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**HallsofIvy** Since the set of real numbers is a subset of the complex numbers, even though <u, v> is defined to be a real number, you can think of it as a function from pairs of vectors to a **complex number**.

For example, $\displaystyle R^2$ with inner product <(a, b), (c, d)>= ac+ bd has inner product always real but the matrix $\displaystyle \begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$ has characteristic equation $\displaystyle \lambda^2+ 1= 0$ which has i and -i as roots. Since we **can't** have Av= iv in $\displaystyle R^2$, the corresponding linear transformation does NOT have any eigevalues. But we can, temporarily switch to "over $\displaystyle C^n$ so that any linear transformation has eigenvalues, show, as you did, that, for a self-adjoint linear transformation, they must be real and so assert that any self-adjoint linear transformation on a vector space over the **real** numbers has eigenvalues.