Orthogonal basis and projection I was hoping someone could explain to me using the vectors below and $V=Span\{v_1,v_2,v_3\}$ how I could determine the orthogonal basis for $V$ and the projection of $y$ onto $V$. I have been googling for examples / explanations but I'm honestly completely stuck. 

Thank you very much :)
 A: Since $dim(V)=3$, you already have a basis of $V$ : $v_1,v_2,v_3$.
You just need to make it orthogonal, using Gram–Schmidt process :
$$u_1=v_1$$
$$u_2=v_2-(v_2,u_1)u_1$$
$$u_3=v_3-(v_3,u_2)u_2-(v_3,u_1)u_1$$
Then :
$$proj_V(y) = \frac{(y,u_1)}{||u_1||^2}u_1+\frac{(y,u_2)}{||u_2||^2}u_2+\frac{(y,u_3)}{||u_3||^2}u_3$$
A: The vectors $\mathbf v_1$ and $\mathbf v_2$ are obviously orthogonal, so Gram-Schmidt orthogonalization seems like the least amount of work, especially since you only have to project one vector. Once you have a $\mathbf v_3'\in V$ with $\mathbf v_1\perp\mathbf v_3'$ and $\mathbf v_2\perp\mathbf v_3'$, then $\operatorname{proj_V}\mathbf y={(\mathbf y,\mathbf v_1)\over(\mathbf v_1,\mathbf v_1)}\mathbf v_1+{(\mathbf y,\mathbf v_2)\over(\mathbf v_2,\mathbf v_2)}\mathbf v_2+{(\mathbf y,\mathbf v_3')\over(\mathbf v_3',\mathbf v_3')}\mathbf v_3'$. The vector $\mathbf v_3'$ is obtained by computing the orthogonal rejection of $\mathbf v_3$ from the other two vectors using the same process: $\mathbf v_3'=\mathbf v_3-{(\mathbf v_3,\mathbf v_1)\over(\mathbf v_1,\mathbf v_1)}\mathbf v_1-{(\mathbf v_3,\mathbf v_2)\over(\mathbf v_2,\mathbf v_2)}\mathbf v_2$.  
Another way to go if you know that the three spanning vectors are linearly independent is to form the matrix $U=\begin{bmatrix}\mathbf v_1&\mathbf v_2&\mathbf v_3\end{bmatrix}$ and compute $\operatorname{proj_V}\mathbf y=U(U^TU)^{-1}U^T\mathbf y$, but that’s a lot more work for this problem. If the $\mathbf v$’s had been an orthonormal basis for $V$, though, this wouldn’t be too bad since it reduces to $UU^T\mathbf y$—the projection formula in matrix form.
