0
$\begingroup$

I'm reading Hoffman and Kunze's linear algebra book and on page 335 they state the spectral theorem for finite-dimensional inner product spaces:

Theorem 9 (Spectral Theorem). Let $T$ be a normal operator on a finite-dimensional complex inner product space $V$ or a self-adjoint operator on a finite-dimensional real inner product space $V$. Let $c_1, \dotsc, c_k$ be the distinct characteristic values of $T$. Let $W_j$ be the characteristic space associated with $c_j$ and $E_j$ the orthogonal projection of $V$ on $W_j$. Then $W_j$ is orthogonal to $W_i$ when $i \neq j$, $V$ is the direct sum of $W_1, \dotsc, W_k$, and $$ T = c_1 E_1 + \dotsb + c_k E_k. $$

Why in the beginning of the statement of this theorem can't we simply say

Let $T$ be a normal operator on a finite-dimensional complex inner product space $V$

and remove this part

or a self-adjoint operator on a finite-dimensional real inner product space $V$

since auto-adjunct operators on a finite-dimensional real inner product space is a normal one on a finite-dimensional complex inner product space.

EDIT

Answering the comments asking why an auto-adjunct operator is a normal one: If an operator $T$ is auto-adjunct, then $T^*=T$ and particularly $T^*T=TT^*$, so it's normal.

$\endgroup$
  • $\begingroup$ I think it would be helpful to explain how exactly an "auto-adjunct operators on a finite-dimensional real inner product space is a normal one on a finite-dimensional complex inner product space". $\endgroup$ – Jendrik Stelzner Oct 1 '16 at 15:09
  • $\begingroup$ @JendrikStelzner If an operator $T$ is auto-adjunct, then $T^*=T$ and particularly $T^*T=TT^*$, so it's normal. $\endgroup$ – user42912 Oct 2 '16 at 1:25
  • $\begingroup$ I should have been more specific: If $T$ is a selfadjoint operator on a real inner product space $V$, then $T$ is in particular a normal operator on this real inner product space $V$; this much (I think) is clear. But you seem to claim that $T$ is (in some way) a normal operator on some complex inner product space, and I don’t see where this complex structure comes from. (At least not without some rather lengthy arguments via extension of scalars.) $\endgroup$ – Jendrik Stelzner Oct 2 '16 at 1:58
  • $\begingroup$ @JendrikStelzner Every real inner product space $V$ is a complex one, no? since $\mathbb R\subset \mathbb C$. $\endgroup$ – user42912 Oct 2 '16 at 2:13
  • $\begingroup$ One can complexify a real inner product space $V$ to get a complex inner product space $V_\mathbb{C}$, and if $T \colon V \to V$ is a self-adjoint operator then we get a self-adjoint operator $T_\mathbb{C} \colon V_\mathbb{C} \to V_\mathbb{C}$. But I don’t think there is a way to regard $V$ itself as a complex inner product space (or even a complex vector space). $\endgroup$ – Jendrik Stelzner Oct 2 '16 at 11:18
2
$\begingroup$

The proof of the theorem requires that V have an orthonormal basis consisting of characteristic vectors (eigenvectors) of T. For a finite-dimensional inner product space over the field of complex numbers, such a basis exists if and only if T is normal. There are some normal operators, however, that have no such basis if the space is over the field of real numbers; for example, the matrix representation of one such operator is $${\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}}.$$ For a finite-dimensional inner product space over the field of real numbers, V has an orthonormal basis consisting of characteristic vectors of T if and only if T is self-adjoint, a more restrictive requirement than normality. Hence, the authors separated the complex and real cases in the hypothesis of the theorem to keep it sharp for each case.

See pages 335–336 in Linear Algebra for the spectral theorem and its (short) proof. See the first two paragraphs of Section 8.5 (pages 311–312) for the source of my answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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