# Why is $W^\perp$ the null space of the basis of $W$?

Let $$W$$ be a subspace of $$\mathbb{R}^n$$ and $$\{w_1, w_2, \cdots, w_k\}$$ form a basis for $$W$$. Form the matrix $$M$$ that has these basis vectors as successive columns. According to my book, $$W^\perp$$ is the null space of $$M^T$$? Why is this?

I would have thought that $$W^\perp$$ is the null space $$M$$ because I believe that the vectors that are orthogonal to $$W$$ will be mapped to the origin by $$M$$. Am I wrong on this and/or drawing a false conclusion?

Let $$x \in W^\perp$$. Then $$(M^Tx)_{i} = w_i \cdot x = 0$$, since $$x$$ is orthogonal to $$w_i$$. This is seen from the definition of transpose and matrix multiplication. Thus $$W^{\perp} \subseteq N(M^T)$$. It isn't necessarily the null space for $$M$$ since matrix multiplication will lead to the dot product of $$x$$ and the $$i^{\mathrm{th}}$$ row of $$M$$, which is comprised of the $$i^{\mathrm{th}}$$ entries of each basis vector, which doesn't really tell us anything about the value.
Showing reverse inclusion: If $$y \in N(M^T)$$, then $$w_i \cdot y = 0$$ for all $$i$$. Thus, $$N(M^T) \subseteq W^\perp$$, and so we have $$N(M^T) =W^\perp$$.