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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?

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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 $N(M^T) =W^\perp$.

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