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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?

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  • $\begingroup$ 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. $\endgroup$ – amd Nov 17 '19 at 21:43
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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 $0$.

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  • $\begingroup$ 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 $0$" ? $\endgroup$ – Mary Star Nov 17 '19 at 12:59
  • $\begingroup$ 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? $\endgroup$ – Mary Star Nov 17 '19 at 13:11
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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$$\begin{vmatrix}1 & 2 & -2 & a \\ -1 & 0 & 1 & b \\ 1 & 3 & 0 & c \\ -1 & -1 & 3 & d\end{vmatrix}\neq0.$$

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  • $\begingroup$ 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? $\endgroup$ – Mary Star Nov 17 '19 at 12:49
  • $\begingroup$ Yes, that is correct. I would take $a=1$ and $b=c=d=0$, but that's a matter of taste. $\endgroup$ – José Carlos Santos Nov 17 '19 at 12:51

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