Thursday, March 14, 2013

Operators on inner product spaces

Today, we'll begin chapter 7: operators on inner product spaces. It's a long chapter, but we won't cover everything - we'll only look at self-adjoint/normal operators and the spectral theorems. And possibly something else that I didn't write down properly.

1Operators on inner product spaces¶

1.1Self-adjoint operators¶

An operator $T \in \mathcal L(V)$ is called self-adjoint or Hermitian if $T = T^*$ (i.e., it is equal to its own adjoint). As we will see, the concept of a self-adjoint operator can be considered to be loosely analogous to that of a number that is equal to its own conjugate (that is, a real number).

1.2Proposition 7.1: Real eigenvalues¶

Every eigenvalue of a self-adjoint operator is real in any vector space.

Proof: Suppose $T = T^*$, and $\lambda$ is an eigenvalue of $T$ with associated eigenvector $v$. Thus $Tv = \lambda v$ for some non-zero $v$. Then

$$\begin{align} \langle v, T^*v \rangle & = \langle Tv, v \rangle \tag{by the definition of adjoint} \\ & = \langle \lambda v, v \rangle \tag{since $\lambda$ is an eigenvalue and $v$ is an eigenvector} \\ & = \lambda \langle v, v \rangle \tag{by homogeneity in the first argument} \\ & = \lambda \lVert v \rVert^2 \tag{by the definition of norm} \\ & \text{also:} \\ \langle v, T^*v \rangle & = \langle v, Tv \rangle \tag{since $T$ is self-adjoint} \\ & = \langle v, \lambda v \rangle \tag{eigenvalue} \\ & = \overline{\lambda} \langle v, v \rangle \tag{by homogeneity and conjugate symmetry} \\ & = \overline{\lambda} \lVert v \rVert^2 \tag{by the definition of norm} \end{align}$$

Thus we have that $\lambda \lVert v \rVert^2 = \overline{\lambda} \lVert v \rVert^2$, which tells us that $\lambda = \overline{\lambda}$ and so every eigenvalue must be real. So a self-adjoint operator is not only analogous to a number that is equal to its own conjugate, it in fact ensures that all of its eigenvalues have that properly. $\blacksquare$

1.3Proposition 7.2: Zero operators¶

If $V$ is a complex (and not real)¹ inner product space, and $T \in \mathcal L(V)$ such that $\langle Tv, v \rangle = 0$ for all $v \in V$, then $T = 0$.

Remark: This is only true if $V$ is not real. For example, if we're working on $\mathbb R^2$, with $T(x, y) = (-y, x)$, then $\langle Tv, v \rangle = \langle (-y, x), (x, y) \rangle = -yx + xy = 0$, and yet $T \neq 0$. Of course, arises from the fact that the eigenvalues are not real, and thus not part of the field we're working over. In a complex vector space, all the eigenvalues are indeed part of the field we're working over (i.e., $\mathbb C$), and so if $\langle Tv, v \rangle = 0$, then it must be that the operator is the zero operator. The proof for this proposition is left as an exercise (it's probably in the textbook).

1.3.1Corollary 7.3: Real numbers and self-adjoint operators¶

If $V$ is a complex (and not real) inner product space, and $T \in \mathcal L(V)$, then $T = T^*$ if and only if $\langle Tv, v \rangle \in \mathbb R$.

Proof: ($\Rightarrow$) First, we do a bit of simplifying:

$$\begin{align} \langle Tv, v \rangle - \overline{\langle Tv, v \rangle} & = \langle Tv, v \rangle - \langle v, Tv \rangle \tag{by conjugate symmetry} \\ & = \langle Tv, v \rangle - \langle T^*v, v \rangle \tag{cuz, self-adjoint} \\ & = \langle Tv - T^*v, v \rangle \tag{by additivity in the first place} \end{align}$$

Now, assume that $T = T^*$. Then $Tv-T^*v = Tv - Tv = 0$. Thus we have $\langle Tv, v \rangle - \overline{\langle Tv, v \rangle} = \langle 0, v \rangle = 0$. So $\langle Tv, v \rangle = \overline{\langle Tv, v \rangle}$ which means that $\langle Tv, v \rangle$ is real (since it's equal to its own conjugate).

($\Leftarrow$): Now for the other direction. Assume that $\langle Tv, v \rangle \in \mathbb R$. Then, $\langle Tv, v \rangle - \overline{\langle Tv, v \rangle} = 0$. By the simplifying above, this means that $\langle (T - T^*)v, v \rangle = 0$. By proposition 7.2, this implies that $T-T^* = 0$, and so $T = T^*$. $\blacksquare$

1.3.2Proposition 7.4: Zero inner-products and self-adjoint operators¶

If $T$ is self-adjoint on $V$ such that $\langle Tv, v \rangle = 0$ for all $v \in V$, then $T = 0$.

Proof: left as an exercise. Probably in the textbook

1.4Normal operators¶

$T \in \mathcal L(V)$ is a normal operator if $TT^* = T^*T$ (i.e., multiplication is commutative).

Self-adjoint operators are trivially normal. However, there are some operators that are normal but not self-adjoint. So normal operators are a superset of self-adjoint operators.

Example: $T \in \mathcal L(F^2)$ such that

$$\mathcal M(T) = \begin{pmatrix} 2 & -3 \\ 3 & 2 \end{pmatrix} \tag{with respect to the standard basis}$$

To find $\mathcal M(T^*)$, we just take the complex conjugate of $\mathcal M(T)$ and transpose it:

$$\mathcal M(T^*) = \begin{pmatrix} 2 & 3 \\ -3 & 2 \end{pmatrix}$$

Clearly, $\mathcal M(T) \neq \mathcal M(T^*)$, so $T$ is not self-adjoint. But the matrix multiplication is indeed commutative (not going to write it out, but it's easy to check this yourself if you aren't convinced). Thus $T$ is normal but not self-adjoint.

1.5Proposition 7.6: Norms and normal operators¶

$T \in \mathcal L(V)$ is normal $\iff$ $\lVert Tv \rVert = \lVert T^* v \rVert$ for all $v \in V$

Proof: $T$ is normal $\iff$ $TT^* - T^*T = 0$ $\iff$ $\langle (TT^* - T^*T)v, v \rangle = 0$² $\iff$ $\langle TT^* v, v \rangle - \langle T^*Tv, v \rangle = 0$ $\iff$ $\langle T^*v, T^*v \rangle - \langle Tv, Tv \rangle = 0$ (by the definition of adjoints) $\iff$ $\lVert T^*v \rVert^2 - \lVert Tv \rVert^2 = 0$ $\iff$ $\lVert T^*v \rVert^2 = \lVert Tv \rVert^2$ $\iff$ $\lVert T^*v \rVert = \lVert Tv \rVert$ as desired. $\blacksquare$

1.5.1Corollary 7.7: Eigenvalues for adjoints of normal operators¶

Suppose $T \in \mathcal L(V)$ is normal. If $v \in V$ is an eigenvector of $T$ with eigenvalue $\lambda \in \mathbb F$, then $v$ is also an eigenvector of $T^*$ with eigenvalue $\overline{\lambda}$.

Proof: Assume $T$ is normal. Since $v$ is an eigenvector, $Tv = \lambda v$. Then $(T-\lambda I)v = 0$. Note that $(T-\lambda I)(T - \lambda I)^* = (T - \lambda I)(T^* - \lambda I^*) = (T-\lambda I)(T^* - \overline{\lambda} I)$ (since $I$ is of course self-adjoint). Then, by distributivity, this is equal to $TT^* - \overline{\lambda}T - \lambda T^* + \lambda\overline{\lambda}I = TT^* - \overline{\lambda}T - \lambda T^* - \overline{\lambda}\lambda I = (T-\lambda I)^*(T - \lambda I)$. Thus $T - \lambda I$ is normal.

Now, $\lVert (T-\lambda I) v\rVert = 0$. By proposition 7.6, then $\lVert (T-\lambda I)v \rVert = (T-\lambda I)^*v \rVert = \lVert (T^* -\overline{\lambda}I)v\rVert = 0$. Thus $(T-\overline{\lambda}I)v = 0$, and so $Tv =\overline{\lambda}I$. This shows that $v$ is an eigenvector of $T^*$ with corresponding eigenvalue $\overline{\lambda}$, as desired.

1.5.2Corollary: Orthogonality of eigenvectors¶

If $T \in \mathcal L(V)$ is normal, then eigenvectors of $T$ corresponding to distinct eigenvalues are orthogonal.

Pretty cool. I'll put the proof in later because I couldn't really follow along with what was written on the board. It's also probably in the textbook.