Find the largest possible number of linearly independent vectors among a given set of vectors

Question

From the book, Linear Algebra by Kuldeep Singh. Miscellaneous Exercise $2,$ Question $2.8:$

Find the largest possible number of linearly independent vectors among

$v_1=\begin{bmatrix} 1 \cr -1 \cr 0 \cr 0 \end{bmatrix} v_2= \begin{bmatrix} 1 \cr 0 \cr -1 \cr 0 \end{bmatrix} v_3= \begin{bmatrix} 1 \cr 0 \cr 0 \cr -1 \end{bmatrix} v_4= \begin{bmatrix} 0 \cr 1 \cr -1 \cr 0 \end{bmatrix} v_5= \begin{bmatrix} 0 \cr 1 \cr 0 \cr -1 \end{bmatrix} v_6= \begin{bmatrix} 0 \cr 0 \cr 1 \cr -1 \end{bmatrix}$

The answer in the back of the book just says, "The first three vectors are linearly independent", which is clearly true. And the answers in the back are very brief and do not give full explanation. But it isn't clear if, behind the scenes, Kuldeep checked that all $\binom{6}{4}=15$ possible combinations of $4$ vectors chosen from these six vectors are linearly dependent. (I have checked this and they are). But is there some reason you do not need to do this? For example, does it suffice to show that the first $3$ vectors are linearly independent, but each of $v_4,\ v_5,$ and $v_6$ are linearly dependent with the first three vectors, and therefore (for some reason) we do not need to check if the other $15-3=12$ possible combinations of four vectors are linearly dependent? If so, why does this suffice?

It is worth mentioning that up until this point in the book, what has been covered is: linear independence and dependence, span, and basis of vectors in $\mathbb{R}^n,$ but things like: kernel, orthogonality, etc. have yet to be covered. I don't think Singh wanted us to use the ideas that haven't yet been covered in the book to answer this question.

Answer 1

Once you know that $\{v_4, v_5, v_6 \} \subseteq \operatorname{span}(v_1, v_2, v_3)$ and that the first three vectors form a linearly independent set, you know that the span of all $6$ vectors is a three-dimensional space. No three-dimensional space can have a set of more than $3$ linearly independent vectors. Since you have identified a subset of $3$ linearly independent vectors, you're done.

Answer 2

Taking the advice of julio_es_sui_glace in the comments to Jean-Claude Arbaut's answer, I try to determine if $\{\mathbf{v_i}\}_{i=1}^{6}$ is a spanning set of $\mathbb{R}^4$. Let $\mathbf{x} = (w,x,y,z)^T.$ The question is: do there exists scalars $k_i,\ 1\leq i\leq 6,$ such that $\displaystyle\sum_{i=1}^{6} k_i\mathbf{v_i} = \mathbf{x}?$ If yes, then $\{\mathbf{v_i}\}_{i=1}^{6}$ is a spanning set of $\mathbb{R}^4;$ if the answer is no for one counter-example $\mathbf{x}$ then $\{\mathbf{v_i}\}_{i=1}^{6}$ is not a spanning set of $\mathbb{R}^4$. We may write the equation $\displaystyle\sum_{i=1}^{6} k_i\mathbf{v_i} = \mathbf{x}$ as an augmented matrix:

$$ \left[\begin{array}{rrrrrr|r} 1 & 1 & 1 & 0 & 0 & 0 & w\\ -1 & 0 & 0 & 1 & 0 & 0 & x\\ 0 & -1 & 0 & -1 & 0 & 1 & y\\ 0 & 0 & -1 & 0 & -1 & -1 & z\\ \end{array}\right]. $$

Applying Gauss-Jordan algorithm to put this into rref (reduced row echelon form) gives:

$$ \left[\begin{array}{rrrrrr|r} 1 & 0 & 0 & -1 & -1 & 0 & -x\\ 0 & 1 & 0 & 1 & 0 & -1 & -y\\ 0 & 0 & 1 & 0 & 1 & 1 & w+x+y\\ 0 & 0 & 0 & 0 & 0 & 0 & w+x+y+z\\ \end{array}\right]. $$

The final row tells us that $w+x+y+z = 0k_1+0k_2+0k_3+0k_4+0k_5+0k_6=0$, so $w+x+y+z$ cannot equal any non-zero real number. Therefore, $ \mathbf{x}$ cannot equal $(1,0,0,0)^T,$ that is, $(1,0,0,0)^T$ is not a linear combination of $\mathbf{v_i},$ and so $\text{Span}(\{\mathbf{v_i}\})\neq\mathbb{R}^4,$ as was desired to prove.

Finally (back to the original question), since all six vectors are not a spanning set, it follows that no four vectors are a spanning set, which implies that no four vectors are a basis, which implies that no four vectors are linearly independent.

Answer 3

All of your vectors are orthogonal to $(1,1,1,1)^T$, hence they lie in a 3-dimensional subspace, and you can't find more than $3$ independent vectors among them.

Answer 4

The coefficients of the vectors are sufficiently simple to notice that $$v_4=v_2-v_1,\qquad v_5=v_3-v_1,\qquad v_6=v_3-v_2.$$ This shows that all vectors are contained in the three-dimensional subspace spanned by $v_1$, $v_2$ and $v_3$. In particular no subset of four of these vectors is linearly independent. It then suffices to verify that $\{v_1,v_2,v_3\}$ is linearly independent.