# Some problem here :(

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• Sep 1st 2009, 05:31 AM
pdnhan
Can anyone please explain some Subspace problem here :(
Let F: R3->R3 be the linear transformation defined by the orthogonal projection of (v belongs to R3) onto the subspace W= {(x,y,z)|x+y+z=0}
a. Find the standard matrix A of F
b. Show that A(A - I) = 0 Why would this be true in general (for any subspace W)?

thanks in advance :D
• Sep 1st 2009, 12:26 PM
Opalg
Quote:

Originally Posted by pdnhan
Let F: R3->R3 be the linear transformation defined by the orthogonal projection of (v belongs to R3) onto the subspace W= {(x,y,z)|x+y+z=0}
a. Find the standard matrix A of F
b. Show that A(A - I) = 0 Why would this be true in general (for any subspace W)?

The vector normal to W is $\displaystyle \mathbf{n}=(1,1,1)$. Given a vector $\displaystyle \mathbf{v} = (x,y,z)$ in $\displaystyle \mathbb{R}^3$, the effect of F on $\displaystyle \mathbf{v}$ will be to add a multiple of $\displaystyle \mathbf{n}$ to $\displaystyle \mathbf{v}$ so as to take it to the subspace W. The condition for $\displaystyle \mathbf{v} + \lambda\mathbf{n} = (x+\lambda,y+\lambda,z+\lambda)$ to belong to W is $\displaystyle x+y+z+3\lambda = 0$. Solve that to see that $\displaystyle F(\mathbf{v}) = (\tfrac23x-\tfrac13y-\tfrac13z, -\tfrac13x+\tfrac23y-\tfrac13z, -\tfrac13x-\tfrac13y+\tfrac23z)$. From that, you should be able to write down the matrix A and verify that $\displaystyle A(A-I)=0$.

For the last part, if A is the matrix of the orthogonal projection onto a subspace, then I–A is the matrix of the orthogonal projection onto the (orthogonal) complementary subspace.
• Sep 1st 2009, 11:36 PM
pdnhan
cheers man
• Sep 2nd 2009, 02:28 PM
pdnhan
hey man, can you please tell me how to get the value of A so I can compare, and your solution to part b) as well?
• Sep 3rd 2009, 12:32 AM
Opalg
Quote:

Originally Posted by pdnhan
hey man, can you please tell me how to get the value of A so I can compare, and your solution to part b) as well?

You tell us yours first, then I'll let you know whether I agree. (Evilgrin)