TITLE: Projection of vectors
QUESTION [1 upvotes]: Compute $:$ $proj_\vec y (\vec x)$ $\vec{x}_1=\begin{bmatrix} 2 \\ 3 \\ 4 \\ 5 \end{bmatrix}, \vec{y}_2 = \begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}$
Since the projection would be $:$ $(-2/0) * \begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}$
and you can't divide by 0.
Does that conclude that there isn't a projection of $:$ $\vec x$ on $\vec y$?

REPLY [0 votes]: Note that the denominator should be the square of the norm of $\vec y$, which is guaranteed to be strictly positive (since $\vec y \neq \vec 0$). It is the dot product of $\vec y$ with itself:
$$
\| \vec y \|^2 = \vec y \cdot \vec y = 1^2 + 0^2 + (-1)^2 + 0^2 = 2
$$
{"set_name": "stack_exchange", "score": 1, "question_id": 1069455}