1. ## Euclidian norm

We have the Euclidean norm (or $l_2-norm$) for a vector on the vector space $R^m$ ; $\parallel{v}\parallel_2 = \sqrt{\sum_{j=1}^m\mid{v_j}\mid^2}$. Given this vector norm on $R^m$ the induced matrix norm is; $\parallel{A}\parallel_2 = \sqrt{max. eigenvalue\ of\ A^T{A}}$. Let $\\ U\ \in\ R^{m\times{m}}$ be orthogonal ( $U^T{U}=I).$ $\\$ I want to show that 1) $\parallel{Uv}\parallel{{_2^2}}=\parallel{v}\parall el{_2^2}$ for all $v\in\ R^m$ and 2) that $\parallel{UA}\parallel_2 = \parallel{AU}\parallel_2$ for any $A \in\ R^{m\times{m}}$

2. Originally Posted by bram kierkels
We have the Euclidean norm (or $l_2-norm$) for a vector on the vector space $R^m$ ; $\parallel{v}\parallel_2 = \sqrt{\sum_{j=1}^m\mid{v_j}\mid^2}$. Given this vector norm on $R^m$ the induced matrix norm is; $\parallel{A}\parallel_2 = \sqrt{max. eigenvalue\ of\ A^T{A}}$. Let $\\ U\ \in\ R^{m\times{m}}$ be orthogonal ( $U^T{U}=I).$ $\\$ I want to show that 1) $\parallel{Uv}\parallel{{_2^2}}=\parallel{v}\parall el{_2^2}$ for all $v\in\ R^m$ and 2) that $\parallel{UA}\parallel_2 = \parallel{AU}\parallel_2$ for any $A \in\ R^{m\times{m}}$
the second part of your problem is trivial because $(UA)^TUA=A^TU^TUA=A^TA$ and $(AU)^TAU=U^TA^TAU=U^{-1}A^TAU$ and we know that similar matrices have the same eigenvalues.

for the first part, suppose $e_j, \ 1 \leq j \leq m,$ is the $j$-th column of $U.$ since $U^{-1}=U^T,$ we have $e_i \cdot e_j = \delta_{ij},$ where $\delta_{ij}$ is the Kronecker's delta. now suppose $v=\begin{bmatrix}v_1 & v_2 & . & . & . & v_m \end{bmatrix}^T.$ then:

$||Uv||_2^2=Uv \cdot Uv=\left(\sum_{i=1}^m v_ie_i \right) \cdot \left(\sum_{j=1}^m v_je_j \right)=\sum_{1\leq i,j \leq m}v_iv_j \delta_{ij}=\sum_{j=1}^mv_j^2=||v||_2^2.$

3. Originally Posted by NonCommAlg
the second part of your problem is trivial because $(UA)^TUA=A^TU^TUA=A^TA$ and $(AU)^TAU=U^TA^TAU=U^{-1}A^TAU$ and we know that similar matrices have the same eigenvalues.

for the first part, suppose $e_j, \ 1 \leq j \leq m,$ is the $j$-th column of $U.$ since $U^{-1}=U^T,$ we have $e_i \cdot e_j = \delta_{ij},$ where $\delta_{ij}$ is the Kronecker's delta. now suppose $v=\begin{bmatrix}v_1 & v_2 & . & . & . & v_m \end{bmatrix}^T.$ then:

$||Uv||_2^2=Uv \cdot Uv=\left(\sum_{i=1}^m v_ie_i \right) \cdot \left(\sum_{j=1}^m v_je_j \right)=\sum_{1\leq i,j \leq m}v_iv_j \delta_{ij}=\sum_{j=1}^mv_j^2=||v||_2^2.$
.

Thanks a lot,
Given a vector norm $||\cdot||$ on $R^m$ the induced matrix norm for m x m matrices A is defined by
$||A||\ =\ max_{v\neq{0}}\frac{||Av||}{||v||}$
That is , $||A||$ is the smallest number $\alpha$ such that $||Av||\leq\alpha||v||\ \forall\ v \in R^m$
So given the Euclidian-norm for a vector as written in the first question why is the induced matrix norm $||A||_2=\sqrt{max.\ eigenvalue A^TA}$
In my book it is given but I donīt see it directly, i think it has something to do with the fact that $A^TA$ is symmetric.