Let W₁ and W₂ be subspaces of an n-dimensional Euclidean space V with dim W₁ < dim W₂. Prove that W₂ contains a nonzero vector orthogonal to every vector of W₁.

Step-by-step solution

1. 1

   Restate the goal as a kernel being nontrivial. "A nonzero $v\in W_2$ orthogonal to all of $W_1$" means exactly

   $W_2\cap W_1^{\perp}\ne\{0\}.$

   Turning an existence claim into a dimension claim is what makes the problem routine: dimensions can be counted, individual vectors cannot.

2. 2

   Build a linear map whose kernel is the set in question. Define

   $T:W_2\to W_1^{*},\qquad T(x)(y)=\langle x,y\rangle\ \ \text{for }y\in W_1.$

   $T$ is linear in $x$ because the inner product is. It sends each vector of $W_2$ to the linear functional it induces on $W_1$.

3. 3

   Identify the kernel. $T(x)$ is the zero functional precisely when $\langle x,y\rangle=0$ for every $y\in W_1$, so

   $\ker T=\{x\in W_2:\;x\perp W_1\}=W_2\cap W_1^{\perp}.$

   Proving $\ker T\ne\{0\}$ now proves the theorem.

4. 4

   Bound the image. The image of $T$ lies inside $W_1^{*}$, and $\dim W_1^{*}=\dim W_1$, so

   $\dim(\operatorname{Im}T)\le \dim W_1.$

   This is the only place the hypothesis enters, and it is where the strict inequality will pay off.

5. 5

   Apply rank-nullity. Since $\dim W_2=\dim(\ker T)+\dim(\operatorname{Im}T)$,

   $\dim(\ker T)=\dim W_2-\dim(\operatorname{Im}T)\ge \dim W_2-\dim W_1\ge 1,$

   using $\dim W_2>\dim W_1$ in the last step. A space of dimension at least $1$ contains a nonzero vector, so there is $v\ne 0$ in $W_2$ with $\langle v,y\rangle=0$ for all $y\in W_1$. $\blacksquare$

6. 6

   Check the bound is sharp and the hypothesis necessary. The argument in fact produces a whole subspace of orthogonal vectors of dimension at least $\dim W_2-\dim W_1$. The strict inequality cannot be dropped: in $\mathbb{R}^{2}$ take $W_1=W_2=\operatorname{span}\{(1,0)\}$, which have equal dimension; the only vector of $W_2$ orthogonal to $W_1$ is $0$.