# An inequality involving spectral norm of a matrix

I have a very simple question that is confusing me at one point. Suppose that $$A$$ is a real, non-negative definite matrix (hence automatically square and symmetric). Then, is it true that: $$|y^\top A x| \leq \|A\|_{\textrm{sp}}|y^\top x|~?$$ Here, $$\|A\|_\textrm{sp}$$ denotes the spectral norm of $$A$$.

I tried along the following lines:

Let $$A = P^\top D P$$ be the spectral decomposition of $$A$$. Let $$X = Px$$ and $$Y=Py$$. Then, $$|y^\top A x| = |Y^\top D X| = \left|\sum_i D_{ii}X_iY_i\right|~.$$ I cannot proceed further from the last step, since the $$X_i$$s and $$Y_i$$s can be positive or negative (although the $$D_{ii}$$s are non-negative). Can anyone help me?

This isn't true. Given any two mutually orthogonal nonzero vectors $$x$$ and $$y$$ in $$\mathbb R^n$$, we can construct an inner product $$\langle\cdot,\cdot\rangle$$ such that $$\langle x,y\rangle\ne0$$. It follows that if $$A$$ is the matrix representation of this inner product with respect to the standard basis, then $$A$$ is positive definite and $$|y^\top Ax|=|\langle x,y\rangle|>0=\|A\||y^\top x|$$.

E.g. suppose that $$u=e_1=(1,0)$$ and $$v=e_1+e_2=(1,1)$$ form an orthonormal basis with respect to the new inner product. Then $$\begin{cases} \langle e_1,e_1\rangle=\langle u,u\rangle=1,\\ \langle e_1,e_2\rangle=\langle u,v-u\rangle=-\langle u,u\rangle=-1,\\ \langle e_2,e_2\rangle=\langle v-u,v-u\rangle=\langle v,v\rangle+\langle u,u\rangle=2. \end{cases}$$ Therefore $$A=\pmatrix{1&-1\\ -1&2}$$ is positive definite and $$|e_2^\top Ae_1|=|-1|>0=\|A\||e_2^\top e_1|$$.