Let $W$ be a subspace of $\mathbb{R}^n$ spanned by $n$ non-zero orthogonal vectors. Show that $W=\mathbb{R}^n$.
My approach:
Suppose span$\{u_1,\dots,u_n\}=W$, where$\{u_1,\dots,u_n\}$ is a set of non-zero orthogonal vectors. Then $u_1,\dots,u_n$ are linearly independent vectors. So rank$(U)=n$, where $U=\begin{pmatrix} u_1 & \cdots &u_n \end{pmatrix}$. Since $U$ has $n$ pivots, by the invertible matrix theorem, $U$ spans $\mathbb{R}^n$, thus, $W=\mathbb{R}^n$.
Is this proof valid? I'm kind of worried here, as I am using the invertible matrix theorem to make the implications, and I am treating $U$ to be an $n\times n$ matrix. Am I overlooking something here?
Thanks
Best Answer
Your approach is fine. If you don't wish to use matrices, here is an alternative.
Suppose $ V $ is finite dimensional over $ F $ and $ W $ a subspace of $ V $ such that $ \dim_F(V) = \dim_F(W) $. Then $ V=W $. Perhaps the quickest way to see this is by quotients, $$ \dim_F(V/W) = \dim_F(V) - \dim_F(W) = 0 $$ meaning $ V/W = (\bar{0}) $ or $ V = W $.
You have $ \dim_{\mathbb{R}}(\mathbb{R}^n) = n $. $ W $ is spanned by an orthogonal set $ S $ of $ n $ vectors. This set is linearly independent. This means $ S $ is a basis for $ W $, so $ \dim_{\mathbb{R}} (W) = n $. By the above, $ V = W $.