7A Self-Adjoint and Normal Operators

Suppose $T \in \mathscr{L}(V,W)$. Then $T^* \in \mathscr{L}(W,V)$.

An intuitive way to prove this would be to observe that $T^*$ is the composition of three functions that preserve vector addition, and it is also the composition of three maps that, though they do not all preserve scaling, two of them conjugate scalars and these two conjugations cancel each other out. On the other hand, we can use the inner product property to prove this. Let $v \in V$ and $w, w_1, w_2 \in W$ and $\lambda \in \mathbb{F}$. Then

and

Suppose $T \in \mathscr{L}(V,W)$. Then a) $(S+T)^* = S^* + T^*.$ b) $(\lambda T)^* = \overline{\lambda} T^*$ for all $\lambda \in \mathbb{F}.$ c) $(T^*)^* = T.$ d) $(ST)^* = T^* S^*.$ e) $I^* = I,$ where $I$ is the identity on $V.$ f) If $T$ is invertible, then $T^*$ is invertible and $(T^*)^{-1} = (T^{-1})^*.$

a) For any $s \in \mathscr{L}(V,W),$

b) For any $\lambda \in \mathbb{F}$,

c) Using the definition of the adjoint, $(T^*)^*$ sends each vector in $v \in V$ to the unique vector $w \in W$ such that $\langle T^*(w), v \rangle = \langle w, (T^*)^*(v) \rangle.$ Using this fact, we observe the following:

$\langle T^*(w), v \rangle = \langle w, T(v) \rangle$ shows that $T(v) = (T^*)^*(v).$ d) This follows from the fact that $(TS)' = S' \circ T'.$ Alternatively, using inner products we have:

for every $v \in V$, $w \in W.$ e) This follows from the fact that the dual of the identity is the identity (becuse precomposing with the identity does not change your functional). Alternatively, we have the following using inner products:

for every $v \in V$, $w \in W.$ f) This also follows from the inverse of the dual being the dual of the inverse for invertible linear maps. This can be shown with inner products, but we would only be using the following equalities, which prove this result (and similarl equalities prove that the dual of the inverse is the inverse of the dual.)

and

Suppose $T \in \mathscr{L}(V,W)$. Then a) $\ker(T^*) = T(V)^\perp.$ b) $T^*(W) = \ker(T)^\perp.$ c) $\ker(T) = T^*(W)^\perp.$ d) $T(V) = \ker(T^*)^\perp.$

We begin by proving part c): $\ker(T^*) = T(V)^\perp$ because for any $v \in V$,

Now, b) is the orthogonal complement of both sides of c), and the remaining two statements are b) and c) with the switching of $T$ and $T^*$.

For any linear map $T \in \mathscr{L}(V,W)$ and orthonormal bases for $V$ and $W$, the corresponding matrix for $T^*$ with respect to these bases is the conjugate transpose of the matrix of $T$ with respect to these bases. Said with symbols, if $\beta$ and $\gamma$ are orthonormal bases for $V$ and $W$ respectively, then

Let $A = M(T, \beta, \gamma)$, $B = \mathcal{M}(T^*, \gamma, \beta)$, $\beta = (e_1, \ldots, e_n)$, and $\gamma = (f_1, \ldots, f_m)$. To find the $j$th column of $B$, we put $f_j$ into $T^*$ and find its coordinates in terms of $\beta$. Because we're working with orthonormal bases, The $i$th coordinate of this column vector can be found by projection: $\langle T^*(f_j), e_i \rangle.$ Thus, $\langle T^*(f_j), e_i \rangle$ is the $i$th entry of the $j$th column of $B$. Now observe what we get from using conjugate symmetry on this inner product:

By the same reasoning, this is the conjugate of the $j$th coordinate of the $i$th column of $A$. Therefore $A$ and $B$ are conjugate transposes of each other.

Every eigenvalue of a self-adjoint operator is a real number.

To prove this, we will show that conjugating an eigenvalue does not change it. Let $T \in \mathscr{L}(V)$ be self-adjoint and let $v \in V$ be an eigenvector with eigenvalue $\lambda$. Then

Because $\Vert v \Vert \neq 0$, $\lambda = \overline{\lambda}$.

Suppose $V$ is a complex inner product space and $T \in \mathscr{L}(V).$ Then $\langle T(v), v \rangle = 0$ for every $v \in V$ if and only if $T = 0$.

This proof uses a generalization of the polarization identity for inner products, and it will first be proved as a claim. Claim: For any $u, w \in V$ and any $T \in \mathscr{L}(V)$,

Proof of claim: Take 4 times the sum and expand each of the inner products using conjugate linearity. You will see that the first terms ($T(u)$ with $u$) cancel for values of $\zeta$ that are negatives of each other, the second terms ($T(u)$ with $w$) remain to equal 4 times the left hand side of the equation, the third terms ($T(w)$ with $u$) cancel for values of $\zeta$ by $(1 \leftrightarrow i)$ and $(-1 \leftrightarrow -i)$, and the fourth terms ($T(w)$ with $w$) cancel similarly to the first terms, for avlues of $\zeta$ that are negatives of each other. There is a version with every term showing and color-coordinated cancellation in my notes, but the unabridged statement of this claim doesn't even fit in a regular computer browser, let alone the expanded versions. With the claim proved, we proceed: On the one hand, if $T$ is the zero map then $\langle T(v), v \rangle = \langle 0, v \rangle = 0$ for all $v \in V$. On the other hand, if $\langle T(v), v \rangle = 0$ for every $v \in V$, then for any $u, w \in V$ we have

Let $V$ be a complex inner product space and $T \in \mathscr{L}(V).$ Then $T$ is self-adjoint if and only if $\langle T(v), v \rangle \in \mathbb{R}$ for every $v \in V.$

$\Rightarrow$: This direction comes from conjugate symmetry. If $T$ is self-adjoint then for any $v \in V$,

$\Leftarrow$: Suppose that for every $v \in V$, $\langle T(v), v \rangle = \overline{\langle T(v), v \rangle}$. Then $\langle T(v), v \rangle = \langle v, T(v) \rangle = \langle T^*(v), v \rangle$, and hence $\langle (T - T^*)(v), v \rangle = 0$ for every $v \in V.$ Because we are working over $\mathbb{F}=\mathbb{C}$, this means $(T-T^*)$ is the zero operator, and therefore $T = T^*$.

Suppose $T$ is a self-adjoint operator on $V$ with the property that $\langle T(v), v \rangle = 0$ for every $v \in V$. Then $T = 0.$

Similarly to the complex case, we first prove a generalization of the real polarization identity. Claim: For every $u, w \in V$ and any self-adjoint $T \in \mathscr{L}(V),$

Proof: There is room to write out the expansion in the real case.

With the claim proven, we can proceed. Let $u, w \in V$ and let $T \in \mathscr{L}(V)$ be self-adjoint. Then similarly to the corresponding proof, we have

Because $T(u)$ is orthogonal to every $w \in V$, $T(u)=0$.

Let $T \in \mathscr{L}(V)$. $T$ is normal if and only if

for every $v \in V.$

Note first that $T$ being normal is equivalent to the difference $(T^*T - TT^*)$ operator being the zero operator. This difference is called the commutator and it is often compared to a trivial element to test whether or not two things commute. We also note that $T^*T$ and $TT^*$ are both self-adjoint, so the commutator $(T^*T - TT^*)$ is also self-adjoint, which we will need for the first implication in the case where $\mathbb{F}=\mathbb{R}$. We are now ready to begin.

Let $T$ be a normal operator on an inner product space $V$. Then a) $\ker(T) = \ker(T^*)$ b) $T(V) = T^*(V)$ c) $V = \ker(T) \oplus T(V)$ d) $T-\lambda I$ is normal for every $\lambda \in \mathbb{F}$. e) If $v \in V$ and $\lambda \in \mathbb{F}$, then $T(v) = \lambda v \iff T^*(v) = \overline{\lambda} v.$

a) $T(v)=0$ if and only if $T^*(v)=0$ because $\Vert T(v) \Vert = \Vert T^*(v) \Vert.$ b) Because their kernels are the same, their ranges must be the same, as range is the orthogonal complement of the kernel. c) This follows from the orthogonal complement decomposition into a direct sum. d) Because $T$ and $T^*$ commute, all of the terms in $\{ T, T^*, \lambda I, \overline{\lambda} I \}$ commute with each other. Therefore $(T-\lambda I)$ and $(T^* - \overline{\lambda} I)$ commute. e) Very similarly to part a), $(T - \lambda I)(v) = 0 \iff (T^* - \overline{\lambda} I)(v) = 0$ because

Let $T$ be a normal operator on $V$. Then eigenvectors of $T$ with distinct eigenvalues are orthogonal.

Let $v_1, v_2$ be eigenvectors of $T$ with distinct eigenvalues $\lambda_1, \lambda_2$. Then

Because $(\lambda_1 - \lambda_2) \neq 0$, $\langle v_1, v_2 \rangle = 0$.

Suppose $T$ is an operator on a complex inner product space $V$. Then $T$ is normal if and only if there exist commuting self-adjoint operators $A$ and $B$ such that $T = A + Bi$.

$\Rightarrow$: Suppose $T$ is normal. Set $A = \frac{T + T^*}{2}$ and $B = \frac{T - T^*}{2i}.$ Then both operators are self-adjoint, and $AB - BA$ simplifies to $\frac{T^*T - TT^*}{2i} = \frac{0}{2i}.$ In summary, $T$ and $T^*$ commuting made the difference $T^*T - TT^*$ zero, which in turn made the difference $AB - BA$ zero, so $A$ and $B$ commute. $\Leftarrow$: Suppose $T = A + iB$ for commuting self-adjoint operators $A$ and $B$. Then $T^* = (A + iB)^* = A^* - iB^* = A - iB.$ Using a similar process to the other direction, we have $A = \frac{T+T^*}{2}$ and $B = \frac{T - T^*}{2i}$. Subsequently, $T^*T - TT^* = \frac{AB - BA}{2} = \frac{0}{2} = 0.$ Therefore $T$ is normal.