Proving that eigenvalues of real self adjoint endomorphism are real

Question

If $\mathbf{A}\colon V \to V$ is selfadjoint for some finite dimensional vector space $V$ over $\mathbb{C}$ then this follows immediately, since for any nonzero eigenvalue $\lambda $ and a corresponding eigenvector $\mathbf{v}$ we have \begin{align*} \lambda \left\langle \mathbf{v}, \mathbf{v}\right\rangle = \left\langle \mathbf{A}\mathbf{v}, \mathbf{v}\right\rangle = \left\langle \mathbf{v}, \mathbf{A}\mathbf{v}\right\rangle = \overline{\lambda }\left\langle \mathbf{v}, \mathbf{v}\right\rangle .\end{align*} Can one argue in a similar way if $V$ is a vectorspace over $\mathbb{R}$?

I thought about just considering the same vector space over $\mathbb{C}$ which, however, gives rise to several questions: Is $\mathbf{A}$ again self adjoint over $\mathbb{C}$? Can we even extend an inner product of $V$ w.r.t. to $\mathbb{R}$ to $\mathbb{C}$?

For the latter question I would say yes, since (if we assumed we already have such an inner product over $\mathbb{C}$) \begin{align*} \left\langle \mathbf{v}, \mathbf{w}\right\rangle _{\mathbb{C}} &= \left\langle \mathbf{a} + i\mathbf{b}, \mathbf{c} + i\mathbf{d}\right\rangle _{\mathbb{C}} \\ &= \left\langle \mathbf{a}, \mathbf{c}\right\rangle _{\mathbb{C}} + \left\langle \mathbf{a},i \mathbf{d}\right\rangle _{\mathbb{C}} + \left\langle i \mathbf{b}, \mathbf{c}\right\rangle _{\mathbb{C}} + \left\langle i \mathbf{b}, i \mathbf{d}\right\rangle _{\mathbb{C}} \\ &\stackrel{\text{(1)}}{=} \left\langle \mathbf{a}, \mathbf{c}\right\rangle _{\mathbb{R}} +i \left\langle \mathbf{a}, \mathbf{d} \right\rangle _{\mathbb{R}} - i \left\langle \mathbf{b}, \mathbf{c}\right\rangle _{\mathbb{R}} + \left\langle \mathbf{b}, \mathbf{d}\right\rangle _{\mathbb{R}} \end{align*} so we can simply use (1) as our definition (which turns out to satisfy all properties of a complex inner product). W.r.t. this inner product $\mathbf{A}$ would also be self adjoint again (so we can borrow the proof from above). Is this argumentation valid?

Answer 1

Here is an alternative that does not involve changing the structure of the original vector space.

Let $M$ be the matrix of $A$ relative to any orthonormal basis of $V$. Then $M$ is a symmetric $n\times n$ matrix over ${\mathbb R}$, where $n$ is the dimension of $V$.

Next we forget about $V$, and consider the linear transformation $B:{\mathbb C}^n\to {\mathbb C}^n$ given by the matrix $M$ (this time viewed as a complex matrix).

Then $B$ is a self-adjoint complex-linear operator on ${\mathbb C}^n$, and the complex theory guarantees that the roots of its characteristic polynomial coincide with the set of eigenvalues of $B$, all of which are real.

Finally, we just have to notice that the characteristic polynomial of $B$ is the same as that for $A$!

Answer 2

Evangelista Torricelli, an Italian physicist and mathematician, who was a student of Galileo and is known for his invention of the barometer, once said

Noi viviamo sommersi nel fondo d'un pelago d'aria. (We live submerged at the bottom of an ocean of air.)

The striking fact about this quotation it that, while it is obviously true, we don't usually feel that way.

Likewise, in mathematical terms, we live surrounded by complex numbers, although we often don't realize it, and this post is perhaps another indication of this fact!

Nevertheless, let me try another answer which doesn't mention complex numbers at all.

Theorem. Let $A$ be a self-adjoint operator acting on a real Hilbert space $V$. Then all of the roots of the characteristic polynomial of $A$ are real.

Proof. The proof will be by induction on the dimension of $V$.

• Case 1: $\text{dim}(V)=1$.

This is obvious!

• Case 2: $\text{dim}(V)=2$.

Let $M$ be the matrix of $A$ relative to an orthonormal basis, so $M$ is symmetric, and hence of the form $$ M=\pmatrix{a & b\cr b & c}. $$ The characteristic polynomial of $A$ is then $$ p(x) = (x-a)(x-c)-b^2 = $$$$ = x^2-(a+c)x+ac-b^2, $$ while its discriminant is $$ \Delta =(a+c)^2-4ac+4b^2 = $$ $$ =a^2+c^2-2ac+4b^2 = (a-c)^2+4b^2, $$ which is positive and hence the roots of $p$ are real.

• Case 3: $\text{dim}(V)\geq 3$.

The characteristic polynomial of $A$ has degree at least three, so it admits a real root, say $\lambda $.

Choose any eigenvector $v$ associated with the eigenvalue $\lambda $, and let $W$ be the space orthogonal to $v$, namely $$ W=\{w\in V:\langle w, v\rangle =0\}. $$ I then claim that $A(W)\subseteq W$. To see this, choose $w\in W$, and notice that $$ \langle Aw, v\rangle = \langle w, Av\rangle = \lambda \langle w, v\rangle = 0. $$ This proves the claim, so we may restrict $A$ to $W$, obtaining an obviously self-adjoint operator $B:W\to W$.

Relative to the decomposition $$ V={\mathbb R}v\oplus W, $$ we then have that $$ A=\lambda 1\oplus B, $$ so it is easy to see that the characteristic polynomial of $A$ coincides with $$ p(x)=(x-\lambda )g(x), $$ where $g$ is the characteristic polynomial of $B$.

Since $\text{dim}(W)=\text{dim}(V)-1$, the induction hypothesis guarentees that all roots of $g$ are real and so we are done!

Answer 3

I came up with another proof which is much more elementary than the attempted argumentation in my initial question: Choosing some basis we know that $\mathbf{A}\in \mathbb{R}^{n, n} $ has some complex eigenvalue, let it be $\lambda $ with corresponding eigenvector $\mathbf{v}$, i.e. $\mathbf{A}\mathbf{v} = \lambda \mathbf{v}$. We can write $\mathbf{v} = \mathbf{v}_{1} + i \mathbf{v}_{2}$ where $\mathbf{v}_{1}, \mathbf{v}_{2} \in \mathbb{R}^{n} $ and $\lambda = \lambda _{1} = i \lambda _{2}$ for $\lambda _{1}, \lambda _{2} \in \mathbb{R}$. This yields the relations \begin{align*} &\mathbf{A}\mathbf{v}_{1} = \lambda _{1} \mathbf{v}_{1} - \lambda _{2} \mathbf{v}_{2} \\[5pt] & \mathbf{A}\mathbf{v}_{2} = \lambda _{2} \mathbf{v}_{1} + \lambda _{1} \mathbf{v}_{2} .\end{align*} Now since $\mathbf{A}$ is self adjoint w.r.t. to the inner product that comes with the real vectorspace $V$ we have $\left\langle \mathbf{A}\mathbf{v}_{1}, \mathbf{v}_{2}\right\rangle = \left\langle \mathbf{v}_{1}, \mathbf{A}\mathbf{v}_{2}\right\rangle $ giving us (plugging in the above relations \begin{align*} &\left\langle \lambda _{1} \mathbf{v}_{1} - \lambda _{2} \mathbf{v}_{2}, \mathbf{v}_{2}\right\rangle = \left\langle \mathbf{v}_{1}, \lambda _{2} \mathbf{v}_{1} + \lambda _{1} \mathbf{v}_{2}\right\rangle \\ & \quad \iff \lambda _{1}\left\langle \mathbf{v}_{1}, \mathbf{v}_{2} \right\rangle - \lambda _{2}\left\langle \mathbf{v}_{2}, \mathbf{v}_{2}\right\rangle = \lambda_{2}\left\langle \mathbf{v}_{1}, \mathbf{v}_{1}\right\rangle + \lambda _{1}\left\langle \mathbf{v}_{2}, \mathbf{v}_{1}\right\rangle \\ & \quad \implies \lambda _{2}( \left\langle \mathbf{v}_{1}, \mathbf{v}_{1}\right\rangle + \left\langle \mathbf{v}_{2}, \mathbf{v}_{2}\right\rangle ) = 0 .\end{align*} Since $\mathbf{v}\neq 0$ by definition it follows that $\lambda _{2} = 0$, as desired.