3
$\begingroup$

my question is this:

I have to prove that for a positive-definite hermitian matrix $M$, i.e. such that $M=(m_{i,j})=(\overline{m_{j,i}})=\overline{M}^T$ of size $n$ x $n$, over the complex numbers, the following operation $$\langle x, y\rangle=\bf x^TM\bar{\bf y}$$ defines an inner product on the vector space. I have shown conjugate symmetry, and linearity in the first argument, and the fact that $\langle x, x\rangle\gt0$ for $x \neq 0$, as well as $x=0 \implies \langle x,x\rangle=0 $. I just need to show that if $$\langle x, x \rangle=0 \implies \bf {x}=0$$

I tried to reason it out simply by writing out $0=\langle x,x\rangle$ and expanding this according to the rule given above, but I don't see why it is that all the entries in the vector $\bf x$ have to be zero. I do know that the matrix must be nonzero, but beyond this I don't really know what to do.

Thanks for your help!

$\endgroup$
0
$\begingroup$

If $M$ is a positive-definite Hermitian matrix then

$$ x^* M x > 0 $$

for all non-zero complex vectors $x$ and $x^*$ means the conjugate-transpose of $x$. That means that $\langle x , x \rangle = 0$, iff $x = 0$

$\endgroup$
  • $\begingroup$ I'm not sure what $\langle x | x\rangle$ means. Also what do you mean by $x^*$? $\endgroup$ – P. Gillich Mar 22 '18 at 0:20
  • $\begingroup$ @P.Gillich I added more details, hopefully it makes more sense now $\endgroup$ – caverac Mar 22 '18 at 0:23
  • $\begingroup$ Ah, alright, this seems clearer now. $\endgroup$ – P. Gillich Mar 22 '18 at 0:25
  • $\begingroup$ So are you saying you get what I'm trying to prove for free simply by definition of positive definite matrix? My textbook defines a matrix as positive definite if $(x_1, x_2, ..., x_n) M(\overline{x_1},\overline{x_2}, ..., \overline{x_n})^T \gt 0 $ for all $\bf x$ not zero $\endgroup$ – P. Gillich Mar 22 '18 at 0:33
  • $\begingroup$ @P.Gillich correct! Also, the key word here is Hermitian $\endgroup$ – caverac Mar 22 '18 at 0:35
0
$\begingroup$

Express $M=NN^*,$

$$\langle x, x \rangle = x^TNN^*\bar{x}=\|N^*x\|^2=0$$

Hence $N^*x=0$ which implies that $x=0$.

$\endgroup$
  • $\begingroup$ How do we know that there exists such an N? Maybe this is a dumb question. $\endgroup$ – P. Gillich Mar 22 '18 at 0:25
  • $\begingroup$ check out the properties of positive-definite matrix. In particular, the cholesky factorization exists. $\endgroup$ – Siong Thye Goh Mar 22 '18 at 0:29
  • $\begingroup$ Thank you, but the result you use I think is beyond the theory we're supposed to use to solve the problem. I do appreciate your help though! $\endgroup$ – P. Gillich Mar 22 '18 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.