Let $V$ be a vector space over a field $K$. If vectors $v_1$, $v_2$, $v_3$, $v_4$ are linearly independent, what can be said about the vectors $2v_1+v_2+v_3$, $v_1-3v_2$, $v_4+v_3$, and $v_3-v_1$?

My guess is that the last four vectors are linearly independent among them. And I tried to prove this: $$v_1, v_2, v_3, v_4 \text{ linearly independent} \Rightarrow \\(\alpha_1v_1+\alpha_2v_2+\alpha_3v_3+\alpha_4v_4 = 0,\ \alpha_1,\alpha_2,\alpha_3,\alpha_4\in K \Rightarrow \alpha_1=\alpha_2=\alpha_3=\alpha_4=0)$$ Here is my first problem. Because I do not know who $K$ is, can I say all those coefficients are equal to $0$? What if the field isn't made of numbers? Or can here $0$ simply denote the null element of $K$?

I then continue by letting $\beta_1,\dots,\beta_4\in K$ and finding that $$\beta_1(2v_1+v_2+v_3) +\beta_2(v_1-3v_2)+\beta_3(v_4+v_3)+\beta_4(v_3-v_1)=\\=v_1(2\beta_1+\beta_2-\beta_4)+v_2(\beta_1-3\beta_2)+v_3(\beta_1+\beta_3+\beta_4) + v_4\cdot\beta_3$$

Then, to check if those 4 vectors are linearly independent among them, I do the following: I equal the last relation to $0$ and, because $v_1,\cdots, v_4$ are linearly independent, that means that each coefficient (each paranthesis) must be $0$. Then I form a system of 4 equations and 4 unknowns and attempt to find that $\beta_1=\dots=\beta_4=0$. Here, again, I'm not sure if the $0$ is the null element of $K$. Indeed, after solving the system, all $\beta=0$.

Is this a correct approach/proof?


2 Answers 2


Your understanding is not so bad, just missing a few key elements.

A field is a special example of a ring, which is a specific kind of algebraic object. All rings have two operations, "addition" $(+)$ and "multiplication" $(\cdot )$ which follow nice rules like distributivity. Also, every ring has a zero element $0$, which is the unique element satisfying $0+r=r+0=r$ for all $r$ in the ring. A vector space is something called a module, which is useful, but not particularly important to your question.

For your first concern, yes, elements of $K$ need not be real or complex numbers at all. They can be elements of an arbitrary set, so long as the set and its operations satisfy the field axioms. When we write "$0$", we don't necessarily mean the number $0$, but rather, the zero element of the field $K$.

The rest of your approach seems fine and pretty standard! Isolate the coefficients of $v_1, v_2, v_3, v_4$, set the whole thing equal to $0$, solve for the coefficients and show that they are equal to $0$. If you've done that, then you're done!


"I do not know who $K$ is"

The set $K$ is a field, then you know that the elements of $K$ satisfy the field axioms.

"What if the field isn't made of numbers?" Your proof does not require that elements of $K$ need to be numbers. They can be anything that satisfies the field axioms.

"Or can here $0$ simply denote the null element of $K$?"

In the context above, $0$ denotes the additive identity of $K$.

"Is this a correct approach/proof?"

Yes, because it uses only the properties of a vector space $V$ over a field $K$ and the assumption that $v_1,v_2,v_3,v_4\in V$ are linearly independent.


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