7
$\begingroup$

Let $V$ be a real vector space and $a,b,c,d,e\in V$.

We have the vectors \begin{align*}&v_1=a+b+c \\ &v_2=2a+2b+2c-d \\ &v_3=a-b-e \\ &v_4=5a+6b-c+d+e \\ &v_5=a-c+3e \\ &v_6=a+b+d+e\end{align*}

I want to justify that these vectors are linearly dependent.

$$$$

These six vectors are linearly dependetn iff we can get the zer0-vector by a linear combination of these vectors, \begin{equation*}\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4+\lambda_5v_5+\lambda_6v_6=0\end{equation*} where at least of the coefficients $\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_4$ or $\lambda_6$ is not equal to zero.

We have the following:

\begin{align*}&\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4+\lambda_5v_5+\lambda_6v_6=0 \\ & \Rightarrow \lambda_1\left (a+b+c\right )+\lambda_2\left (2a+2b+2c-d\right )+\lambda_3\left (a-b-e\right )+\lambda_4\left (5a+6b-c+d+e\right )+\lambda_5\left (a-c+3e\right )+\lambda_6\left (a+b+d+e\right )=0 \\ & \Rightarrow \lambda_1a+\lambda_1b+\lambda_1c+2\lambda_2 a+2\lambda_2 b+2\lambda_2 c-\lambda_2d+\lambda_3 a-\lambda_3 b-\lambda_3 e+5\lambda_4 a+6\lambda_4 b-\lambda_4 c+\lambda_4 d+\lambda_4 e+\lambda_5 a-\lambda_5 c+3\lambda_5 e+\lambda_6a+\lambda_6b+\lambda_6d+\lambda_6e=0 \\ & \Rightarrow a\left (\lambda_1+2\lambda_2+\lambda_3+5\lambda_4+\lambda_5+\lambda_6\right )+b\left (\lambda_1+2\lambda_2-\lambda_3+6\lambda_4+\lambda_6\right )+c\left (\lambda_1+2\lambda_2 -\lambda_4-\lambda_5\right )+d\left (-\lambda_2+\lambda_4+\lambda_6\right ) +e\left (-\lambda_3 +\lambda_4 +3\lambda_5 +\lambda_6\right )=0\end{align*}

Is everything correct? How could we continue?

Do we use the fact that $a,b,c,d,e\in V$ ? But how?

$\endgroup$
3
  • 2
    $\begingroup$ I didn't check the computations, but the reasoning is correct. Yo are already using those are elements of $V$ since you are summing them and multiplying by $\lambda_i$. $\endgroup$
    – Exodd
    Jan 2, 2018 at 14:44
  • 3
    $\begingroup$ By the way, if you know a bit about dimensions in vector spaces, you can prove those are dependent without computations $\endgroup$
    – Exodd
    Jan 2, 2018 at 14:46
  • $\begingroup$ Thank you!! :-) @Exodd $\endgroup$
    – Mary Star
    Jan 2, 2018 at 15:41

2 Answers 2

Reset to default
13
$\begingroup$

The space spanned by $5$ vectors is at most (if a,b,c,d,e happen to be linearly independent) $5$ dimensional. You are asking if $6$ vectors in such a space are linearly dependent, but they must be by something called the exchange lemma.

$\endgroup$
5
  • $\begingroup$ So, we have the following: The dimension of a vector space id the number of the basis vectors und so the maximal number of linearly independent vectors. We consider the space that is spanned by the vectors $a,b,c,d,e$. The dimension of this vector space is at most $5$. The vectors $v_1, v_2, v_3, v_4, v_4, v_5, v_6$ belong to the span of $a,b,c,d,e$, since the span of $a,b,c,d,e$ is the set of all linear combonations of these $5$ vectors. $\endgroup$
    – Mary Star
    Jan 2, 2018 at 15:39
  • $\begingroup$ The vectors $v_1, v_2, v_3, v_4, v_4, v_5, v_6$ cannot be linear independent, since the dimension of the span is at most $5$ und so the maximal number of independent vectors is at most $5$. Have I understood everything correctly? $\endgroup$
    – Mary Star
    Jan 2, 2018 at 15:39
  • 1
    $\begingroup$ yup! good work ! $\endgroup$
    – operatorerror
    Jan 2, 2018 at 15:40
  • $\begingroup$ Great!! Thank you so much!! :-) $\endgroup$
    – Mary Star
    Jan 2, 2018 at 15:41
  • 1
    $\begingroup$ certainly, good luck $\endgroup$
    – operatorerror
    Jan 2, 2018 at 15:49
3
$\begingroup$

If you select the set $\{v_k \}_{k=1}^{\color{red}{5}}$ you can write

$$ \left(\begin{array}{c}v_1\\ v_2 \\ v_3 \\v_4 \\ v_5 \end{array}\right) = \left(\begin{array}{ccccc} 1 & 1 & 1 & 0 & 0 \\ 2 & 2 & 2 & -1 & 0 \\ 1 & -1 & 0 & 0 & -1 \\ 5 & 6 & -1 & 1 & 1 \\ 1 & 0 & -1 & 0 & 3 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c \\ d \\ e\end{array}\right) $$

Which can be solved for $\{a, \cdots, e \}$ since the array as determinant different from zero (-44). This means it is possible to write $a, \cdots, e$ as a linear combination of $v_1, \cdots, v_5$.

This means that $v_6$ can be written as a linear combination of $\{v_k \}_{k=1}^{\color{red}{5}}$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .