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Distance to a column space

Linear algebraHard

Let $A = 111123$ and $b = (1, 2, 6)$ . What is the squared distance from $b$ to the column space of $A$ ? Give a fraction in lowest terms.

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Hint

The closest point of the column space is the least squares fit, and what is left over is orthogonal to every column.

Solution

The closest point is $A (c, m)$ , where $(c, m)$ solves the normal equations $3 c + 6 m = 9$ and $6 c + 14 m = 23$ .
From the first, $c = 3 - 2 m$ ; substituting into the second gives $18 + 2 m = 23$ , so $m = \frac{5}{2}$ and $c = - 2$ .
The closest point is $(- 2 + \frac{5}{2}, - 2 + 5, - 2 + \frac{15}{2}) = (\frac{1}{2}, 3, \frac{11}{2})$ , so the residual is $(\frac{1}{2}, - 1, \frac{1}{2})$ .
Its squared length is $\frac{1}{4} + 1 + \frac{1}{4} = \frac{3}{2}$ .

Answer: $\frac{3}{2}$