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I am referencing Chad's answer from this link: https://math.stackexchange.com/a/2997916/

I have also added an image from the referenced link: enter image description here

If b is being projected..shouldn't the math be the other way around?

Theo Bendit
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  • What do you mean by "the other way around"? Do you mean that the two vectors should be swapped, i.e.$$p = \frac{\begin{bmatrix} 5 \ 1 \end{bmatrix} \cdot \begin{bmatrix} 4 \ 6 \end{bmatrix}}{\begin{bmatrix} 5 \ 1 \end{bmatrix} \cdot \begin{bmatrix} 5 \ 1 \end{bmatrix}} \begin{bmatrix} 5 \ 1 \end{bmatrix}?$$ – Theo Bendit Jul 05 '21 at 17:13
  • Yes. That is what I meant. –  Jul 05 '21 at 17:14
  • Then no, the original verion is correct. Note that projecting onto $C(A)$ should produce something parallel to $\begin{bmatrix}4\6\end{bmatrix}$, whereas the above is parallel to $b$ instead. You should be able to take the final result $\begin{bmatrix}2\3\end{bmatrix}$ and show that $b-\begin{bmatrix}2\3\end{bmatrix}$ is perpendicular to $b$. This will establish that $\begin{bmatrix}2\3\end{bmatrix}$ is indeed the perpendicular projection onto the span of $\begin{bmatrix}4\6\end{bmatrix}$. – Theo Bendit Jul 05 '21 at 17:18
  • Ah that is true. Isn't the projection not just parallel to C(A) but directly on the same line as C(A)? –  Jul 05 '21 at 17:26
  • Yes, but we're talking about vectors anchored at the origin (as we usually are when we represent them simply by coordinates). So, two vectors being parallel is equivalent to the lying in the same line containing the origin. – Theo Bendit Jul 05 '21 at 17:29
  • I only know vectors when represented by coordinates. What are the other ways vectors are represented? –  Jul 05 '21 at 17:40
  • Sometimes by a pair of points, like $\overrightarrow{AB}$. – Theo Bendit Jul 05 '21 at 17:41
  • Let us continue this discussion in chat. –  Jul 05 '21 at 17:44

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