1. ## Eigenvector

In the attached example, how to reduce to -v1+2v(subscript 2)=0 ? Thank you

2. Originally Posted by jasonmark
In the attached example, can do I reduce to the equation -v1+2v2=0 ? Thank you

To get from here

$\bigg[\begin{array}{cc}2&2\\-1&5\end{array}\bigg]\bigg[\begin{array}{c}v_1\\v_2\end{array}\bigg]=\bigg[\begin{array}{c}3v_1\\3v_2\end{array}\bigg]$

to here

$-v_1+2v_2=0$

We do this:

First multiply the matrices together:

$\bigg[\begin{array}{cc}2&2\\-1&5\end{array}\bigg]\bigg[\begin{array}{c}v_1\\v_2\end{array}\bigg]=\bigg[\begin{array}{c}3v_1\\3v_2\end{array}\bigg]=\bigg[\begin{array}{cc}2v_1&2v_2\\-1v_1&5v_2\end{array}\bigg]$

Now set the two matrices equal to each other:

$\bigg[\begin{array}{cc}2v_1&2v_2\\-1v_1&5v_2\end{array}\bigg]=\bigg[\begin{array}{c}3v_1\\3v_2\end{array}\bigg]$

We can write the matrices as a system of equations:

$\left\{\begin{array}{rcl} 2v_1+2v_2 & = & 3v_1\\-v_1+5v_2 & = & 3v_2\end{array}\right.$

Thus, if we set each equation equal to zero, we have

$\left\{\begin{array}{rcl} -v_1+2v_2 & = & 0\\-v_1+2v_2 & = & 0\end{array}\right.$

Now, how do we determine $v_1$ and $v_2$?

There are infinitely many solutions.

I'll pick a simple one: $v_1=1$ and $v_2=\tfrac{1}{2}$

These two values, and any integer multiple of them will satisfy the equation we came up with.

Does this make sense?

--Chris

3. Thanks a lots!