Science:Math Exam Resources/Courses/MATH221/April 2013/Question 10/Solution 1

We first note that there is a non-trivial linear relationship between the three vectors, namely $\left[{\begin{array}{cccc}1&2&3&4\\\end{array}}\right]=\left[{\begin{array}{cccc}1&0&1&0\\\end{array}}\right]+2\left[{\begin{array}{cccc}0&1&1&2\\\end{array}}\right]$ Thus, we get that $W$ has dimension 2 and thus $W^{\bot }$ is also 2 dimensional. Let $V$ be the matrix of basis vectors to $W$,

$V={\begin{bmatrix}1&0\\0&1\\1&1\\0&2\\\end{bmatrix}}.$

If $x=[a,b,c,d]^{T}$ is a basis vector to $W^{\bot }$ then orthogonality tells us that

${\begin{aligned}V^{T}x&=0\\{\begin{bmatrix}1&0&1&0\\0&1&1&2\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}&=\mathbf {0} \end{aligned}}$

Computing this we get the following relationships

${\begin{aligned}a+c&=0\\b+c+2d&=0\end{aligned}}$

Since the dimension (or rank) of V is two we will have two free parameters. In particular, this means that the basis is not unique (it never is). We choose $a$ and $d$ as free parameters to get

${\begin{aligned}c&=-a\\b&=-2d-c=-2d+a\end{aligned}}$

and finally that

$x={\begin{bmatrix}a\\-2d+a\\-a\\d\end{bmatrix}}=a{\begin{bmatrix}1\\1\\-1\\0\end{bmatrix}}+d{\begin{bmatrix}0\\-2\\0\\1\end{bmatrix}}$

Therefore the two orthogonal vectors $\left[{\begin{array}{cccc}1&1&-1&0\\\end{array}}\right],\left[{\begin{array}{cccc}0&-2&0&1\\\end{array}}\right]$ are a basis for $W^{\bot }$. Notice that the dimension of $W^{\bot }$ is two, which happens to be the same dimension as that of $W$.