I am trying to solve following problem

Let $S$ be the subspace of $\mathbb{R}^4$ which is the intersection of two planes \begin{align} &H_1 = \{(x,y,z,w) | x+ 3y - 2z + w =0\} \\ &H_2 = \{(x,y,z,w)| 2x-2y + z-w =0 \} \end{align} Find an orthogonal basis of $S$ and find the point in $S$ which is closest to the point $(1,2,2,1)$ (with respect to the norm defined by the usual dot product.)

Based on my previous posts Finding basis from intersection of two planes , I know how to calculate orthogonal basis of $S$.

Finding a orthogonal basis is the finding nullspace of $A$ \begin{align} A = \begin{pmatrix} 1 & 3 & -2 & 1 \\ 2 & -2 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0& 0 & 0 & 0 \end{pmatrix} \end{align} i.e. $B_A = \{(1,-3,0,8)^T, (1,5,8,0)^T\}$. Now perform Gram-Schmidt process, \begin{align} u_1 = (1, -3, 0, 8)^T \quad u_2 = v_2 - \frac{(u_1, v_2)}{(u_1, u_1)}u_1 = \left( \frac{44}{37}, \frac{164}{37}, 8, \frac{56}{37} \right)^T \end{align}

Now the problem is finding the point in $S$ which is closest to the point $(1,2,2,1)$.

The point in $S$ can be written as a linear combination of its basis, so $s = a u_1 + b u_2$, and then distance from $(1,2,2,1)$ is given as follows

\begin{align} l^2 =\left( a + \frac{44}{37} b -1 \right)^2 + \left( -3a + \frac{164}{37} b -2\right)^2 + \left( 8b - 2 \right)^2 + \left( 8a + \frac{56}{37}b - 1 \right)^2 \end{align}

then the length is described by two parameters $a,b$.

By using calculus[finding $l_x=l_y=0$ , I tried to obtain maximum length but the computation is not good.

Let $f=l^2$, then from $\frac{\partial f}{\partial a} =0$ we have $a=\frac{3}{74}$ and from $\frac{\partial f}{\partial b} =0$ we have $b = \frac{255}{808}$

Are there any other ways?

  • $\begingroup$ You can solve this by the discrete least-square method. $\endgroup$ – Malkoun Feb 1 at 8:09

$S$ is a $2$-dimensional subspace $W$ of $\mathbb{R}^4$. Let us say you found a basis $w_1$ and $w_2$ of $W$ (not necessarily orthonormal). Let $A = (w_1 w_2)$ be the $4$ by $2$ matrix containing $w_1$ its first column and $w_2$ as its second. Let $\mathbf{x} = (x_1,x_2)^T$ denote the relevant unknown coefficients. Let $\mathbf{b} = (1,2,2,1)^T$.

The problem is equivalent to solving $A\mathbf{x} = \mathbf{b}$ in the least square sense. In other words, you want to find $\mathbf{x}$ such that

$E(\mathbf{x}) = \|A\mathbf{x} - \mathbf{b}\|^2$

is minimized. This is equivalent to solving

$$A^T A \mathbf{x} = A^T \mathbf{b},$$

which is a (non-singular) linear system of two equations and two unknowns. In order to convince you that this is true, an arbitrary point of $W$ is of the form $A\mathbf{x}$ for some $\mathbf{x}$. Geometrically, if $\mathbf{x} = \mathbf{\hat{x}}$ is the point at which the distance (or distance squared) is minimized, then it is clear geometrically that $\mathbf{b} - A\mathbf{\hat{x}}$ must be orthogonal to $W$, which is equivalent to requiring that it must be orthogonal to $w_1$ and $w_2$. In other words, we must have

$$ A^T(\mathbf{b} - A \mathbf{\hat{x}}) = \mathbf{0}, $$

which is equivalent to

$$A^T A \mathbf{\hat{x}} = A^T \mathbf{b}.$$

Edit $1$: taking

$A = \left( \begin{array}{cc} 1 & 1 \\ -3 & 5 \\ 0 & 8 \\ 8 & 0 \end{array} \right)$

we get

$A^T A \mathbf{x} = A^T \mathbf{b}$

which is equivalent to

$\left( \begin{array}{cc} 74 & -14 \\ -14 & 90 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right) = \left( \begin{array}{c} 3 \\ 27 \end{array} \right)$

whose solution is

$\left( \begin{array}{c} x_1 \\ x_2 \end{array} \right) = \left( \begin{array}{c} \frac{81}{202} \\ \frac{255}{202} \end{array} \right).$

So the point on $W$ closest to $\mathbf{b}$ is thus

$\frac{81}{202} w_1 + \frac{255}{202} w_2,$

where $w_1$ and $w_2$ are the first and second columns of $A$.

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