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Given $u_1,u_2,u_3,u_4$ linearly independent vectors in $R^5$. $v_1,v_2,v_3$ $\in R^5$ .

They are defined like that:

$v_1 = u_1 + u_2 - u_4$

$v_2 = au_1 - u_3 + u_4$

$v_3 = u_2 + au_3 - 6u_4$

How can I find for which $a$ , vectors $v_1,v_2,v_3$ will be linearly dependent?

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3 Answers 3

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We show that $v_1,v_2,v_3\in \mathbb{R}^5$ are always linearly independent. First suppose $a=0$. Now let $\lambda_1,\lambda_2,\lambda_3\in \mathbb{R}$ such that $\lambda_1v_1+\lambda_2v_2+\lambda_3v_3=0$. Then $$\lambda_1u_1+2(\lambda_1+\lambda_3)u_2-\lambda_2u_3-6(\lambda_1+\lambda_2+\lambda_3)u_4=0 \implies \lambda_1=\lambda_2=\lambda_3=0$$ Now suppose $a\not=0$. We express the $v_i$ as column vectors now and perform elementary matrix operations on \begin{bmatrix} 1 & a & 0 \\ 1 & 0 & 1 \\ 0 & -1 & a \\ -1 & 1 & -6 \\ 0 & 0 & 0 \end{bmatrix} to obtain \begin{bmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & 1-\frac{1}{a^2} \\ 0 & 0 & -1 -\frac{1}{a}+\frac{6}{a^2} \\ 0 & 0 & 0 \end{bmatrix} Note that since $a=1,-1$ both do not yield the zero vector in the third column, we can conclude that $v_1,v_2,v_3\in \mathbb{R}^5$ are always linearly independent

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It is immaterial that the vectors are in $\mathbb{R}^5$ or any other vector space. The vectors $v_1$, $v_2$ and $v_3$ belong to the span of $\{u_1,u_2,u_3,u_4\}$ of which the set is a basis.

A set of vectors is linearly independent if and only if their coordinate vectors with respect to a basis are linearly independent. Thus consider the matrix having as columns the coordinate vectors with respect to the given basis, that is $$ \begin{bmatrix} 1 & a & 0 \\ 2 & 0 & 1 \\ 0 & -1 & a \\ -1 & 1 & -6 \\ \end{bmatrix} $$ Let's try Gaussian elimination: \begin{align} \begin{bmatrix} 1 & a & 0 \\ 2 & 0 & 1 \\ 0 & -1 & a \\ -1 & 1 & -6 \\ \end{bmatrix} &\to \begin{bmatrix} 1 & a & 0 \\ 0 & -2a & 1 \\ 0 & -1 & a \\ 0 & 1+a & -6 \\ \end{bmatrix} &&\begin{aligned}R_2&\gets R_2-2R_1\\ R_4&\gets R_4+R_1\end{aligned} \\[6px] &\to \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & -a \\ 0 & -2a & 1 \\ 0 & 1+a & -6 \\ \end{bmatrix} &&\begin{aligned}R_2&\leftrightarrow R_3\\ R_2&\gets -R_2\end{aligned} \\[6px] &\to \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & -a \\ 0 & 0 & 1-2a^2 \\ 0 & 0 & -6+a+a^2 \\ \end{bmatrix} &&\begin{aligned}R_3&\gets R_3+2aR_2\\ R_4&\gets R_4-(1+a)R_2\end{aligned} \end{align} For no value of $a$ we have both $1-2a^2=0$ and $a^2+a-6=0$, so the rank of this matrix is $3$ and therefore the vectors $v_1$, $v_2$ and $v_3$ are linearly independent for any value of $a$.

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$v_1, v_2, v_3$ will be linearly dependent if there are $\alpha_1, \alpha_2, \alpha_3$ not all $0$ such that $\alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = 0$

For your $v_1, v_2$ and $v_3$ we have:

$0 = \alpha_1v_1 + \alpha_2v_2 + \alpha_3v_3 = \alpha_1(u_1 + u_2 - u_4) + \alpha_2(au_1 - u_3 + u_4) + \alpha_3(u_2 + au_3 - 6u_4) = u_1(\alpha_1 + a\alpha_2) + u_2(\alpha_1 + \alpha_3) + u_3(-\alpha_2 + a\alpha_3) + u_4(-\alpha_1 + \alpha_2 -6\alpha_3)$

As you have that $u_1, u_2, u_3, u_4$ are linearly independent, you have that

\begin{cases} \alpha_1 + a\alpha_2 = 0 \\ \alpha_1 + \alpha_3 = 0\\ -\alpha_2 + a\alpha_3 = 0 \\ -\alpha_1 + \alpha_2 -6\alpha_3 = 0 \end{cases}

$\implies$

\begin{cases} a = \frac{-\alpha_1}{\alpha_2} \\ \alpha_1 = -\alpha_3\\ a = \frac{\alpha_2}{\alpha_3} \\ -\alpha_1 + \alpha_2 -6\alpha_3 = 0 \end{cases}

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  • $\begingroup$ What is $\alpha_4$? $\endgroup$
    – almagest
    Commented May 12, 2016 at 18:45
  • $\begingroup$ Are you claiming that $v_1,v_2,v_3$ are linearly dependent? I cannot see any $a$ for which that is true. $\endgroup$
    – almagest
    Commented May 12, 2016 at 18:47
  • $\begingroup$ Now you have to compute a little bit more! You won't find a specific value for a... You will find something in terms of $\alpha_1, \alpha_2, \alpha_3$ $\endgroup$ Commented May 12, 2016 at 18:59

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