- #1

- 1,860

- 0

This is a pretty basic question, but I just want to make sure. The question is to find the least square solution for

[tex] \newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}

\left(

\begin{array}{cc}

1 & 2\\

2 & 4

\end{array}

\right) x = \colv{2}{2}[/tex]

I can just find the orthogonal projection of the two vectors, right? In other words, use the find [tex]w = a_1\mathbf{v_1} + a_2\mathbf{v_2}[/tex]

[tex]\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}

a_1 {\colv{1}{2}} + a_2 {\colv{2}{4}} = {\colv{w_1}{w_2}}[/tex]

where

[tex] a_1 = \frac{<b,v_1>}{||v_1||^2}[/tex]

and

[tex] a_2 = \frac{<b,v_2>}{||v_2||^2}[/tex]

I don't really want to use the method with positive definite because the numbers turn out sticky.

Then solution can be expressed as [tex]\mathbf{z} = \mathbf{b} - \mathbf{w}[/tex], such that z is orthogonal to the range of K?

[tex] \newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}

\left(

\begin{array}{cc}

1 & 2\\

2 & 4

\end{array}

\right) x = \colv{2}{2}[/tex]

I can just find the orthogonal projection of the two vectors, right? In other words, use the find [tex]w = a_1\mathbf{v_1} + a_2\mathbf{v_2}[/tex]

[tex]\newcommand{\colv}[2] {\left(\begin{array}{c} #1 \\ #2 \end{array}\right)}

a_1 {\colv{1}{2}} + a_2 {\colv{2}{4}} = {\colv{w_1}{w_2}}[/tex]

where

[tex] a_1 = \frac{<b,v_1>}{||v_1||^2}[/tex]

and

[tex] a_2 = \frac{<b,v_2>}{||v_2||^2}[/tex]

I don't really want to use the method with positive definite because the numbers turn out sticky.

Then solution can be expressed as [tex]\mathbf{z} = \mathbf{b} - \mathbf{w}[/tex], such that z is orthogonal to the range of K?

Last edited: