Let $\newcommand{\<}{\langle}$ $\newcommand{\>}{\rangle}$ $T \in L(V)$ be a linear operator on an n-dimensional real inner-product space of $(V,<.,.>)$ whose singular value decomposition is given by two orthonormal bases $(u_{1},u_{2},...,u_{n})$, $(v_{1},v_{2},...,v_{n})$ of $V$ and singulr values $\sigma_{1,},\sigma_{2},...,\sigma_{n} \geq 0$ such that $Tx=\sum_{j=1}^{n}\sigma_{j}<x,v_{j}>u_{j}$. Prove that for any $n <m$ we have $\| Tx - \sum_{j=1}^{m}\sigma_{j}<x,v_{j}>u_{j}\| \leq \sigma_{m+1}\|x\|$.
My solution
$$\| Tx - \sum_{j=1}^{m}\sigma_{j}<x,v_{j}>u_{j}\|=\\ \|\sum_{j=m+1}^{n}\sigma_{j}<x,v_{j}>u_{j}\|=\\ \|\sigma_{m+1}<x,v_{m+1}>u_{m+1}+...+ \sigma_{n}<x,v_{n}>u_{n}\|=\\ \|\sigma_{m+1}<x,v_{m+1}>u_{m+1}\|+....+\|\sigma_{n}<x,v_{n}>u_{n}\|=\\ |\sigma_{m+1}<x,v_{m+1}>|\|u_{m+1}\|+...+|\sigma_{n}<x,v_{n}>|\|u_{n}\|=\\ |\sigma_{m+1}<x,v_{m+1}>|+...+|\sigma_{n}<x,v_{n}>|=\\ \sigma_{m+1}\|x\|\|v_{m+1}\|+...+\sigma_{n}\|x\|\|v_{n}\|=\\ \sigma_{m+1}\|x\|+...+\sigma_{n}\|x\| $$ can anyone help?
A reasonable attempt, but this wasn't quite the correct trick. $$ \begin{align} \left\| Tx - \sum_{j=1}^m \sigma_j \<x,v_j\>u_j \right\|^2 &= \left\| \sum_{j=m+1}^n \sigma_j \<x,v_j\>u_j \right\|^2 = \sum_{j=m+1}^n \sigma_j^2 |\<x,v_j\>|^2 \\ & \leq \sum_{j=m+1}^n \sigma_{m+1}^2 |\<x,v_j\>|^2 = \sigma_{m+1}^2 \sum_{j=m+1}^n |\<x,v_j\>|^2 \\ & \leq \sigma_{m+1}^2 \sum_{j=1}^n |\<x,v_j\>|^2 = \sigma_{m+1}^2 \|x\|^2 \end{align} $$ The conclusion follows.