# Linear independent vectors - Span - Basis

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?

• If you had instead written the vectors as rows of a matrix for the first part, you could then read a solution for the third part from the reduced matrix. – amd Nov 17 '19 at 21:43

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 $$0$$.

• About point 4, do you mean to write the definition that the vectors $v_1, \ldots , v_k$ are linear independent? Or what do you mean by "here is the non-trivial linear combination $v_i−v_j$ evaluating to $0$" ? – Mary Star Nov 17 '19 at 12:59
• When $v_i=v_j$ we get that $v_i-v_j=0$, that means that the vectors $v_i$ and $v_j$ are not linear independent, since at $\lambda_iv_i+\lambda_jv_j=0$ we have non-zero values for the coefficients. Since $v_i, v_j$ are linear dependent, then the vectors $v_1, \ldots , v_i, v_j, \ldots , v_k$ cannot be linear independent, a contradiction. Is everything correct and complete? Or did you mean something else? – Mary Star Nov 17 '19 at 13:11

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$$\begin{vmatrix}1 & 2 & -2 & a \\ -1 & 0 & 1 & b \\ 1 & 3 & 0 & c \\ -1 & -1 & 3 & d\end{vmatrix}\neq0.$$

• Ok! So I found that it should hold that $7a-b-3c+5d\neq 0$, so we can take arbitrary values so that the result is different from zero, or not? For example we can take $a=b=c=d=1$ then we get the determinant $7-1-3+5=8\neq 0$. Is that correct? – Mary Star Nov 17 '19 at 12:49
• Yes, that is correct. I would take $a=1$ and $b=c=d=0$, but that's a matter of taste. – José Carlos Santos Nov 17 '19 at 12:51