I've encountered a little trouble while solving this problem:
Let $\overrightarrow{u}=(\lambda,1,0),\space\overrightarrow{v}=(1,\lambda,1),\space\overrightarrow{w}=(0,1,\lambda)$.
Find all values of $\lambda$ which make {$\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$} a linearly dependent subset of $\mathbb{R}^3$.
My solution:
As the linear dependent condition, I rewrite those vectors as a linear combination:
$\alpha\overrightarrow{u}+\beta\overrightarrow{v}=\overrightarrow{w}$
And I obtained these linear systems:
$\alpha\lambda+\beta=0\\
\alpha+\lambda\beta=1\\
\beta=\lambda$
Lastly, when I solved these linear systems, I found that $\lambda=0, \lambda=-1, \lambda=1$
I am confused with my own method and solution. Is someone can explain to me where did I go wrong? I will appreciate it. Sorry for my poor English.
Best Answer
You made some mistake while solving the system, because its solutions are $(\alpha,\beta,\lambda)=(1,0,0)$, $(\alpha,\beta,\lambda)=\left(-1,\sqrt2,\sqrt2\right)$, and $(\alpha,\beta,\lambda)=\left(-1,-\sqrt2,-\sqrt2\right)$. Besides, this will give you only the cases in which $\vec w$ is a linear combination of $\vec u$ and $\vec v$. But it could also happen that $\vec u$ is a linear combination of $\vec v$ and $\vec w$ or that $\vec v$ is a linear combination of $\vec u$ and $\vec w$.