2
$\begingroup$

Let $V$ be a subspace of $\mathbb{R}^n$ with $V \ne \mathbb{R}^n$ and $V\ne \{0\}$. Let $A$ be the matrix of the linear transformation $\text{proj}_V : \mathbb{R}^n\to\mathbb{R}^n$ that is the projection onto $V$.

Calculate $A$ in the case where $n=3$ and $V$ is given by the equation $x+y+z=0$.

I am not really sure how to approach this question. I am aware that the normal vector will be $(1,1,1)$. And I know that the orthogonal projection onto the plane $V$ given by the equation $x+y+z=0$ is equal to the identity minus the orthogonal projection onto the orthogonal complement. I believe the orthogonal complement will be in the span of the normal vector (not entirley sure why?) but I am also unsure how to find the orthogonal complement.

$\endgroup$
1
$\begingroup$

You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.

Now, since$$(1,0,0)=\frac13(1,1,1)+\frac13(1,-1,0)+\frac13(1,0,-1),$$you must have$$A.(1,0,0)=\frac13(1,-1,0)+\frac13(1,0,-1)=\left(\frac23,-\frac13,-\frac13\right).$$So, the entries of the first column of the matrix of $\operatorname{proj}_V$ with respect to the standard basis will be $\frac23$, $-\frac13$ and $-\frac13$. Can you take it from here?

$\endgroup$
  • $\begingroup$ Where are you getting the 1/3 from? $\endgroup$ – Molly Jul 27 '18 at 20:50
  • $\begingroup$ From solving the system $(1,0,0)=\alpha(1,1,1)+\beta(1,-1,0)+\gamma(1,0,-1)$. $\endgroup$ – José Carlos Santos Jul 27 '18 at 21:11
1
$\begingroup$

Hint:

The projection of a vector $\vec u$ onto the normal to the plane $x+y+z=0$, with normal vector $\vec n=(1,1,1)$, is given by the formula: $$p_{\vec n}(\vec u)=\vec u-\frac{\vec n\cdot\vec u}{\vec n\cdot\vec n}\,\vec n.$$ Apply this formula to each vector of the canonical basis to find the column vectors of the projection matrix.

$\endgroup$
0
$\begingroup$

As you say, $u:=(1,1,1)^T$ is a normal vector to the plane, thus it will span the orthogonal complement.

Its projection is given by $\varphi:v\mapsto\frac{u^Tv}{u^Tu}u$.
You can check that it fixes $u$, but gives $0$ whenever $v\perp u$.
To obtain its matrix in the standard basis, just calculate $\varphi(e_i)$ for the standard basis $e_1,e_2,e_3$.

$\endgroup$
0
$\begingroup$

$\left\{\frac1{\sqrt{2}}(1,-1,0), \frac1{\sqrt{6}}(1,1,-2)\right\}$ is an orthonormal basis for $V$ so the projection is given by

\begin{align} P(x,y,z) &= \left\langle (x,y,z), \frac1{\sqrt{2}}(1,-1,0)\right\rangle\frac1{\sqrt{2}}(1,-1,0)+\left\langle (x,y,z), \frac1{\sqrt{6}}(1,1,-2)\right\rangle\frac1{\sqrt{6}}(1,1,-2) \\ &= \frac12 (x-y, -x+y,0) + \frac16(x+y-2z,x+y-2z,-x-y+4z)\\ &= \frac16 (4x-2y-2z, -2x+4y-2z, -2x-2y+4z) \end{align}

so $$A = \frac16\begin{bmatrix} 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \end{bmatrix}$$

$\endgroup$
0
$\begingroup$

The orthogonal projection of $(x,y,z)$ onto the plane through the origin with normal $(1,1,1)$ is the unique $(x',y',z')$ on the plane such that $$ ((x,y,z)-(x',y',z')) = c(1,1,1), \\ x'+y'+z'=0. $$ Dotting the first equation with $(1,1,1)$ and using $x'+y'+z'=0$ gives $$ c = \frac{1}{3}(x+y+z). $$

So the projection of $(x,y,z)$ onto the given plane is

\begin{align} (x',y',z') &= (x,y,z)-\frac{1}{3}(x+y+z)(1,1,1) \\ &= \frac{1}{3}(2x-y-z,-x+2y-z,-x-y+2z). \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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