# Eigenvalues

• Mar 6th 2009, 09:57 PM
Chandru1
Eigenvalues
Let $\displaystyle A$ be an $\displaystyle n \times n$ real matrix such that $\displaystyle A^{2}=I, A \neq \pm{I}$
(where I denotes the n × n-identity matrix). Show that
(i) A has two eigenvalues $\displaystyle \lambda_{1},\lambda_{2}$.
(ii) Every element $\displaystyle x \in \mathbb{R}^{n}$ can be expressed uniquely as $\displaystyle x_{1}+x_{2}$,
where $\displaystyle Ax_{1}=\lambda_{1}x_{1}$ and $\displaystyle Ax_{2}=\lambda_{2}x_{2}$
• Mar 7th 2009, 07:43 AM
Hardwarista
Quote:

Originally Posted by Chandru1
Let $\displaystyle A$ be an $\displaystyle n \times n$ real matrix such that $\displaystyle A^{2}=I, A \neq \pm{I}$
(where I denotes the n × n-identity matrix). Show that
(i) A has two eigenvalues $\displaystyle \lambda_{1},\lambda_{2}$.
(ii) Every element $\displaystyle x \in \mathbb{R}^{n}$ can be expressed uniquely as $\displaystyle x_{1}+x_{2}$,
where $\displaystyle Ax_{1}=\lambda_{1}x_{1}$ and $\displaystyle Ax_{2}=\lambda_{2}x_{2}$

Take two endomorphisms of $\displaystyle \mathbb{R}^{n}$ :
$\displaystyle u = A - I$
$\displaystyle v = A + I$

You have

• $\displaystyle u \circ v = v \circ u = 0$
• $\displaystyle u \neq 0$
• $\displaystyle v \neq 0$

according to the problem.

You see that

• $\displaystyle Ker (u) \cap Ker (v) = \oslash$
• $\displaystyle \mathbb{R}^{n} = Ker (u \circ v) = Ker (v) \cup v^{-1}(Ker (u) \cap Im (v))$

and, after a little manipulation and comaprison of dimensions,
dim Ker (u) + dim Ker (v) = n

Which is sufficient to say that $\displaystyle Ker(u) \oplus Ker(v) = \mathbb{R}^{n}$

$\displaystyle dim Ker (u) \neq 0$, cause it would mean v = 0
$\displaystyle dim Ker (v) \neq 0$, cause it would mean u = 0

So there are effectively 2 eigenvalues for A ( 1 and -1),
and the associated eigenspaces are in direct sum (Ker u and Ker v),
which guarantees the existance and unicity if the decomposition.