# [SOLVED] Projections question

• Apr 7th 2010, 06:21 PM
jass10816
[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 \$\displaystyle V\$ is a subspace and \$\displaystyle V_1\$ is a subspace of \$\displaystyle V\$, does \$\displaystyle 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.
• Apr 7th 2010, 11:58 PM
Failure
Quote:

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 \$\displaystyle V\$ is a subspace and \$\displaystyle V_1\$ is a subspace of \$\displaystyle V\$, does \$\displaystyle 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.
• Apr 8th 2010, 06:37 AM
jass10816
Okay, but as long as the projection of y onto V is non-null, then the equality holds?
• Apr 8th 2010, 06:54 AM
Failure
Quote:

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 \$\displaystyle \mathbb{R}^3\$: a vector y that has positive \$\displaystyle x_1, x_2\$ and \$\displaystyle x_3\$ components, say y=(1,1,1), gets first projected onto the \$\displaystyle x_2x_3\$-plane (your V), this reduces its \$\displaystyle x_1\$ component to 0. Then you project that projection (0,1,1) onto the \$\displaystyle x_1\$-axis (i.e. the direction of a vector that points in the positive direction of the \$\displaystyle x_1\$-axis): you get the null vector (0,0,0).
But if you project that vector directly onto the \$\displaystyle x_1\$-axis, you get a non-null vector, namely (1,0,0).