Suppose we are given a vector space $V$ with $dim(V)=n$. Then if we have a subspace $W\subseteq V$ with $dim(W)<n$, is $W^{\perp}$, the orthogonal complement of $W$, necessarily non-empty (discounting the zero vector)? It seems to follow implicitly from the fact that $dim(V) = dim(W) + dim(W^{\perp})$, but I am not quite sure.
Is the orthogonal complement always non-empty
linear algebra
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Best Answer
Yes, assuming $V$ is an inner-product space (such as $\mathbb{R}^n$). Choose an orthonormal basis $B = \{v_1, v_2, \ldots, v_k\}$ of $W$, and extended it to a basis $A = \{v_1, v_2, \ldots, v_n\}$ of $V$. Then perform Gram-Schmidt on $A$. Check that in Gram-Schmidt, if the first $k$ basis vectors are already orthonormal, they will not be changed; so we get a new basis $A' = \{v_1, v_2, \ldots, v_k, w_{k+1}, \ldots, w_n\}$. By definition Gram-Schmidt gives us an orthonormal basis, so the $(k+1)$-st basis element $w_{k+1}$ is orthogonal to the first $k$, which are the basis vectors for $B$. So $w_{k+1}$ is orthogonal to $W$. In fact, we get a basis $\{w_{k+1}, \ldots, w_n\}$ for $W^\perp$.