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Given $A$ and $B$, $n\times n$ complex matrices. If $\langle x,y\rangle =y^{*}x$ for all $x,y\in \mathbb C^{n}$, then the following are equivalent:

(1) $\langle Ax,y\rangle=\langle Bx,y\rangle$, for all $x,y\in \mathbb C^{n}$.

(2) $\langle Ax,x\rangle=\langle Bx,x\rangle$, for all $x,y\in \mathbb C^{n}$.

(1) implies (2) is easy, how to prove (2) implies (1)?

Jonas Meyer
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Casey
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2 Answers2

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Hint: Use the algebraic properties of the inner product to expand each side of the following equations:

  • $\langle A(x+y),x+y\rangle=\langle B(x+y),x+y\rangle$
  • $\langle A(x+iy),x+iy\rangle=\langle B(x+iy),x+iy\rangle$
Jonas Meyer
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    +1: Why is it though, apart from the algebra, if the problem were to state $A,B \in \mathbb{R}^{n \times n}$ and $x,y \in \mathbb{R}^{n \times 1}$, the statement turns out to be false? –  Aug 14 '11 at 04:45
  • Dear @Sivaram: Why is the statement false over $\mathbb R$? – Pierre-Yves Gaillard Aug 14 '11 at 04:52
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    @Sivaram: Good question, but I don't have a good answer. Perhaps a good example in the real case is $A=\begin{bmatrix}0&-1\1&0\end{bmatrix}$ and $B=\begin{bmatrix}0&1\-1&0\end{bmatrix}$. Going from the reals to complexes often yields more than you'd expect, and I'm reminded of analyticity of complex differentiable functions, and of existence of complex zeros of polynomials, and of nonemptiness of spectra in complex Banach algebras. – Jonas Meyer Aug 14 '11 at 04:58
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    @Pierre: For instance, as I jave mentioned in the earlier comment that if the problem were to state $A,B \in \mathbb{R}^{n \times n}$ and $x,y \in \mathbb{R}^{n \times 1}$, then letting $A = \begin{pmatrix} 1 & 2 \ 2 & 1\end{pmatrix}$ and $A = \begin{pmatrix} 1 & 3 \ 1 & 1\end{pmatrix}$, we will that $2$ doesn't imply $1$ –  Aug 14 '11 at 05:01
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    @Sivaram: In fact, all examples stem from the fact that over real inner product spaces there are nonzero linear transformations that map each vector to a perpendicular vector. To simplify my earlier example, take $A=\begin{bmatrix}0&-1\1&0\end{bmatrix}$ (a rotation by $\pi/2$) and $B=0$. In your example, one of the matrices is obtained from the other by adding a rotation by $\pi/2$. – Jonas Meyer Aug 14 '11 at 05:15
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    @Sivaram: Thanks for answering my silly question! I think that over a field whose characteristic is not 2, two square matrices define the same quadratic form iff they have the same symmetrization. What Jonas’s argument shows is that sesquilinear forms are determined by their restriction to the diagonal (even if they are not symmetric). Is this right, or am I making another confusion? – Pierre-Yves Gaillard Aug 14 '11 at 05:52
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    @Pierre-Yves: Well put. In the real case, condition (2) translates to $A+A^\mathrm{T}=B+B^\mathrm{T}$. – Jonas Meyer Aug 14 '11 at 05:56
  • @Jonas: Thanks a lot! So, sesquilinear forms are determined by their restriction to the diagonal (even if they are not symmetric). I was completely unaware of this fact. (I find this very interesting, and upvoted the question, your answer, and all the comments that are not mine.) – Pierre-Yves Gaillard Aug 14 '11 at 05:59
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    @Pierre-Yves: In case you're interested, there is the related concept of numerical radius that shows up sometimes in operator theory. – Jonas Meyer Aug 14 '11 at 06:05
  • I explained the qualitative difference between the real and complex cases in a separate answer. – tparker Jul 07 '18 at 05:47
  • @tparker - It seems to me the sesquilinear map $f:\mathbb C^2\times\mathbb C^2\to\mathbb C$ attached to your matrix satisfies $f((1,i),(1,i))=-2i$, and thus is nonzero on the diagonal. - If $S$ is a set, by "the diagonal of $S\times S$" I mean the set ${(s,s)\ |\ s\in S}$. – Pierre-Yves Gaillard Jul 07 '18 at 15:38
  • @Pierre-YvesGaillard Oh, I see what you mean, I thought you just meant the diagonal of the matrix $A$. Yes, the polarization identity determines an arbitrary sesquilinear form, but only a symmetric bilinear (i.e. quadratic) form, from the diagonal for a field of characteristic greater than 2, as discussed here. My answer below answers your first question in the comments. – tparker Jul 08 '18 at 02:04
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A very similar but slightly easier approach is to move everyone over to one side of the equations, use linearity to combine the operators, and consider the difference $C := A - B$. Then your first equation implies that $\forall x, y \in \mathbb{C}^n, \langle C x, y \rangle = 0$, which is clearly equivalent to the operator equation $C = 0$. The statement to be shown then becomes $$\forall x \in \mathbb{C}^n, \langle C x, x \rangle = 0 \iff C = 0.$$ You can then do the same trick that Jonas Meyer suggested and expand out $\langle C(x + y), x + y \rangle$ and $\langle C(x + i y), x + i y \rangle$, and the algebra is slightly simpler.

This simple trick also helps explain the qualitatively different behavior between the real and complex cases: it's just that the complex numbers are algebraically closed and the real numbers are not. Any nonzero matrix $C$'s characteristic polynomial is $n$-th order and nontrivial (by which I mean it isn't just $\lambda^n$, but has subleading terms as well). Any nontrivial polynomial has at least one nonzero complex root, so any nonzero complex matrix $C$ will have at least one nonzero eigenvalue $\lambda$. Since the numerical range of any matrix contains its eigenvalues, the numerical range of any nonzero complex matrix is nontrivial, i.e. is not simply $\{ 0 \}$. (Another way to see this is that the numerical radius of any complex matrix is at least one-half its operator norm, which is positive for a nonzero matrix.)

On the other hand, a nonzero real matrix has a nontrivial real characteristic polynomial that may not have any real roots, like the example $$\left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right]$$ given in the comments to Jonas Mayer's answer, so it may not have any eigenvectors. That's why its numerical radius can be $0$.

tparker
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