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How to show that if $A, B \in M_{n}(\mathbb{C})$, and $x,y\in \mathbb{C}^{n}$ then the following two conditions are equivalent,

1.) $\langle Ax,y\rangle =\langle Bx,y\rangle$

2.) $\langle Ax,x \rangle= \langle Bx,x \rangle$

for all $x,y$?

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1  
I don't understand the question. You mean $x,y\in\mathbb C^n$? But why should this hold for all $A,B,x,y$? –  Stefan Geschke Sep 24 '11 at 18:15
    
@Stefan, sorry about the confusion, I have edited the question. –  user12354 Sep 24 '11 at 18:20
    
1. => 2. should be trivial then, by taking $y = x$. –  TMM Sep 24 '11 at 18:30
    
@Thijs, yes, 1--> is trivial. But I don't know how to go from 2-->1 –  user12354 Sep 24 '11 at 18:31

2 Answers 2

up vote 2 down vote accepted

Both (1) and (2) are equivalent to $A=B$.

By taking differences, it is equivalent to show that the following are equivalent:

(1) $A=0$.

(2) $\left<Ax,y\right>=0$ for all $x,y$.

(3) $\left<Ax,x\right>=0$ for all $x$.

Obviously 1 implies 2 implies 3. Also 2 implies 1 by considering $y=Ax$. Assume 3 holds. Then compute, for arbitrary $\lambda\in\mathbb{C}$:

$0=\left<A(\lambda x+y),\lambda x+y\right>=\lambda \left<Ax,y\right>+\bar{\lambda}\left<Ay,x\right>.$

Hence for $\lambda\neq 0$ we get

$\left<Ax,y\right>=-\frac{\bar{\lambda}}{\lambda}\left<Ay,x\right>$.

For fixed $x,y$, we can vary $\frac{\bar{\lambda}}{\lambda}$, so that $\left<Ax,y\right>=0$, proving 2.

Note that we really used that the field is $\mathbb{C}$! If the field was $\mathbb{R}$, then $\frac{\bar{\lambda}}{\lambda}$ is constant equal to 1 so there is nothing to vary; in $\mathbb{C}$ one could e.g. take $\lambda=1$ and $\lambda=i$. Indeed in the real case it is not true: take $n=2$ and $A$ a rotation of 90 degrees, then $A$ is not zero, but $\left<Ax,x\right>=0$ for all $x$.

However, if in the real case we add the assumption that $A=A^*$ (symmetric), then it remains true, as can be easily seen form the above proof noting that $\left<Ax,y\right>=\left<x,A^*y\right>$. Or alternatively, if $A=A*$ then $(x,y)\mapsto B(x,y):=\left<Ax,y\right>$ is a symmetric bilinear form, and such forms are uniquely determined by their associated quadratic form $x\mapsto q(x):=B(x,x)$ since $2B(x,y)=q(x)+q(y)+q(x+y)$.

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If $\langle Ax,x\rangle=\langle Bx,x\rangle$ for all $x$, then from $\langle A(x+y),x+y\rangle = \langle B(x+y),x+y\rangle$ it follows that $$ \langle Ax,y\rangle +\langle Ay,x\rangle = \langle Bx,y\rangle +\langle By,x\rangle. $$ Similarly from $\langle A(x+iy),x+iy\rangle = \langle B(x+iy),x+iy\rangle$ it follows that $$ i\langle Ax,y\rangle -i\langle Ay,x\rangle = i\langle Bx,y\rangle -i\langle By,x\rangle. $$ (Here I am with the physicists: $\langle u,v\rangle=u^*v$.) From these two equations we get $\langle Ax,y\rangle=\langle Bx,y\rangle$.

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