Answer to Question #328755 in Linear Algebra for Nahyijb

Question

For each of the linear operators below, determine whether it is normal, self adjoint, or neither

(a) T / (R ^ 2) -> R ^ 2 defined by T(x, y) = (2x - 2y, - 2x + 5y)

(b) T / (C ^ 2) -> C ^ 2 defined by T(x,y)=(2x+iu.x+2y)

Answer 1

Remind the definitions of normal and self-adjoint operators. The operator $N$ is called normal if and only if it is a continuous linear operator and it commutes with its Hermitian adjoint. Namely, $NN^*=N^*N$. The operator $N$ is called self-adjoint if and only if $N=N^*$. The operator is called adjoint if and only if $<Na,b>=<a,N^*b>$ for any vectors $a$ and $b$.

a). The operator can be presented as a matrix: $(x,y)\left(\begin{array}{ll}2&-2\\-2&5\end{array}\right)$. Denote $T=\left(\begin{array}{ll}2&-2\\-2&5\end{array}\right)$. The operator is linear, since $(\alpha a+\beta b)T=\alpha(aT)+\beta(bT)$ for any numbers $\alpha,\beta\in{\mathbb{R}}$ and vectors $a,b\in{\mathbb{R}}^2$. Suppose that we have vectors $(x_1,y_1)$ and $(x_2,y_2)$. Consider the scalar product: $<(2x_1-2y_1,-2x_1+5y_1),(x_2,y_2)>(2x_1-2y_1)x_2+(-2x_1+5y_1)y_2=x_1(2x_2-2y_2)+y_1(-2x_2+5y_2)=<(x_1,y_1),(2x_2-2y_2,-2x_2+5y_2)>$ As we can see, the operator is self-adjoint. It is also normal, since $NN^*=(N^*)^2=N^*N$

b). The operator is not linear, since $T(x_1+x_2,y_1+y_2)=(2x_1+2x_2+iu,x_1+x_2+2(y_1+y_2))$ and $T(x_1,y_1)=(2x_1+iu,x_1+2y_1)$, $T(x_2,y_2)=(2x_2+iu,x_2+2y_2)$. As we can see: $T(x_1+x_2,y_1+y_2)\neq T(x_1,y_1)+T(x_2,y_2)$. Thus, the operator is not normal. Compute the scalar product $<T(x_1,y_1),(x_2,y_2)>=<2x_1+iu,x_1+2y_1,(x_2,y_2)>=2(x_1+iu)x_2+(x_1+2y_1)y_2=x_1(2x_2+y_2)+y_1(2y_2)+2iux_2$

From the scalar product we see that it can not be presented as $<(x_1,y_1),H(x_2,y_2)>$. Thus, it is not possible to find the adjoint operator. Therefore, the operator is not self-adjoint.

Answer: a). The operator is normal and self-adjoint. b). The operator is not normal and not self-adjoint.