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**Tekken** Let **w** = $\displaystyle \left(\begin{array}{ccc}-1\ \-1\ \-1\end{array}\right)$$\displaystyle

$ Find the projection of w on space V.

$\displaystyle V = $ $\displaystyle <\left(\begin{array}{ccc} 1\ \ 4\ \ 0\end{array}\right)\left(\begin{array}{ccc} 1\ \ -1\ \ 2\end{array}\right)>$

I've already found $\displaystyle V^{perp} $ $\displaystyle = $ the span of $\displaystyle \left(\begin{array}{ccc} -4\ \ 1\ \ -1\end{array}\right)$

This can't possibly be correct since $\displaystyle (1\,-\!\!1\,\,2)\cdot (-4\,\,1\,-\!\!1)\neq 0$ ....check your work: it must be $\displaystyle V^{\perp}=Span\{(-8\,2\,5)\}$

I worked out the projection matrix i.e. $\displaystyle A^T(A.A^T)^{-1}.A $ to be 3x3 matrix where $\displaystyle A = \left(\begin{array}{ccc} -4\ \ 1\ \ -1\end{array}\right) $

?? How a matrix with one single row is a 3x3 matrix? What did you actually mean here?

Tonio

However i dont now where to bring **w** into the problem