# Thread: Prove that eigenspace is a T-invariant subspace

1. ## Prove that eigenspace is a T-invariant subspace

Let T: V -> V, V is a finite-dimensional vector space
T^2 is identity operator.
lamda be a scalar.
The eigenspace V^(lamda) is the set of eigen-vectors of T with eigen-value lamda, together with zero.

Prove that V^(lamda) is a T-invariant subspace.

Prove that for all v in V, v - Tv is either an eigen-vector with eigen-value -1 or zero vector

Prove that V is direct sum of the eigenspaces V^(1) and V^(-1)

2. Originally Posted by mathbeginner
Let T: V -> V, V is a finite-dimensional vector space
T^2 is identity operator.
lamda be a scalar.
The eigenspace V^(lamda) is the set of eigen-vectors of T with eigen-value lamda, together with zero.

Prove that V^(lamda) is a T-invariant subspace.

Prove that for all v in V, v - Tv is either an eigen-vector with eigen-value -1 or zero vector

Prove that V is direct sum of the eigenspaces V^(1) and V^(-1

Let $E_\lambda$ be the eigenspace with eigenvalue lambda. Then if
$v \in E_\lambda$ then

$T(v)=\lambda v; T^2(v)=T(T(v))=T(\lambda v)=\lambda^2v$

Notice that both $T(v) \in E_\lambda$ and $T^2(v) \in E_{\lambda}$

w is an eigenvector if $T(w)=\lambda w$ and $w \ne 0$

Calculate

$T(v-T(v))=T(v)-T^2(v)=...$

This should get you started.

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# show that eigenspace is invariant

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