# $A\vec u =2\vec u$ and $A\vec{v}= 3\vec{v}$, are $\vec{u}$, $\vec{v}$ linearly dependent?

The Question:

Let $A$ be an $n$ x $n$ matrix and $\vec{u}\mbox{,} \vec{v} \in \mathbb{R}^n$ such that $A\vec{u} = 2\vec{u}$ and $A\vec{v} = 3\vec{v}$. $\vec{u} \ne \vec{0}$ and $\vec{v} \ne \vec{0}$. Are $\vec{u}$, $\vec{v}$ linearly dependent?

My gut is that they are independent, but I'm not sure how to prove it. I believe this is how to start...

if $2\vec{u} + 3\vec{v} = 0 \rightarrow \vec{u}$ & $\vec{v}$ are linearly dependent.

\begin{align*} 2\vec{u} + 3\vec{v} = 0 \\ A\vec{u} + A\vec{v} = 0 \\ A(\vec{u} + \vec{v}) = 0 \end{align*}

... now i'm not really sure where to go. If $\vec{u} = -\vec{v}$ then the above is true (and the vectors are linearly dependent). But I also can see that if:

\begin{equation*} A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \\ \end{pmatrix} \end{equation*} \begin{equation*} \vec{u} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{equation*} \begin{equation*} \vec{v} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{equation*} \begin{equation*} A\vec{u} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} = 2\vec{u} \end{equation*} \begin{equation*} A\vec{v} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \\ \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \end{pmatrix} = 3\vec{v} \end{equation*}

Which shows (by observation) that the two vectors are linearly independent and satisfy the conditions of the question.

So how can I finish this problem (which I believe is to show that $\vec{u} \ne -\vec{v}$)?

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there is a mix up of $u$ and $v$ on title and first line. – Maesumi Feb 22 '13 at 3:49
Thanks. Fixed it. – equant Feb 22 '13 at 3:49
If they were dependent then they would have the same eigenvalue. $Au=au$ and $v=cu$ results in $Av=av$. So they are independent. – Maesumi Feb 22 '13 at 3:53

Assume $au+bv=0$.

Then $0=A(au+bv)=a(Au)+b(Av)$.

But you know something about $Au$ and $Av$, so you can do something to that last equation.

You'll get two equations in $u$ and $v$, and you can do something with them.

Can you take it from there?

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I think so. I end up multiplying one equation by a scalar so that I can subtract one equation from the other, leaving me with something like $b\vec{v} = 0$ which is enough to demonstrate that they are independent because the only way the final equation can be true is if $b=0$. Does that work? – equant Feb 22 '13 at 4:27
Pretty much. After you get $b=0$ you go back to show that also $a=0$, and that shows $u$ and $v$ are linearly independent. – Gerry Myerson Feb 22 '13 at 4:54