MHB Linearly Independent: Why ${}\left\{X+Y, Y+Z, Z+W, W+X\right\}$ Isn't

[Dethrone]

Let ${}\left\{X, Y, Z, W\right\}$ be an independent set in $\Bbb{R}^n$, is the following set independent?
${}\left\{X+Y, Y+Z, Z+W, W+X\right\}$

My textbook says it isn't, but I'm not sure why. Let $\lambda_1, \lambda_2, \lambda_3, \lambda_4$ be scalars, then $\lambda_1(X+Y)+\lambda_2(Y+Z)+\lambda_3(Z+W)+\lambda_4(W+X)=0$. Expanding an simplifying, we get $(\lambda_1+\lambda_4)X+(\lambda_1+\lambda_2)Y+(\lambda_2+\lambda_3)Z+(\lambda_3+\lambda_4)W=0$.

Since we are given that ${}\left\{X, Y, Z, W\right\}$ is independent, then $(\lambda_1+\lambda_4)=0$, $(\lambda_1+\lambda_2)=0$, $(\lambda_2+\lambda_3)=0$, and $(\lambda_3+\lambda_4)=0$. This implies that $\lambda_1=\lambda_3=-\lambda_2=-\lambda_4=0$. Why is it not independent, then?

 

[Euge]

Hello again Rido12,

The set $\{X + Y, Y + Z, Z + W, W + X\}$ is dependent because $$(X + Y) - (Y + Z) + (Z + W) - (W + X) = 0.$$ This relation has nothing to do with the independence of $\{X, Y, Z, W\}$; the above identity holds for all $X, Y, Z, W \in \Bbb R^n$.

 

[Dethrone]

Hi Euge! That makes complete sense, but how come I wasn't able to distill that solution when wrote it as a linear combination? What is wrong with the reasoning?
Starting with this line:
$\lambda_1(X+Y)+\lambda_2(Y+Z)+\lambda_3(Z+W)+\lambda_4(W+X)=0$, a solution is clearly $\lambda_1=1, \lambda_2=-1, \lambda_3=1, \lambda_4=-1$, as you have pointed out.
What is wrong with rewriting that as $(\lambda_1+\lambda_4)X+(\lambda_1+\lambda_2)Y+(\lambda_2+\lambda_3)Z+(\lambda_3+\lambda_4)W=0$ and using ${}\left\{X, Y, Z, W\right\}$'s independence as a tool? Could the question be purposely trying to mislead, as that fact was clearly given as an assumption?

 

[Euge]

You're making mistakes in your calculations. The conditions $\lambda_1 + \lambda_4 = \lambda_1 + \lambda_2 = \lambda_2 + \lambda_3 = \lambda_3 + \lambda_4$ imply $\lambda_1 = \lambda_3$ and $\lambda_2 = \lambda_4$. Since $\lambda_1 + \lambda_2 = 0$, $\lambda_1 = -\lambda_2$. The general solution will be $(\lambda_1, \lambda_2, \lambda_3, \lambda_4) = \lambda(1,-1,1,-1)$, $\lambda \in \Bbb R$.

 

[bwpbruce]

A set of vectors $\{\textbf{W}, \textbf{X}, \textbf{Y}, \textbf{Z}\}$ are $\textbf{Linearly Independent}$ if there exists scalars $a,b,c,d$ such that $a\textbf{W} + b\textbf{X} + c\textbf{Y} + d\textbf{Z} = \textbf{0}, a = b = c = d = \textbf{0}$.

A set of vectors $\{\textbf{W}, \textbf{X}, \textbf{Y}, \textbf{Z}\}$ are $\textbf{Linearly Dependent}$ if there exists scalars $a,b,c,d$ such that $a\textbf{W} + b\textbf{X} + c\textbf{Y} + d\textbf{Z} = \textbf{0}$ and $a - d$ are not all $\textbf{0}$.

Let $W = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, X = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, Y = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, Z = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

Then

$X + Y = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}
\\Y + Z = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}
\\Z + W = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}
\\W + X = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

And
$\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix} - \left(\begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}\right)
= \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} - \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} = \textbf{0}$

Therefore:
$(Z + W) + (X + Y) - ((Y + Z) + (W + Z)) = \textbf{0}$
and $(a, b, c, d) = (1, 1, -1, -1)$

Conclusion:
The set $\{X + Y, Y + Z, W + Z, Z + W\}$ is not linearly independent.

 

[Dethrone]

Thanks Euge and bwpbruce for the help!It turns out that I actually had the answer from the beginning when I stated $\lambda_1=\lambda_3=-\lambda_2=-\lambda_4$ in the first post, as it not only implies that $\lambda_1=\lambda_2=\lambda_3=\lambda_4=0$, but $\lambda_1=\lambda_3=1$ and $\lambda_2=\lambda_4=-1$ and vice-versa, which does satisfy the equation. Needless to say I probably stayed up too late that night.