MHB How to Extend These Vectors to a Basis in $\mathbb{R}^4$?

[mathmari]

Hey!

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

I want to determine a maximal linearly independent subset of $\{v_1, v_2, v_3\}$ and to extend these to a basis of $\mathbb{R}^4$.
I have done the following:
$$\begin{pmatrix}
0 & t & 2 \\
1 & 2 & 2 \\
-1 & 0 & 2 \\
1 & 1 & 0
\end{pmatrix}\rightarrow \begin{pmatrix}
1 & 2 & 2 \\
0 & t & 2 \\
-1 & 0 & 2 \\
1 & 1 & 0
\end{pmatrix} \rightarrow \begin{pmatrix}
1 & 2 & 2 \\
0 & t & 2 \\
0 & 1 & 2 \\
0 & 1 & 2
\end{pmatrix}\rightarrow \begin{pmatrix}
1 & 2 & 2 \\
0 & t & 2 \\
0 & 1 & 2 \\
0 & 0 & 0
\end{pmatrix}\rightarrow \begin{pmatrix}
1 & 2 & 2 \\
0 & t & 2 \\
0 & 0 & 2t-2 \\
0 & 0 & 0
\end{pmatrix}$$

If $t\neq 1$ we get the following system: $$\lambda_1+2\lambda_2+2\lambda_3=0 \\ t\lambda_2+2\lambda_3=0 \\ (2t-2)\lambda_3=0$$
So, $\lambda_1=\lambda_2=\lambda_3=0$ and so the vectors are linearly independent.

If $t=1$ we get the system: $$\lambda_1+2\lambda_2+2\lambda_3=0 \\ \lambda_2+2\lambda_3=0=0$$
So, $(\lambda_1, \lambda_2, \lambda_3)=\lambda_3 (2, -2, 1), \lambda_3\in \mathbb{R}$.
Therefore, $v_3=-2v_1+2v_2$ and $v_1$ and $v_2$ are linearly independent. Is everything correct so far? (Wondering)

So do we get a different maximal linearly independent subset for $t=1$ and a different for $t\neq 1$ ? (Wondering) So, do we have to take cases for $t$ to extend the vectors to a basis? (Wondering)

 

[mathmari]

Or, since we are looking for the maximal linearly independent subset of $\{v_1,v_2,v_3\}$, we take the case $t\neq 1$, because then the linearly independent subset is the whole set? (Wondering)

 

[mathmari]

There is a typo at the first post...

The vectors are\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 \begin{pmatrix}
1 & 2 & 2 \\
0 & t & 2 \\
0 & 2 & 4 \\
t & 1 & 0
\end{pmatrix} \rightarrow \begin{pmatrix}
1 & 2 & 2 \\
0 & t & 2 \\
0 & 1 & 2 \\
0 & 1-2t & -2t
\end{pmatrix}\rightarrow \begin{pmatrix}
1 & 2 & 2 \\
0 & t & 2 \\
0 & 1 & 2 \\
0 & 2t & 2+2t
\end{pmatrix}\rightarrow \begin{pmatrix}
1 & 2 & 2 \\
0 & 1 & 2 \\
0 & t & 2 \\
0 & t & 1+t
\end{pmatrix}\\ &\overset{ t\neq 1}{\rightarrow } \begin{pmatrix}
1 & 2 & 2 \\
0 & 1 & 2 \\
0 & t & 2 \\
0 & 0 & t-1
\end{pmatrix}\rightarrow \begin{pmatrix}
1 & 2 & 2 \\
0 & 1 & 2 \\
0 & 0 & t-1 \\
0 & 0 & 0
\end{pmatrix}\end{align*} So, for $t\neq 1$ the $3$ vectors are linearly independent.

For $t=1$ we get fpr example $v_3=-2v_1+2v_2$, where $v_1$ and $v_2$ are linearly independent.

About extending the basis I have done the following:
When $t\neq 1$ :
We need one more vector. We write the given vectors as lilnes of a matrix and we add also a zero line:
\begin{equation*}\begin{pmatrix}
0 &
1&
-1 &
t
\\ t& 2& 0& 1 \\ 2&
2&
2&
0\\ 0&0&0&0
\end{pmatrix}\end{equation*}We have to bring the matrix into a row-echelon form \begin{equation*}\begin{pmatrix}
2 &
2&
2 & 0
\\ 0& 1& -1&t \\ 0 &
0&
4-4t&
2t^2-4t+2\\ 0&0&0&0
\end{pmatrix}\end{equation*}At the diagonals we write the element $1$ :
\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*}

So, the vectors \begin{equation*}\begin{pmatrix}
0 \\
1\\
-1 \\ t\end{pmatrix}, \begin{pmatrix}
t\\ 2\\ 0\\ 1 \end{pmatrix} , \begin{pmatrix} 2 \\
2\\
2\\
0\end{pmatrix}, \begin{pmatrix}0\\ 0\\ 0\\ 1
\end{pmatrix}\end{equation*} form a basis.

Is this correct? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

Let's bring the matrix in column echelon form instead of row echelon form. (Thinking)

$$\begin{array}{}
\quad \times \quad \phantom{0} \quad -t \\
\begin{pmatrix}
1 & 0 & t \\
1 & 1 & 2 \\
1 & -1 & 0 \\
0 & 1 & 1
\end{pmatrix} \to
\end{array}

\begin{array}{}
\quad \phantom{0} \quad \times \quad t-2 \\
\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 2-t \\
1 & -1 & -t \\
0 & 1 & 1
\end{pmatrix} \to
\end{array}

\begin{array}{}
\phantom{0} \\
\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & -1 & -2(t-1) \\
0 & 1 & t-1
\end{pmatrix}
\end{array}
$$

The columns are now our independent vectors.
For $t=1$ the third column is the 0 vector, showing that we have a dependent set, and furthermore the first 2 columns form an independent basis.
For $t\ne 1$, the third column is an independent vector with non-zero entries in the 3rd and 4th rows.

That means we can complete the basis by adding a 4th column that completes the column echelon form:
$$
\to\left(\begin{array}{ccc|c}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
1 & -1 & -2(t-1) & 0 \\
0 & 1 & t-1 & 1
\end{array}\right)
$$

So indeed, we can pick \begin{pmatrix}0\\0\\0\\1\end{pmatrix} as the 4th basis vector. (Happy)

 

[mathmari]

Ah ok... Thank you very much! (Smile)