Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the $\operatorname{Proj}_wv$ for the given vector $v$ and subspace $W$. Let $V$ be the Euclidean space $\mathbb{R}^4$, and $W$ the subspace with basis $[1, 1, 0, 1], [0, 1, 1, 0], [-1, 0, 0, 1]$

(a) $v = [2,1,3,0]$

ans should be - $[7/5,11/5,9/5,-3/5]$

My attempt at the solution was basically we can find the basis perpendicular to $W$ as $[ 1,-2,2, 1]$ then, $[2, 1, 3, 0] = a[1, 1, 0, 1] + b[0, -1, 1, 0] + c[0 ,2, 0,3] + d[1,-2,2,1]$ We solve for $a,b,c,d$ and get $a = 16/3,b=29/3,c=-2/3,d=-10/3$ now the problem is what do I do from here?

share|cite|improve this question
To type $\mathbb{R}^4$, type $\mathbb{R}^4$. – Arturo Magidin Nov 29 '11 at 22:54
If you find a basis for $W^\perp$ that is orthogonal to $W$, say with Gram-Schmidt, finding the projecting is much easier. – Joe Johnson 126 Nov 29 '11 at 23:10
Or, finding an orthonormal basis for $W$ itself makes finding the projection very easy. – Arturo Magidin Nov 29 '11 at 23:15
up vote 2 down vote accepted

You can do it that way (though you must have an arithmetical error somewhere; the denominator of $3$ cannot be right), and the remaining piece is then simply to take $a[1, 1, 0, 1] + b[0, -1, 1, 0] + c[0 ,2, 0,3]$, forgetting the part perpendicular to $W$.

However, it is much easier to normalize your $[1,-2,2,1]$ to $n=\frac{1}{\sqrt{10}}[1,-2,2,1]$ -- then the projection map is simply $v\mapsto v - (v\cdot n)n$. (If you write that out fully, the square root even disappears).

share|cite|improve this answer
I did that and I end up getting $(9/5,3/5,9/5,0)$. Just redid it here to see if I had made an error - on WolframAlpha – eWizardII Nov 29 '11 at 23:33
Ah, nevermind figured it out I forgot to take the dot product - that was probably my error earlier with the other approach - thanks alot! – eWizardII Nov 29 '11 at 23:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.