Which is the extended basis?

Question

Let $t\in \mathbb{R}$ and the vectors \begin{equation*}v_1=\begin{pmatrix} 0\\ 1\\ -1\\ t \end{pmatrix}, v_2=\begin{pmatrix} t\\ 2\\ 0\\ 1 \end{pmatrix}, v_3=\begin{pmatrix} 2\\ 2\\ 2\\ 0 \end{pmatrix}\end{equation*} in $\mathbb{R}^4$.

We have the following: \begin{align*}&\begin{pmatrix} 0 & t & 2 \\ 1 & 2 & 2 \\ -1 & 0 & 2 \\ t & 1 & 0 \end{pmatrix}\rightarrow\ldots \rightarrow \begin{pmatrix} 1 & 2 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & t-1 \\ 0 & 0 & 0 \end{pmatrix}\end{align*}

If $t\neq 1$ the three vectors are linealy independent.

I want to extend these to a basis:

• $t\neq 1$ :

  We need one more vector. We write the given vectors as lines of a matrix and write one zero line:

$\begin{pmatrix} 0 & 1& -1 & t \\ t& 2& 0& 1 \\ 2& 2& 2& 0\\ 0&0&0&0 \end{pmatrix}$

We have to write in a row-echelon form, so we get the following, or not? $\begin{pmatrix} 2 & 2& 2 & 0 \\ 0& 1& -1&t \\ 0 & 0& 4-4t& 2t^2-4t+2\\ 0&0&0&0 \end{pmatrix}$

At the diagonal we write $1$: \begin{equation*}\begin{pmatrix} 2 & 2& 2 & 0 \\ 0& 1& -1&t \\ 0 & 0& 4-4t& 2t^2-4t+2\\ 0&0&0&1 \end{pmatrix}\end{equation*}

Which is now the basis? The three initial vectors and the last line of that matrix? Or all the lines of the matrix?

Answer 1

You can take advantage of the fact that $$ \mathbb{R}^4=C(A)\oplus N(A^T) $$ so you just need to find a basis of the null space of $A^T$.

\begin{align} A^T=\begin{bmatrix} 0 & 1 & -1 & t \\ t & 2 & 0 & 1 \\ 2 & 2 & 2 & 0 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & 1 & 0 \\ t & 2 & 0 & 1 \\ 0 & 1 & -1 & t \end{bmatrix} && R_3\leftrightarrow R_1, R_1\gets\frac{1}{2}R_1 \\[6px]&\to \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 2-t & -t & 1 \\ 0 & 1 & -1 & t \end{bmatrix} && R_2\gets R_2-R_1 \\[6px]&\to \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & -1 & t \\ 0 & 2-t & -t & 1 \end{bmatrix} && R_3\leftrightarrow R_2 \\[6px]&\to \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & -1 & t \\ 0 & 0 & 2-2t & t^2-2t+1 \end{bmatrix} && R_3\gets R_3+(2-t)R_2 \end{align} If $t=1$, the matrix is $$ \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \qquad R_1\gets R_1-R_2 $$ so a basis for $N(A^T)$ is given by $$ \begin{bmatrix} -2 \\ 1 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ -1 \\ 0 \\ 1 \end{bmatrix} $$ by solving the system $A^T=0$ and isolating two linearly independent solutions.

For $t\ne1$ you can go further in the elimination. I leave to you ending it.

Answer 2

Consider the matrix

$$ A = \begin{bmatrix} 1 & 2 & 2 & 1\\ 0 & t & 2 & a\\ -1 & 0 & 2 & b\\ t & 1 & 0 & c \end{bmatrix} $$

Applying elementary row operations to $A$ you end up with

$$ \begin{bmatrix} 1 & 2 & 2 & 1\\ 0 & 2 & 4 & b + 1\\ 0 & 0 & 4-4t & 2a - t(b + 1)\\ 0 & 0 & 0 & 2a + 2c-b - 1+t(b-1) \end{bmatrix} $$

So in order to have a basis you need to choose $a, b, c, t$ such that $t\neq 1$ and $2a + 2c-b - 1+t(b-1) \neq 0$. For an even simpler approach observe that appending the vector $(0,0,0,1)$ as the fourth column of $A$ results in a full rank matrix if $t\neq 1$.

Answer 3

We can compute the vector that is perpendicular to the other three vectors by computing the generalized cross product of the first three colums: $$ \begin{bmatrix} 0&t&2&4-4t\\ 1&2&2&2t^2-2\\ -1&0&2&-2(t-1)^2\\ t&1&0&4-4t \end{bmatrix} $$ Each element in the fourth column is the cofactor at that position.

In addition, the determinant of the matrix is $$ \left(4-4t\right)^2+\left(2t^2-2\right)^2+\left(2(t-1)^2\right)^2+\left(4-4t\right)^2 $$ Since the only common factor of the terms is $1-t$, the determinant is $0$ only when $t=1$.