Let $V$ be a $\mathbb{K}$-vector space with a dot product. Let $T \in L(V)$ (endomorphism) be a self-adjoint operator ($\forall x, y \in V, T(x) \cdot y = x \cdot T(y)$)
Let $W$ be an invariant subspace of $V$ by $T$.
Let $W^{\perp}=\{y \in V , y\cdot w = O, \forall w \in W\}$ be a subspace of $V$ (we know it to be invariant by $T$).
- Then the union of the orthonormal basis of $W$ and $W^{\perp}$ is an orthonormal basis of $V$
The proof goes like this :
"Using Gram-Schmidt's method, we can orthonormalize any basis of $W$ and $W^{\perp}$. The reunion of both orthonormalized basis is an orthonormal basis of V"
I hav no further information
Could someone clarify as to why the reunion of both orthonormalized basis is an orthonormal basis of $V$?
Best Answer
Suppose you have a basis $V_W$ for $W$ and a basis $V_{W^{\perp}}$ for $W^{\perp}$. Using Gram-Schmidt's method, you find a orthonormal basis $V^{'}_W$ for $W$ and a orthonormal basis $V^{'}_{W^{\perp}}$ for $W^{\perp}$.
Now, you know that $W \cap W^{\perp} = \emptyset$, so given $u_1 \in V^{'}_{W}$ and $u_2 \in V^{'}_{W^{\perp}}$, you have that they are orthogonal. So, $V^{'}_{W^{\perp}} \cup V^{'}_{W}$ is an orthogonal basis for V.
It is trivial that each vector in $V^{'}_{W^{\perp}} \cup V^{'}_{W}$ has norm 1.