Calculate basis of $U^\perp$ with $U=\langle \left(\begin{array}{c} 1\\ 1\\ 1\\ \end{array}\right) \rangle$

Question

I used the approach $\mathbb{L}((1,1,1),0)$ and therefore got $(\left(\begin{array}{c} -1\\ 1\\ 0\\ \end{array}\right), \left(\begin{array}{c} -1\\ 0\\ 1\\ \end{array}\right))$

However in the solution the basis is $(\left(\begin{array}{c} 1\\ -1\\ 0\\ \end{array}\right), \left(\begin{array}{c} 0\\ -1\\ 1\\ \end{array}\right))$

Does my approach have an mistake or are both basis correct?

Answer 1

Here's a result you can rely on to verify whether or not your candidate basis is valid.

Proposition. Suppose that $\{v_1,\dotsc, v_d\}$ is a basis of a vector subspace $V\subset\Bbb R^n$ and let $\{w_1,\dotsc, w_d\}\subset\Bbb R^n$. Let $A$ and $B$ be the matrices whose columns are $\{v_1,\dotsc,v_d\}$ and $\{w_1,\dotsc,w_d\}$ respectively, so \begin{align*} A &= \begin{bmatrix}v_1 & \dotsb & v_d\end{bmatrix} & B &= \begin{bmatrix}w_1 & \dotsb & w_d\end{bmatrix} \end{align*} Then $\{w_1,\dotsc,w_d\}$ is a basis of $V$ if and only if there is an invertible matrix $P$ satisfying $AP=B$.

In your situation, the matrices $A$ and $B$ are given by \begin{align*} A &= \left[\begin{array}{rr} -1 & -1 \\ 1 & 0 \\ 0 & 1 \end{array}\right] & B &= \left[\begin{array}{rr} 1 & 0 \\ -1 & -1 \\ 0 & 1 \end{array}\right] \end{align*}

Is there an invertible $P$ satisfying $AP=B$?

Hint. Using the notation from the proposition, we have $w_1=-v_1$ and $w_2=-v_1+v_2$. How might this help us construct $P$?

Answer 2

There is no such thing as the basis for a vector space, except for the trivial case when the vector space has dimension $0$ (that is, $V=\{0\}$).

The author of the solution just made a different choice of the free variables when solving $x+y+z=0$. Writing the equation as $y=-x-z$, yields the basis $$ \left\{\,\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}\,\right\} $$ However, from $x=-y-z$, as you did, you can also get the basis $$ \left\{\,\begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}\,\right\} $$ The important thing is that when you remove the row(s) corresponding to the nonfree variable(s), the resulting vectors make a basis of $\mathbb{R}^k$ ($k$ the number of free variables). Choosing the standard basis is the most obvious choice, of course.