Finding the Orthogonal Projection from a Matrix where $a_{ij} = \vec v_i\cdot \vec v_j$

I am working on a problem that is asking to find $$\operatorname{proj}_Vv_3$$ where the matrix $$A=\begin{bmatrix} 3&5&11\\ 5&9&20\\ 11&20&49\end{bmatrix}$$ has entries $$a_{ij}=\vec v_i\cdot \vec v_j$$. I though that the simplest way to answer this question was to use the formula $$(u_1,x)u_1+(u_2,x)u_2 = proj_vx .$$

Since the question also asks to express the solution as a linear combination of $$v_1$$ and $$v_2$$, I assumed that I could just plug the numbers in. I knew I had to make the vectors unit vectors so my equation looked something like this:

$$3(5)(v2)+ 1/11(v3)$$ This answer is not the correct answer but I am wondering if I am reading the matrix in the wrong way or if I am confused about the usage of the formula. Does someone have a better idea?

• What is $v$ here? – amd Apr 1 '19 at 1:16

The formula as you wrote it makes no sense. The correct version, since $$V=\operatorname{span}\{v_1,v_2\}$$, is $$\operatorname{proj}_V(v_3)=\frac{v_3\cdot v_1}{v_1\cdot v_1}\,v_1+\frac{v_3\cdot v_2}{v_2\cdot v_2}\,v_2.$$ Reading the four numbers from the matrix we get $$\operatorname{proj}_V(v_3)=\frac{a_{13}}{a_{11}}\,v_1+\frac{a_{23}}{a_{22}}\,v_2=\frac{11}{3}\,v_1+\frac{5}{9}\,v_2.$$