Find the projection of W onto v for

  • Thread starter Thread starter Jamin2112
  • Start date Start date
  • Tags Tags
    Projection
Jamin2112
Messages
973
Reaction score
12

Homework Statement



the given vector v and subspace W.

(a)
Let W be the subspace with basis {(1 1 0 1)T, (0 1 1 0)T, (-1 0 0 1)T} and v = (2 1 4 0)T.


Homework Equations



ProjWv = (<W, v> / <W, W>) * W

The Attempt at a Solution



So I'm trying to wrap my head around this problem by imaging a simpler setup, say, v = (2 1)T and W = span{(1 0)T, (0 1)T}. Visually I see a plane in R2 with the vector (2 1)T sticking out and then any other vector in R2 projected upon it. As for the formula for projection, I'm not sure how I can input a subspace W, an infinite set of vectors. What's up with that? Where do I go?
 
Physics news on Phys.org
That projection formula only holds for vectors since, like you've noted, there's not an inner product for subspaces.

However, you CAN use that formula to project onto a subspace if you have an orthogonal basis. So, generate an orthogonal basis for W and then project v onto each of those vectors. Then add up the results.
 
stringy said:
That projection formula only holds for vectors since, like you've noted, there's not an inner product for subspaces.

However, you CAN use that formula to project onto a subspace if you have an orthogonal basis. So, generate an orthogonal basis for W and then project v onto each of those vectors. Then add up the results.

Will do. By the way, do you know of any orthonormal basis calculators/applets? I couldn't find any through Google.
 
Yeah, calculating ONBs is not fun. I know some computer systems can do it, like Mathematica. I don't know of anything on the web though. However, if you're using the projection formula as you have it written, the basis vectors don't have to be unit length. If you wrote

proj_w \ v = &lt;v,w&gt;,

with basis vector w, THEN they'd have to be unit length.

So you only need an orthogonal basis. And those basis vectors that you wrote are already almost an orthogonal basis!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top