Let $x, y$ be vectors in $\mathbb{R}^n$. Can it be shown that if the inner product satisfies $\langle x, y\rangle > 0$, then $\langle Ax, y\rangle > 0$ whenever $A$ is positive-definite?
Weighted inner product
0
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linear-algebra
inequality
inner-product-space
2 Answers
1
In fact, every inner product can be written in the form $$ \langle x,y \rangle = y^TMx $$ for some positive definite matrix $M$. It follows that for any inner product (and its associated $M$), we have $$ \langle Ax, y \rangle = y^TMAx $$ Your question can therefore be rephrased as follows: if $M$ and $A$ are positive definite, then does $y^TMx > 0 \implies y^TM(Ax) > 0$?
The answer to this is no. For example, take: $$ M = I, \quad y = (1,-1), \quad x = (2,1), \quad A = \pmatrix{1&0\\0&3} $$
2
No, this won't work. Take $x = \binom 2 {-1}$ and $y = \binom 1 {1}$ in $\mathbb R^2$ and $$A = \left(\begin{matrix} 1 & 0 \\ 0 & 10 \end{matrix} \right).$$ Certainly $A$ is positive definite and $\langle x , y \rangle = 1 > 0$ but $\langle Ax, y \rangle = -8 < 0.$ Higher dimensional counterexamples are easy to construct as well.