mbox{diag }(\lambda_1,\ldots,\lambda_n)$ where $\lambda_i\geq 0$ are the eigenvalues of $A^hA$.