Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have been told that the adjoint of an operator behaves much like complex conjugation, and so self-adjoint operators are like real numbers. Can someone explain this analogy more in depth, or give a reference?

For example, a corollary on p. 129 of Axler's "Linear Algebra Done Right" says that the proposition

$T$ is self-adjoint iff $(Tv, v) \in \mathbb{R}$ for every $v \in V$.

is an example of how self-adjoint operators behave like real numbers. I don't see what he's talking about here, and am wondering if there are other propositions that illustrate parallels between self-adjoint operators and real numbers.

share|improve this question
    
Did you inspect the properties of adjoint operators? What more do you want? –  Jonas Teuwen Aug 24 '11 at 22:49
    
In regards to the Corollary: let $(,)$ be the Hermitian inner product on $\mathbb{C}$, so $(w,z) = \bar{w}z$. Let $u$ be a complex number, consider $(uz,z) = \bar{u}\bar{z}z = \bar{u}|z|^2$, so $(uz,z)\in\mathbb{R}$ iff $u$ is real. –  Willie Wong Aug 24 '11 at 23:11
1  
Note that if $\lambda$ is a complex number, then $(\lambda T)^*=\overline\lambda T^*$, and in particular the adjoint coincides with complex conjugation on scalar multiples of the identity. –  Jonas Meyer Aug 24 '11 at 23:15
add comment

4 Answers

They are both *-operations.

share|improve this answer
add comment

Firstly, if $T$ is a linear operator on a finite-dimensional inner product space, and $B$ is an orthonormal basis for this space, then $[T^{\ast}]_B = [T]_B^{\ast}$, that is, the matrix representation of the adjoint is the conjugate transpose of the matrix representation of the original operator.

Secondly, observe that if $\lambda \in \mathbb{R}$, then $\langle \lambda v,v\rangle \in \mathbb{R}$ for all $V$, whereas if $\lambda \notin \mathbb{R}$, then $\langle \lambda v,v\rangle$ is never real (except when $v = 0$). This is part of the analogy between real numbers and self-adjoint operators.

Continuing with the analogy between complex conjugation and adjoints, there are many ways in which unitary operators (those operators $U$ satisfying $UU^{\ast} = \mathrm{id}$) behave like unit-length numbers (those complex numbers $z$ satisfying $zz^{\ast} = 1$).

For further details, you can check out this handout from the linear algebra course I taught this summer:

http://math.berkeley.edu/~akgupta/MAT110-Su11/hand4.pdf

share|improve this answer
add comment

Let $T$ be a normal operator on a finite-dimensional Hilbert space $H$. By the spectral theorem, $T$ admits an orthonormal basis of eigenvectors. If $\lambda_1, ... \lambda_n$ denote its eigenvalues, then $T^{\ast}$ has eigenvalues $\overline{\lambda_1}, ... \overline{\lambda_n}$. In particular, as Jonas mentions in the comments, the adjoint of scalar multiplication by $\lambda$ is scalar multiplication by $\bar{\lambda}$, and if $T$ is self-adjoint, all of its eigenvalues are real.

Moreover, any normal operator $T$ can be written as the sum of its "real part" and "imaginary part"

$$T = \frac{T + T^{\ast}}{2} + \frac{T - T^{\ast}}{2}.$$

The first term is self-adjoint and has eigenvalues $\text{Re}(\lambda_1), ... \text{Re}(\lambda_n)$. The second term is skew-adjoint (analogous to purely imaginary) and has eigenvalues $\text{Im}(\lambda_1), ... \text{Im}(\lambda_n)$. Both have the same eigenvectors as $T$.

There are deeper levels on which to understand this analogy, but I'm not sure how useful they would be at this point.

share|improve this answer
    
Any operator can be written that way. –  Ricky Demer Aug 25 '11 at 1:04
1  
@Ricky: yes, but the statement about real and imaginary parts of the eigenvalues isn't true unless the real and imaginary parts of $T$ commute. –  Qiaochu Yuan Aug 25 '11 at 1:06
add comment

If an element $T$ of the algebra $M_n(\mathbb C)$ of $n$-by-$n$ matrices with complex entries acts as a linear transformation on the vector space $\mathbb C^n$ by left multiplication of column vectors, and if $\mathbb C^n$ has the standard inner product, then $T^*$ is obtained by transposing $T$ and taking the complex conjugate of each entry. In particular, if $T$ is a diagonal matrix, then $T$ is self-adjoint if and only if each of its entries is real. This carries over to operators on any inner product space if you look at matrices with respect to an orthonormal basis.

Here's a related fact that will not be mentioned in an introductory linear algebra course, but might be worth mentioning here. Suppose that $A$ is a commutative algebra of continuous linear operators on an inner product space such that $T^*$ is in $A$ whenever $T$ is in $A$. Then $A$ is isomorphic to an algebra of complex-valued functions on some set, in such a way that the adjoint in $A$ becomes pointwise complex conjugation of functions.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.