Real Symmetric Endomorphism: Diagonalizability and Eigenvalues Explained

[penguin007]

Hi,

We know that if u is a real symetric endomorphism, then u has a real eigenvalue and that u is diagonalizable.
But can we say that u is diagonalizable with only real eigenvalues?

 

[HallsofIvy]

Yes. Since you are talking about eigenvalues, I take it that u is an endomorphism on some vector space (a linear transformation from vector space V to itself). Specifically, if u is symmetric and \lambda is an eigenvalue, then there exist non-zero x such that ux= \lambda x. Further, we can take x to have magnitude 1: <x, x>= 1. Then \lambda= \lambda&lt;x, x&gt;= &lt;\lambda x, x&gt;= &lt;ux, x&gt;= &lt;x, ux&gt; (because u is symmetric). &lt;x , ux&gt;= \overline{&lt;ux, x&gt;}= \overline{\lambda x, x}= \overline{\lambda}\overline{&lt;x, x&gt;}= \overline{\lambda}.
(The overline indicates complex conjugation.)

Since \lambda= \overline{\lambda}, \lambda is real. That is all eigenvalues are real for a symmetric endomorphism.

 

[penguin007]

I think you forgot an overline in the end, but I can't understand why we have:
<x,ux>=conjugate(<ux,x>) ?

 

[HallsofIvy]

I think you forgot an overline in the end, but I can't understand why we have:
<x,ux>=conjugate(<ux,x>) ?

One of the requirements in the definition of "inner product" is that &lt;u, v&gt;= \overline{&lt;v, u&gt;}. Of course, if the vector space is over the real numbers, that is the same as "<u, v>= <v, u>" but then your question is trivial.