Find the orthogonal complement of W

Question

Consider the finite inner product space $\mathbb{R}^3$ where the inner product is the normal dot product.

Find the orthogonal complement of W where W is

$W = {(x,y,x+y) : x,y \in \mathbb{R}} \subset V$

Further find an orthonormal basis for W.

Then if $f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be the linear map given by $f(x,y,z)=(x,y,x+y)$. Show that there is no orthonormal basis of $\mathbb{R}^3$ for which this map is diagonal$.

I cannot see how to do this as when I take a general dot product it doesnt seem to lead anywhere.

Answer 1

Hints:

$$(a,b,c)\in W^\perp\iff \forall\,(x,y,z)\in W\,,\;\;(a,b,c)\cdot(x,y,z)=ax+by+cz=0$$

But we know that $\;z=x+y\;$ , so the rightmost condition becomes

$$ax+by+c(x+y)=0\iff (a+c)x+(b+c)y=0$$

Now, how and with how many degrees of freedom (i.e., how can you independently) choose $\;a,b,c\;$ so that the above equality is true for all $\;x,y\in\Bbb R\;$ ? This will give you not only $\;W^\perp\;$ but also its dimension...

Answer 2

You can write $$ \left[\begin{array}{c} x \\ y \\ x+y \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]. $$ Now you can see that you're looking for a vector which is orthogonal to column space of the above 3x3 matrix $M$. An easy-to-spot vector orthogonal to the column space is $(1,1,-1)$. The second part is addressed by showing that you cannot diagonalize $M$ with an orthonormal basis.