# Find all "Inner Products" such that <x, Tx> = 0.

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• November 20th 2011, 10:57 AM
TaylorM0192
Find all "Inner Products" such that <x, Tx> = 0.
Hello,

I came across this problem while preparing for one of my midterms, and it erks me that I still can't find a solution.

Given an linear operator T: R^2 -> R^2 rotation by pi/2 (i.e. T(x, y) = T(-y, x)), find all inner products < , > such that for all vectors x in R^2, <x, Tx> = 0.

It is obvious that the standard inner (dot) product is one such inner product. A few others could certainly be imagined (i.e. scalar multiples of the dot product). But we are, of course, asked to find all possible cases.

I know that all inner products on finite dimensional inner product spaces are represented by a Hermitian matrix G (G = G*) which satisfies the additional condition [x]*G[x] > 0 for all vectors x in V with respect to some basis.

Conversely, when we have such a G satisfying the above, it always represents an inner product on V with respect to a certain basis B.

The inner product is of course given as <x, y> = [y]*[G][x], where [y]* is the conjugate transpose of the coordinate vector of y represented in the basis B, likewise for [x] without the conjugation/transpose.

In the question, the conjugation can be removed, so we have <x, y> = [y]G[x].

The condition imposed on the inner product is <x, Tx> = [Tx]G[x] = 0.

Thus, it seems to be, that to solve the problem we must find all possible G which satisfies this equation.

But the problem I face is that even if I was able to use this equation to find such G, how do I account for the representation of x and Tx in the corresponding basis that G represents an inner product? Needless to say, my attempts down this path of reasoning haven't led me to a solution.

If anyone could help extend this process to get the solution, or propose a different approach, I would appreciate it!
• November 20th 2011, 02:02 PM
Jose27
Re: Find all "Inner Products" such that <x, Tx> = 0.
$\langle x,Ay\rangle= \langle A^*x,y \rangle = \langle x, y \rangle _1$ characterizes all inner products, when $A$ varies over positive (symmetric) operators in $V$. Your hypothesis say $\langle Ax, Tx \rangle =0$ for all $x$, which means that $Ax$ must be orthogonal to the othogonal complement of $x$, and since you're in two dimensions this is just $Ax=\lambda_x x$. By linearity of $A$ it's not difficult to see that $\lambda_x=\lambda_y$ for all $x,y\in \mathbb{R}^2$. Since $A$ must be positive, this means $\lambda >0$, and so $A=\lambda I$ and all inner products are multiples of the usual.
• November 20th 2011, 02:21 PM
Deveno
Re: Find all "Inner Products" such that <x, Tx> = 0.
since we are dealing with R^2, G is a symmetric matrix. moreover, G must be positive-definite, that is, if (x,y) is not (0,0), and G =

[a b]
[b c],

$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}a&b\\b&c\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} > 0$

so $ax^2 + 2bxy + cy^2 > 0$

since (x,0) and (0,y) are not (0,0) for x,y ≠ 0, we must have a,c > 0.

moreover, by "completing the squares" we see that:

$ax^2 + 2bxy + cy^2 = a\left(x+\frac{b}{a}y\right)^2 + \left(\frac{ac - b^2}{a}\right)y^2$,

so we must also have, in addition to a,c > 0, that $|b| < \sqrt{ac}$.

now, in addition, we require that <x,Tx> = 0, that is:

$x^TGTx = 0$ so that:

$(c - a)xy + b(x^2 - y^2) = 0$ for all x,y in R. if we choose x = 1, y = 0, we see that b = 0.

if we choose x = y, we see that a = c, so the only possible candidates are matrices of the form aI, for a > 0, that is to say: POSITIVE scalar multiples of the usual inner product.
• November 20th 2011, 07:55 PM
TaylorM0192
Re: Find all "Inner Products" such that <x, Tx> = 0.
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