The problem below has been taken from Klaus Janich's Linear Algebra textbook.

Let $V$ be a vector space over $\mathbb{R}$ and let $a,b,c,d \in V$. Suppose that:

$v_1 = a + b + c + d$

$v_2 = 2a + 2b + c -d$

$v_3 = a + b + 3c - d$

$v_4 = a - c + d$

$v_5 = -b + c - d$

Show that $(v_1,v_2,\ldots,v_5)$ is linearly dependent. There is an elegant way to do this.

I've come up with an argument and I just need someone to check if it works or not.

My Proof Attempt:

Suppose that the given list of vectors, which we denote by $A$ is linearly independent. Then, the list of vectors $B = (a,b,c,d)$ is also linearly independent. This follows from the fact that any linear combination of the vectors in $A$ can be written as a linear combination of the vectors in $B$.

So, $A$ is linearly independent and clearly spans $L(A)$. That is the linear hull of $A$. Hence, the vectors in $A$ form a basis for $L(A)$. However, it is also true that the vectors in $B$ form a basis for $L(A)$ as well. Therefore, $L(A)$ has two bases of different lengths.

That is a contradiction. Hence, the vectors in $A$ are not linearly independent.

Please give detailed feedback on the way I've written my proof up as well. If there's any way I can improve my literary style to match modern standards, I'd gladly take that way. Thank you in advance.

  • $\begingroup$ Bump haha. Could anyone help? $\endgroup$ – Abhi Feb 23 at 3:52

Yeah, that proof is fine. You could also note that you have 5 vectors in a space with dimension at most 4. Hence, they must be linearly dependent.

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  • $\begingroup$ Thank you so much! $\endgroup$ – Abhi Feb 24 at 9:08

You can prove a generalized version of this result. Precisely, given $n$ vectors $\{v_{1},v_{2},\ldots,v_{n}\}$ from the linear space $V$, any set of $n+1$ vectors $\{w_{1},w_{2},\ldots,w_{n},w_{n+1}\}$, where each $w_{j}\in\text{Span}\{v_{1},v_{2},\ldots,v_{n}\}$, is linear dependent.

Let $\{v_{1},v_{2},\ldots,v_{n}\}\subset V$. Since each $w_{j}\in\text{Span}\{v_{1},v_{2},\ldots,v_{n}\}$, the set $\{w_{1},w_{2},\ldots,w_{n+1}\}$ must be LD. Indeed, we have that $\dim\text{Span}\{v_{1},v_{2},\ldots,v_{n}\} \leq n$. If the vectors $\{w_{1},w_{2},\ldots, w_{n+1}\}$ were LI, then we would have that $\dim\text{Span}\{w_{1},w_{2},\ldots,w_{n+1}\} = n+1$, which is impossible, because $$\text{Span}\{w_{1},w_{2},\ldots,w_{n+1}\}\subseteq\text{Span}\{v_{1},v_{2},\ldots,v_{n}\}$$

Hopefully this helps.

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  • $\begingroup$ Thank you for the more general result. I'll certainly look at your proof, it looks very interesting. $\endgroup$ – Abhi Feb 24 at 9:08

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