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Prove that there does not exist a linear self-adjoint operator T on R3 with the standard Euclidean scalar product such that T((1, 2, 3)) = (3, 2, 1) and T((4, 5, 6)) = (4, 5, 6).

Where do I begin? The first thing I think of is representing the transformation as a matrix but I think that even if the transformation is not self-adjoint the matrix can be.

Thanks!

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Note that is $\mathbb{R}^3$ a self-adjoint operator is represented by a symmetric matrix. Moreover, symmetric matrix over the reals is orthogonally diagonalizable, that is, there exists an orthogonal matrix $U$ (i.e. $UU^t = I$) such that $U^{-1}TU = D$ is diagonal.

Assume $T$ is self-adjoint. Then there is a diagonal matrix $D$ and an orthogonal matrix $U$ such that $T = U D U^{-1} = U D U^t$. As $D$ is symmetric $$ T^t = (U D U^t)^t = (U^t)^t D^t U^t = U D^t U^t = U D U^t = T $$ and hence $T = T^t$ and $T$ must be symmetric.

Now change and complete $(1,2,3)^t, \ (4,5,6)^t$ the basis into an orthogonal basis (say the standard basis) and show that $T$ is not symmetric.

Note that $T$ is not a symmetric matrix since $$ T = \left[ \begin{matrix} 3 & 4 & a \\ 2 & 5 & b \\ 1 & 6 & c \end{matrix} \right] $$ and $4 \ne 2$.

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  • $\begingroup$ You need to be careful here. $(1, 2, 3)$ and $(4, 5, 6)$ are not orthogonal to each other, so it is not possible to complete these two vectors into an orthogonal basis. $\endgroup$ – user99914 Nov 10 '14 at 23:04
  • $\begingroup$ That's why I wrote "change and complete" and not just "complete". $\endgroup$ – LinAlgMan Nov 12 '14 at 16:57
  • $\begingroup$ @LinAlgMan: Thanks for the reply. But it is now not clear to me why $T$ is written in that form... $\endgroup$ – user99914 Nov 13 '14 at 8:58
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Let's assume that $T$ is symmetric and see what would happens. We know that there is an orthongonal basis $\{v_1, v_2, v_3\}$ so that $$Tv_i = \lambda_i v_i.$$ Given that $T(4, 5, 6) = (4, 5, 6)$, we see that $(4, 5, 6)$ is an eigenvector of $T$. Thus we can let

$$v_1 = \frac{(4, 5, 6)}{||(4, 5, 6)||},\ \ \lambda _1= 1.$$

Let $V = \text{span}\langle v_2, v_3\rangle$. Then for all $w = a_2 v_2 + a_3 v_3 \in V$,

$$Tw \cdot v_1 = (a_2Tv_2 + a_3 Tv_3) \cdot v_1 = 0.$$

(As $v_2\cdot v_1 = v_3 \cdot v_1 = 0$). This mean that $Tw \cdot (4, 5, 6) = 0$ whenever $w\in V$. Now consider

$$w = (1, 2, 3) - \big((1, 2, 3)\cdot v_1\big) v_1= (1, 2, 3) - \frac{32}{77} (4, 5, 6)$$

Then $w \in V$ and

$$Tw = T(1, 2, 3) - \frac{32}{77} T(4, 5, 6) = (3, 2, 1) - \frac{32}{77}(4, 5, 6)$$

Thus $Tw \cdot (4, 5, 6) = -4 \neq 0$. This is a contradiction and so $T$ is not symmetric.

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