# Simplification of the statement of the spectral theorem

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.

• 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". – Jendrik Stelzner Oct 1 '16 at 15:09
• @JendrikStelzner If an operator $T$ is auto-adjunct, then $T^*=T$ and particularly $T^*T=TT^*$, so it's normal. – user42912 Oct 2 '16 at 1:25
• 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.) – Jendrik Stelzner Oct 2 '16 at 1:58
• @JendrikStelzner Every real inner product space $V$ is a complex one, no? since $\mathbb R\subset \mathbb C$. – user42912 Oct 2 '16 at 2:13
• 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). – Jendrik Stelzner Oct 2 '16 at 11:18

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.