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Is it true that the kernel and range of a complex $n\times n$ Hermitian matrix are always independent? If yes, can we prove it without using inner product? Thanks!

velut luna
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  • First you say you're considering a "general vector space which may not having an inner product", then you contradict that by specifying $\mathbb C^n$ in particular. Which is right? – hmakholm left over Monica Feb 22 '15 at 07:54
  • See, perhaps, http://math.stackexchange.com/a/467256/27978. – copper.hat Feb 22 '15 at 08:02
  • @Henning Makholm. Since I am talking about Hermitian matrix, I think I'd better assume the field is $\mathbb{C}$. But I don't want to equip it with an inner product. – velut luna Feb 22 '15 at 11:08
  • @Kyson: I have trouble seeing how you can have a Hermitian matrix without having chosen a (finite) basis for your vector space first. That's going to give you an inner product whether you want it or not. It's not clear to me how to even define what it would mean for a proof to pretend it doesn't exist. – hmakholm left over Monica Feb 22 '15 at 14:24
  • @Henning Makholm. I don't think what I am asking is basis dependent. Just choose any basis you like. I also don't see why choosing a basis has anything to do with inner product. – velut luna Feb 22 '15 at 14:34
  • @Kyson: You can't even write down a matrix without having a basis for the vector space. And once you have a basis, that derives a standard inner product. – hmakholm left over Monica Feb 22 '15 at 14:39
  • I don't agree that whenever we say something about a matrix, we need to specify a basis. Also proving independence does not require inner product. If what you mean is that the problem is due to "Hermitian" (which is basis dependent and requires an inner product?), I am not sure. – velut luna Feb 22 '15 at 14:57
  • What do you mean by "kernel and range are independent"? Have zero intersection? – TZakrevskiy Feb 22 '15 at 15:22
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    for a linear transformation, you dont need to choose a basis. but for a matrix representation of it, you will need two sets of basis: one in the domain and one in the range space. – abel Feb 22 '15 at 15:24
  • @TZakrevskiy. Yes, zero intersection. – velut luna Feb 22 '15 at 15:27
  • My question comes from the the proof of the Jordan form in wikipedia http://en.wikipedia.org/wiki/Jordan_normal_form, in which the author said "$\text{Ran}(A-\lambda I)\cap \text{Ker}(A-\lambda I) = {0}$,... this would be the case, for example, if $A$ was Hermitian". It's just an aside though. – velut luna Feb 22 '15 at 15:40

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Suppose a vector $x\in\mathbb{C}^n$ is in both the kernel and range of a Hermitian matrix $A$. Then $Ax=0$, and we may write $x=Ay$ for some vector $y$. Then $$\|x\|^2= x^*x = (Ay)^*x = y^*A^*x = y^*Ax = y^*0 = 0$$ Since $x$ has zero length, we must have $x=0$. This proves that the kernel and range of a Hermitian matrix $A$ have trivial intersection.

Brent Kerby
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