If S is a set of $7$ vectors in $\mathbb{R}^6$. Is $S$ linearly dependent/independent? Or does it depend on the vectors chosen?

Question

Struggling with this question a little bit. I thought it depends on the vectors chosen, but then I realized I was wrong. If I were to to guess, I'd definitely go with $S$ is linearly dependent, but I'm not sure why. Can anyone provide some reasoning, or correct me if I'm wrong?

Answer 1

Intuitively and in bare words, you can see that as a matter of degrees of freedom. In $R^n$ you have $n$ DOF: everytime you create a new vector independent from the others you shall "consume" one DOF.
To realize that, consider you have placed a first vector: then placing a vector independent from the first, means that you shall place it "not-aligned" with the first. i.e. $\mathbf v_2 \ne \lambda \mathbf v_1$. Two vectors not aligned, define a plane ($\mu \mathbf v_2 + \lambda \mathbf v_1$) = 2 DOF.
The third shall not be in that plane: 2 ==> 3 DOF.
Then if you put more than $n$ vectors, they should be dependent from the first $n$ ones, or some of these were not independent already.

Answer 2

It is linear dependent. The maximum number of linear independent vectors in a vector space of dimension $n$ is $n$.

Justifying this needs more or less work depending on how much linear algebra you know or trust in.

Let us assume there were seven linear independent vectors $v^{(i)}$ in $\mathbb{R}^6$.

Then already $v^{(1)}, \dotsc, v^{(6)}$ would be linear independent. See appendix A.

Six linear independent vectors form a basis of $\mathbb{R}^6$. (This is a fact of linear algebra I base this argument on.)

So as $v^{(7)} \in \mathbb{R}^6$ we have $$ v^{(7)} = \sum_{i=1}^6 \alpha_i v^{(i)} $$ but this means $$ 0 = \sum_{i=1}^6 \alpha_i v^{(i)} - 1 \cdot v^{(7)} = \sum_{i=1}^7 \beta_i v^{(i)} $$ so we found a non-trivial linear combination with coefficients $\beta = (\alpha_1, \dotsc, \alpha_6, -1) \ne 0$ of the null vector and the seven vectors are linear dependent. Contradiction.

Appendix A

If seven vectors $v^{(i)}$ are linear independent then already $v^{(1)}, \dotsc, v^{(6)}$ would be linear independent:

Otherwise $v^{(1)}, \dotsc, v^{(6)}$ were linear dependent which means there are $\lambda_i$ with $$ 0 = \sum_{i=1}^6 \lambda_i v^{(i)} $$ and not all $\lambda_i = 0$. But this would mean $$ 0 = \sum_{i=1}^6 \lambda_i v^{(i)} + 0 \cdot v^{(7)} = \sum_{i=1}^7 \mu_i v^{(i)} $$ where $\mu = (\lambda_1, \dotsc, \lambda_6, 0)$ is a non-trivial vector of coefficients, which means the seven vectors would be linear dependent. Contradiction.

Answer 3

Use the pigeon-hole principle:

Say you have found six linearly independent vectors in $\mathbb{R}^6$. Since they already span $\mathbb{R}^6$, you cannot find a seventh vector that lies in $\mathbb{R}^6$ that is also linearly independent to all the others.