# Spanning $\mathbb {C}^4$ with a system of vectors

I have the following set of column vectors: $$v_1 = (1, 1, 0, 0)^T\\ v_2 = (1, 0, 1, 0)^T\\ v_3 = (0, 0, 1, 1)^T\\ v_4 = (0, 1, 0, 1)^T$$ I was asked:

(a) Are the vectors linearly independent?

(b) Do they span $\mathbb{R}^4$?

(c) Do they span $\mathbb{C}^4$?

I solved parts (a) and (b) be writing the augmented matrix that corresponds to the vector equation $av_1 + bv_2 +cv_3 +dv_4 = 0$. I found that the system had free variables, and thus the system of vectors is not linearly independent because there exist non-trivial solutions to the above vector equation. I used the fact the there did not exist a pivot in every row of the coefficient matrix to reason that the system of vectors does not span $\mathbb{R}^4$.

However, I want to ensure that I understand part (c) correctly: I do not believe that these vectors span $\mathbb{C}^4$ for the same reason that they do not span $\mathbb{R}^4$. Is this correct? I may be missing something. Btw, I am an engineering student.

Best

• How could they span $\mathbb{C}^{4}$ if they do not span $\mathbb{R}^{4}$, when $\mathbb{C} \cong \mathbb{R}^{2}$? – Thomas Russell Sep 24 '15 at 21:27
• If the vectors are linearly dependent over $\Bbb R$, they're linearly dependent over $\Bbb C$. – Arthur Sep 24 '15 at 21:38
• So, the system of vectors not spanning R^4 implies that they do not span C^4? If so, why? – Dane C-Rape Sep 24 '15 at 21:44

You're not missing anything. It can definitely be a little subtle to pin down exactly why "dependent in $\Bbb R^4$ implies dependent in $\Bbb C^4$," though. Note that I'm implicitly using dimension here, since four vectors will only span a four-dimensional space if they're linearly independent (that is, not linearly dependent).

It's essentially because $\Bbb R$ is a subfield of $\Bbb C$. Let's call your set of vectors $S = \{v_1, v_2, v_3, v_4\}$, and consider the two spans

$$\operatorname{Span}_\Bbb R(S) = \left\{\sum \alpha_i v_i : \alpha_i \in \Bbb R,\ v_i \in S \right\}$$ and

$$\operatorname{Span}_\Bbb C(S) = \left\{\sum \alpha_i v_i : \alpha_i \in \Bbb C,\ v_i \in S \right\}.$$

Now, since $\Bbb R \subset \Bbb C$, we have $\operatorname{Span}_\Bbb R(S) \subset \operatorname{Span}_\Bbb C(S)$; if you can use real coefficients to achieve a particular linear combination, then you can certainly use complex coefficients (by continuing to use the same real coefficients).

To say the set $S$ is linearly dependent in $\Bbb R^4$ is to say that $0 \in \operatorname{Span}_\Bbb R(S)$ (requiring at least one $\alpha_i$ to be nonzero), which implies that $0 \in \operatorname{Span}_\Bbb C(S)$ (again, with not all $\alpha_i = 0$), hence the set is linearly dependent in $\Bbb C^4$ as well.

• Note that I'm not implying your answer should replicate anything like this; I'm merely explaining why you're correct. – pjs36 Sep 24 '15 at 22:13
• "To say the set S is linearly dependent in R4 is to say that 0∈SpanR(S)", would'nt the zero vector be an element of SpanR(S) even if the vectors were linearly independent (ie. the trivial solution is still in the span)? I am a bit confused by this, but I feel that I am almost there. – Dane C-Rape Sep 24 '15 at 23:07
• @DaneC-Rape That's a very good point; I should have specified that we need to use at least one nonzero coefficient. What I wrote is wrong, it does need tweaked, thank you. – pjs36 Sep 24 '15 at 23:09

The vectors $v_1,v_2,v_3,v_4$ are, indeed, linearly dependent. For example,

$$v_1=\begin{bmatrix} 1\\ 1\\ 0\\ 0 \end{bmatrix}= v_2- v_3+v_4= \begin{bmatrix} 1\\ 0\\ 1\\ 0 \end{bmatrix} -\begin{bmatrix} 0\\ 0\\ 1\\ 1 \end{bmatrix}+ \begin{bmatrix} 0\\ 1\\ 0\\ 1 \end{bmatrix} .$$

So, $v_1,v_2,v_3,v_4$ cannot span $R^4$, a fortiori, they cannot span $C^4.$

$v_2,v_3,v_4$ are linearly independent because, as far as the first component, there are no constants $\gamma$ and $\delta$ such that $\gamma0+\delta0=1.$

As a result $v_2,v_3,v_4$ span a three dimensional real or complex space.