$\dim W + \dim W^\perp = n$ Let $W$ be a subspace of $\mathbb{R}^n$. We know that $W^\perp$ is also a vector subspace of $\mathbb{R}^n$. How can I show that $\dim W + \dim W^\perp = n$?
I know that for a matrix $A$, $(Lgn A)^\perp = Ker A$ and $(ImaA)^\perp = Ker A^T$. I think it might help, but I am really not sure how I can use it.
 A: You should prove that given $V$ vector space over $F$ then
$V=W \oplus W^{\perp}$  this is  you should prove that
$$\forall v\in V v=w+w^{\prime}$$ where $w\in W$ and $w^{\prime}\in W^{\perp}$
$$W \cap W^{\perp}= \lbrace 0 \rbrace$$
Of this fact we can deduce
$\dim V= \dim W + \dim W^{\perp}$
A: Let $A$ be a matrix whose columns are a basis for $W$. If $W = \operatorname{Col}(A)$ then $W^\perp = \operatorname{Null}(A^\top)$.
By Rank-Nullity, we know that
$$\dim \operatorname{Null}(A^\top) + \dim \operatorname{Col}(A^\top) = n$$
We also know that
$$\dim \operatorname{Null}(A^\top) = \dim W^\perp, \text{ and } \dim \operatorname{Col}(A^\top) = \dim \operatorname{Col}(A) = \dim W$$
since $A$ and $A^\top$ have the same rank. The result follows.
A: Start with an orthonormal basis of $W$, say $\{v_1, v_2, ... v_k \}$ and extend it to an orthonormal basis of $\mathbb{R}^n$. Since it has dimension $n$, this would then be $\{v_1, ... v_k, v_{k+1}, ..v_n \}$. Then by definition we have that $\{v_{k+1}, .. v_n\}$ is a linearly independent set in $W^{\perp}$. And since $\{v_1 .. v_n \}$ is a basis for $\mathbb{R}^n$, we simply have that for any $v \in \mathbb{R}$, $v = a_1 v_1 + ... a_kv_k + a_{k+1}v_{k+1} + ... a_{n}v_n  = (a_1v_1 + .... a_kv_k) + (a_{k+1}v_{k+1} + ... a_nv_n)$. The former is a vector in $W$, and the latter in $W^{\perp}$. Thus every vector in $\mathbb{R}^n$ is the sum of vectors in $W$ and $W^{\perp}$. Thus $\mathbb{R}^n = W + W^{\perp}$, and clearly $W \cap W^{\perp} = \emptyset$ since if $v \in W \cap W^{\perp}$, then $<v, v> = 0$ which means $v = 0$.
So we have $\mathbb{R}^n = W + W^{\perp}$ and $W \cap W^{\perp} = 0$, from which the result follows.
