# Operators on Odd-Dimensional Real Vector Spaces Eigenvalue Existence Proof

• November 25th 2010, 02:07 PM
Anthonny
Operators on Odd-Dimensional Real Vector Spaces Eigenvalue Existence Proof
I'm having trouble seeing how induction is used in this proof for this theorem which states that every operator on an odd-dimensional real vector space has an eigenvalue.

The proof starts by supposing V is a real vector space with odd dimension. Then the proof states that it will be using induction in step sizes of 2 on the dimension of V (which I understand since we're proving this for odd dimensions).

The induction starts with when dimV=1 then the operator on V has an eigenvalue which I understand.

Then the proof supposes that dimV is an odd number greater than 1. Using induction we can assume that the desired result holds for dimensions 2 less than dimV.

I can't clearly see how this is true i.e. how, say, if dimV=3 then the operator on V has an eigenvalue.

Thank you.
• November 25th 2010, 03:36 PM
tonio
Quote:

Originally Posted by Anthonny
I'm having trouble seeing how induction is used in this proof for this theorem which states that every operator on an odd-dimensional real vector space has an eigenvalue.

Hint (the size of an elephant): every real polynomial of odd degree has a real root.

Oh, and no induction at all needed.

Tonio

The proof starts by supposing V is a real vector space with odd dimension. Then the proof states that it will be using induction in step sizes of 2 on the dimension of V (which I understand since we're proving this for odd dimensions).

The induction starts with when dimV=1 then the operator on V has an eigenvalue which I understand.

Then the proof supposes that dimV is an odd number greater than 1. Using induction we can assume that the desired result holds for dimensions 2 less than dimV.

I can't clearly see how this is true i.e. how, say, if dimV=3 then the operator on V has an eigenvalue.

Thank you.

.
• November 25th 2010, 10:17 PM
Anthonny
Quote:

Originally Posted by tonio
.

I haven't really learned eigenvalues using the characteristic polynomial, but rather solving them directly using linear transformations.

However, I believe from this hint the characteristic polynomial of an operator on an odd-dimensional real vector space is of odd-degree which from this hint implies that it has a real root and therefore an eigenvalue, although I'm not completely sure.

Although this is a completely valid proof, my text presents the proof using induction

Thanks anyway
• November 25th 2010, 11:38 PM
FernandoRevilla
Quote:

Originally Posted by Anthonny
Although this is a completely valid proof, my text presents the proof using induction

Suppose $\dim V=2n+1$ and consider a basis on $V$:

$B=\left\{e_1,\ldots,e_{2n-1},e_{2n},e_{2n+1} \right\}$

Denote: $F_1=,\;F_2=
$
( so, $V=F_1\oplus{F_2}$ )

Consider a linear map: $T:V\rightarrow{V}$ . Then:

$\widehat{T}:F_1\rightarrow{F_1},\;\widehat{T}(x)=T (x)$ is a linear map.

Could you continue?.

Regards.

Fernando Revilla