Linear independent vectors - Span - Basis

Question

Let \begin{equation*}x_1:=\begin{pmatrix}1 \\ -1 \\ 1 \\ -1\end{pmatrix}, \ x_2:=\begin{pmatrix}2 \\ 0 \\ 3 \\ -1\end{pmatrix}, \ x_3:=\begin{pmatrix}-2 \\ 1 \\ 0 \\ 3\end{pmatrix}, \ y:=\begin{pmatrix}2 \\ 3 \\ 7 \\ 2\end{pmatrix}\in \mathbb{R}^4\end{equation*}

• Show that $x_1, x_2, x_3$ are linearly independent.
• Show that $y\in \text{Lin}(x_1, x_2, x_3)$.
• Give a vector $x\in \mathbb{R}^4$ such that $(x, x_1, x_2, x_3)$ is a basis of $\mathbb{R}^4$.
• Let $v_1, \ldots , v_k\in \mathbb{R}^n$ be linear independent vectors. Then $v_1, \ldots , v_k$ are pairwise different.

I have done the following:

• We write the given vectors are columns of a matrix and then we apply the Gaussian elimination algorithm.

  The number of non-zero rows is $3$ and this is equal to the number of vectors. This means that the vectors $x_1, x_2, x_3$ are linearly independent.

• $y\in \text{Lin}(x_1, x_2, x_3)$ mean that $y$ can be written as a linear combination of $x_1, x_2, x_3$, so we have to show that there exist $c_1, c_2, c_3$ such that $y=c_1x_1+c_2x_2+c_3x_2$.

  For that we have to show that the system $Ac=y$ has a solution with $A=\begin{pmatrix}1 & 2 & -2 \\ -1 & 0 & 1 \\ 1 & 3 & 0 \\ -1 & -1 & 3\end{pmatrix} $ and $c=\begin{pmatrix}c_1 \\ c_2 \\ c_3\end{pmatrix}$, right?

• How can we find a $4$th vector? Could you give me a hint?

• If the vectors weren't pairwise different, they wouldn't be linearly independent, would they? But how can we prove that formally? Do we assume that they are not pairwise different?

Answer 1

Point 1 and Point 2: you're right.

Point 3: Consider your Gaussian elimination from Point 1. Add a fourth vector to the reduced matrix in a way that makes it obviously independent from the other three, then do the Gaussian elimination steps backwards and see what your fourth vector becomes.

Point 4: again, correct. To prove it, simply say that $v_i=v_j$ implies that there is the non-trivial linear combination $v_i-v_j$ evaluating to $0$.

Answer 2

You proved correctly that $x_1$, $x_2$, and $x_3$ are linearly independent. So, $y$ is a linearly combination of them if and only if the matrix whose columns are $x_1$, $x_2$, $x_3$, and $y$ is singular.

In order to find fourth vector, take $(a,b,c,d)^T$ such that$$\begin{vmatrix}1 & 2 & -2 & a \\ -1 & 0 & 1 & b \\ 1 & 3 & 0 & c \\ -1 & -1 & 3 & d\end{vmatrix}\neq0.$$