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

1. ## 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.

2. 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.
.

3. 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

4. Originally Posted by Anthonny
Although this is a completely valid proof, my text presents the proof using induction
Suppose $\displaystyle \dim V=2n+1$ and consider a basis on $\displaystyle V$:

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

Denote: $\displaystyle F_1=<e_1,\ldots,e_{2n-1}>,\;F_2=<e_{2n},e_{2n+1}>$ ( so, $\displaystyle V=F_1\oplus{F_2}$ )

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

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

Could you continue?.

Regards.

Fernando Revilla