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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!

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2 Answers 2

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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 $x = 0$

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  • $\begingroup$ I'm not sure what $\langle x | x\rangle$ means. Also what do you mean by $x^*$? $\endgroup$
    – P. Gillich
    Commented Mar 22, 2018 at 0:20
  • $\begingroup$ @P.Gillich I added more details, hopefully it makes more sense now $\endgroup$
    – caverac
    Commented Mar 22, 2018 at 0:23
  • $\begingroup$ Ah, alright, this seems clearer now. $\endgroup$
    – P. Gillich
    Commented Mar 22, 2018 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
    Commented Mar 22, 2018 at 0:33
  • $\begingroup$ @P.Gillich correct! Also, the key word here is Hermitian $\endgroup$
    – caverac
    Commented Mar 22, 2018 at 0:35
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Express $M=NN^*,$

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

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

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  • $\begingroup$ How do we know that there exists such an N? Maybe this is a dumb question. $\endgroup$
    – P. Gillich
    Commented Mar 22, 2018 at 0:25
  • $\begingroup$ check out the properties of positive-definite matrix. In particular, the cholesky factorization exists. $\endgroup$ Commented Mar 22, 2018 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
    Commented Mar 22, 2018 at 0:56

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