How Far Is y From the Column Space?

A growth team ran four campaigns. Spend is recorded in the second column of the design matrix; the first column of ones is the intercept.

$X = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ 1 & 4 \end{pmatrix} \qquad \mathbf{y} = \begin{pmatrix} 2 \\ 4 \\ 5 \\ 9 \end{pmatrix}$

No vector $\boldsymbol{\beta}$ satisfies $X\boldsymbol{\beta} = \mathbf{y}$ exactly — $\mathbf{y}$ does not lie in the two-dimensional column space of $X$. Least squares therefore returns the $\hat{\boldsymbol{\beta}}$ that minimises

$\lVert \mathbf{y} - X\boldsymbol{\beta} \rVert_2$

and $X\hat{\boldsymbol{\beta}}$ is the orthogonal projection of $\mathbf{y}$ onto $\operatorname{col}(X)$.

How far is $\mathbf{y}$ from that column space? That is, report the residual norm

$\lVert \mathbf{y} - X\hat{\boldsymbol{\beta}} \rVert_2$

Round your answer to 2 decimal places.