I have points $(1,1) (2,2)$ and $(3,2)$. Now I wish use least-squares to fit a line through it.

The equations I get, $$ C+D=1$$ $$ C+2D=2$$ $$ C+3D=2$$ Now, of course there exists no solution to this system($Ax=b$). We are obtaining the best fit hyperplanein $\mathbb R^{2}$(,i.e., a line) for the points. This plane has the closest projections of vector $b$.


  1. I can't understand what the plane in $\mathbb R^{3}$ represents on which we are projecting on. What's does it signify?
  2. How the the plane in $\mathbb R^{3}$ relate to the best-fit line we obtain?

The linear system is $$ \begin{align} % \mathbf{A} c &= b \\[2pt] % \left[ \begin{array}{cc} 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{array} \right] % \left[ \begin{array}{c} C \\ D \end{array} \right] % & = % \left[ \begin{array}{c} 1 \\ 2 \\ 2 \end{array} \right] % \end{align} $$ The solution is $$ \left[ \begin{array}{c} C \\ D \end{array} \right]_{LS} = \frac{1}{6} \left[ \begin{array}{c} 4 \\ 3 \end{array} \right] $$ The matrix $\mathbf{A}$ is a map which takes vectors from the domain $\mathbb{C}^{n}$ and maps them to vectors in the image $\mathbb{C}^{m}$.

The matrix $\mathbf{A}$ is a map which connects vectors from the solution space $\mathbb{C}^{n}$ and to vectors in the data space $\mathbb{C}^{m}$.

If you are doing a linear regression with 10 data points, the solution space is $\mathbb{C}^{2}$ where your slope and intercept live. Your data space is $\mathbb{C}^{10}$ where the measurements live.

More about the fundamental projectors: Least squares solutions and the orthogonal projector onto the column space

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.