Help with projection matrix!

• Jul 11th 2009, 02:19 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 $\displaystyle \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$

• Jul 11th 2009, 05:48 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 $\displaystyle \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$

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

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

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

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

However, the matrix

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

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