# Invariant subspace of orthogonal transformation

$V$ is Euclidean space, $W$ is it's subspace.
$T$ is an orthogonal transformation T: V $\rightarrow$ V
$W$ is $T$ invariant.
Is W equal to the range of $T(w)$, $w \in W$?

I thought about it visually:
Orthogonal transformation can be viewed as a (rigid or improper) rotation.
Given a unique vector the result of the transformation will also be unique, so the range of $T(w)$ must be the same size than that of W.
Given that W is $T$ invariant: $T(w) \in W$, $w \in W$, then they must be equal.

Is this correct?

Thanks.

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## 1 Answer

Every orthogonal transformation is invertible. This means in particular that it is injective, and then its kernel is zero. Considering $T$ restricted to $W$, you have (by the rank-nullity theorem) $$\dim\Big(\ker \left(T\mid_W\right)\Big) + \dim\Big(T(W)\Big) = \dim( W ).$$ Since $\ker\left(T\mid_W\right) = \{0\}$, we must have $T(W) = W$.

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One should also add that $V$ is finite-dimensional, which OP seems to forget to include. – awllower Jul 13 '13 at 15:32
@awllower You're right, but I think the term "Euclidean space" usually means some $\mathbb{R}^n$, as opposed to "Inner product space", which could mean an infinite-dimensional vector space. – student Jul 13 '13 at 18:32
I agree. In any case, thanks for this good answer. – awllower Jul 13 '13 at 23:35