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For a finite-dimensional inner product space over $\mathbb{C}$, it is clear that every linear transformation is diagonalisable. In my lecture notes, the lecturer claims that:

For a finite-dimensional inner product space, every self-adjoint transformation has at least one eigenvector.

This fact is then used in a proof of the Spectral Theorem for Hermitian matrices.

Over $\mathbb{R}$ for example, we have - in general - no reason to assume that a linear map has an eigenvalue, so why is it the case that a self-adjoint map must have one?

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    $\begingroup$ It is not clear, nor even true, that every linear transformaton is diagonalisable. $\endgroup$ Commented Jun 11, 2013 at 12:47

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Note: over a finite-(nonzero)-dimensional complex vector space, every linear transformation has at least one eigenvalue, since the characteristic polynomial splits. It follows by induction that it is upper trianguarizable. But not every linear transformation, even over $\mathbb{C}$, is diagonalizable. Typically, consider the Jordan block $$ T=\pmatrix{0&1\\0&0}. $$

Key point: for a self-adjoint transformation, every eigenvalue is actually real. Indeed, if $T^*=T$ and if $Tx=\lambda x$ with $x\neq 0$, then $$ \lambda (x,x)=(x,\lambda x)=(x,Tx)=(Tx,x)=(\lambda x,x)=\overline{\lambda}(x,x)\quad\Rightarrow\quad \lambda=\overline{\lambda} $$ where I took an inner-product antilinear in the first variable.

Conclusion: it suffices to prove the result for matrices, up to taking the matrix $A$ of $T$ in an orthonormal basis. Since the characteristic polynomial of $A$ splits over $\mathbb{C}$ and since every root is real, it splits over $\mathbb{R}$. And the characteristic polynomial of $A$ is the same, whether you look at it as an element of $M_n(\mathbb{R})$, or as an element of $M_n(\mathbb{C})$. So the charateristic polynomial of $T$ splits over $\mathbb{R}$. And in general, for a linear transformation $T$ over a finite-dimensional $K$-vector space, every root $\lambda$ in $K$ of the characteristic polynomial in $K[X]$ is an eigenvalue of $T$. And conversely. Since both are equivalent to $T-\lambda \mbox{Id}$ not being invertible. Via the determinant for the root side. And via the rank-nullity theorem on the eigenvalue side.

Another way to reach the conclusion: now take $A$ the matrix of $T$ in an orthonormal basis. Then $T$ is self-adjoint if and only if $A^*=A$ is Hermitian (equal to its transconjugate). Over $\mathbb{R}$, this is just $A^T=A$ (i.e. $A$ symmetric, equal to its transpose). But we can see $A$, even if it is in $M_n(\mathbb{R})$, as an element of $M_n(\mathbb{C})$. So it has at least an an eigenpair $(\lambda,x)$ in $\mathbb{C}\times \mathbb{C}^n$. But as we have just shown, $\lambda $ must be real. Therefore, writing $x=y+iz$ where $y$ and $z$ are the real and imaginary parts of $x\in\mathbb{C}^n$ in $\mathbb{R}^n$, we get $$ Ay+iAz=A(y+iz)=Ax=\lambda x=\lambda (y+iz)=\lambda y+i\lambda z\quad \Rightarrow \quad Ay=\lambda y\quad Az=\lambda z. $$ Si $x\neq 0$, one of $y,z$ must be nonzero, giving a real eigenpair for $A$. Whence for $T$ going back from the matrix to the operator.

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    $\begingroup$ Nice explanation. It would seem to me however that the fact implicitly used in the "Conclusion" paragraph (namely that is a linear transformation of a real vector space has a real eigenvalue then it also has an actual (i.e., real) eigenvector) could have been used in the "Another way" paragraph as well. Now reading the last paragraph gives the impression something was incomplete in the previous one. The mentioned fact is just that finding $\ker (A-\lambda I)$ is linear algebra, can can be done over the reals. $\endgroup$ Commented Jun 11, 2013 at 13:33
  • $\begingroup$ @MarcvanLeeuwen Right. I'll add the conclusion to the conclusion. For the last paragraph, I added it because I like this real/imaginary part argument. $\endgroup$
    – Julien
    Commented Jun 11, 2013 at 13:41
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This answer assumed the question was about complex inner product spaces (it does not say the contrary); probably wrongly so though. I'll leave it until the question is more explicit about this.

Self-adjointness is a red herring (but instead you need your space to have nonzero dimension). Every linear transformation of a finite nonzero dimensional complex vector space has at least one eigenvalue, and hence an eigenvector (in fact infinitely many of the latter). Just take a root of the characteristic polynomial for the eigenvalue (the minimal polynomial will do just as well).

This does not of course mean such a linear transformation is diagonalisable.

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    $\begingroup$ I think the OP was concerned about the case of a real inner-product space. $\endgroup$
    – Julien
    Commented Jun 11, 2013 at 13:14
  • $\begingroup$ @julien: Yes, upon rereading it would seem so. I was put on the wrong footing by the opening sentence that mentions a complex inner product space; one might assume it remains complex until further notice, but indeed probably the cited sentence does not suppose that. $\endgroup$ Commented Jun 11, 2013 at 13:18

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