1. ## [SOLVED] Projections question

Can you obtain the same vector by first projecting y onto V, then projecting this vector on x as by projecting y directly on x or more generally , if $V$ is a subspace and $V_1$ is a subspace of $V$, does $p(p(\textbf{y}|V)|V_1) = p(\textbf{y}|V_1)$

From the way the question is phrased it seems like its true, but I was wondering if there was an intuitive explanation as to why.

2. Originally Posted by jass10816
Can you obtain the same vector by first projecting y onto V, then projecting this vector on x as by projecting y directly on x or more generally , if $V$ is a subspace and $V_1$ is a subspace of $V$, does $p(p(\textbf{y}|V)|V_1) = p(y|V_1)$

From the way the question is phrased it seems like its true, but I was wondering if there was an intuitive explanation as to why.
I'm not sure about your question nor your notation. But let me guess: suppose we are given vectors x and y such that the direct projection of y on x gives a non-null vector. Now, I assume that, given a vector y like this, it is possible to find a subspace V such that y, when projected onto V, gives the null-vector. If so: we have rejected the above proposition, because no amount of further projecting that projection of y will ever give something other than the null-vector.

3. Okay, but as long as the projection of y onto V is non-null, then the equality holds?

4. Originally Posted by jass10816
Okay, but as long as the projection of y onto V is non-null, then the equality holds?
Of course not. Consider a quite elementary case in $\mathbb{R}^3$: a vector y that has positive $x_1, x_2$ and $x_3$ components, say y=(1,1,1), gets first projected onto the $x_2x_3$-plane (your V), this reduces its $x_1$ component to 0. Then you project that projection (0,1,1) onto the $x_1$-axis (i.e. the direction of a vector that points in the positive direction of the $x_1$-axis): you get the null vector (0,0,0).
But if you project that vector directly onto the $x_1$-axis, you get a non-null vector, namely (1,0,0).