# Help with projection matrix!

• Jul 11th 2009, 03:19 AM
ypatia
Help with projection matrix!
Let V subspace of R3 with inner product V={(x,y,z) in R3 : x+z=0, y=2x}

1. how can we find the projection P onto V

2. how can we find the projection matrix P as of the normal basis of R3

3. how can we find a basis of R3 for which the matrix P will be $\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$

• Jul 11th 2009, 06:48 AM
hatsoff
Quote:

Originally Posted by ypatia
Let V subspace of R3 with inner product V={(x,y,z) in R3 : x+z=0, y=2x}

1. how can we find the projection P onto V

2. how can we find the projection matrix P as of the normal basis of R3

3. how can we find a basis of R3 for which the matrix P will be $\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$

Clearly, we have a $\beta=\{(1,2,-1)\}$ a basis for $\mathsf{V}$, and $\mathsf{T}:\mathbb{R}^3\to\mathbb{R}^3$ with $\mathsf{T}((x,y,z))=(x,2x,-x)$ a projection. $\mathsf{T}$ is probably not an orthogonal projection, however.

If we want to normalize this, the set $\gamma=\{(\sqrt{6}/6,\sqrt{6}/3,-\sqrt{6}/6)\}$ is a normal basis for $\mathsf{V}$. For ease of notation, let $\alpha$ refer to the standard basis of $\mathbb{R}^3$. Then

$[\mathsf{U}]^{\gamma}_{\alpha}=\left(\begin{array}{ccc}\sqrt{6 }/6&0&0\\\sqrt{6}/3&0&0\\-\sqrt{6}/6&0&0\end{array}\right)$

where $\mathsf{U}:\mathbb{R}^3\to\mathsf{V}$ has the same definition as $\mathsf{T}$, other than its range. So $[\mathsf{U}]^{\gamma}_{\alpha}$ is the projection matrix of $\mathsf{T}$. Please note again that this is probably not orthogonal.

However, the matrix

$P=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$

will never be a projection matrix for any projection of $\mathbb{R}^3$ along $\mathsf{V}$. For consider $(x,y,z)\in\mathbb{R}^3$ with $x\neq 0$. Then $P\cdot (x,y,z)^T=(x,0,0)^T$. Since $x\neq 0$, we have $2x,-x\neq 0$, so $(x,0,0)\neq (x,2x,-x)$ $\Rightarrow$ $(x,0,0)\notin \mathsf{V}$.