Latex test post
Sam_Lai
\\ 3) \\ \begin{matrix} 2/3 & -1/2 & 1/3 \\ -8/3 & 2 & -1/3 \\ 1 & -1/2 & 0 \end{matrix} \\ 4) Linear, not linear, not linear, linear
CarlHarmon: May 31, 2015, 10:24 a.m.
There is another way to approach this problem. You can calculate the orthonormal basis and then use it to find the "matrix of orthogonal projection onto V" using the equation A($A^TA)^-1A^T$. Then you multiply the given vector by this matrix. While this takes some time, you could then use that same matrix for multiple vectors. It's essentially a more general method.
CarlHarmon: May 31, 2015, 10:41 a.m.
\newline Um... this is definitely not correct. At the very least, the answer at the end needs to have three terms. I'm not sure where that definition of inner product at the beginning is coming from. \hfill \break \newline The way I approached this problem by using {$1,x,x^2$} as a basis for $P_2(R)$. Then I tried caculating $u_1, u_2$ and $u_3$ using these vectors. But by the time you get to $u_3$ the math becomes hideous, so hope there is a better way.\hfill \break
CarlHarmon: May 31, 2015, 10:42 a.m.
Um... this is definitely not correct. At the very least, the answer at the end needs to have three terms. I'm not sure where that definition of inner product at the beginning is coming from. \\The way I approached this problem by using {$1,x,x^2$} as a basis for $P_2(R)$. Then I tried caculating $u_1, u_2$ and $u_3$ using these vectors. But by the time you get to $u_3$ the math becomes hideous, so hope there is a better way.