Eigenvector Help: Solving Problems 3 & 4

[Quadruple Bypass]

http://orion.math.iastate.edu/vika/cal3_files/lec33267.pdf

i searched eigenvectors on google and this showed up. here are some problems i need further explaining for example 3 and 4.

3. where do they get e1+e2= 0 equation from? then where did they get e= (1,-1)

4. where do they get the e1+2e2+e3= 0 eq from?

any help would really be appreciated...i understand how to get the eigenvalues but i have no clue what to do to get the vectors

 

[chanvincent]

3.
Write out the vector \vec{e} explicitly, i.e. (e_1,e_2), then substitude into (A-3E) \vec{e} = 0, you will see the result immediately.

4.
Same as above...

 

[HallsofIvy]

The first matrix is
\left(\begin{array}{cc}-2 & 1 \\ -1 & 4\end{array}\right)
which, by solving the "characteristic equation" is determined to have the single eigenvalue -3.

The definition of "eigenvalue" is that there exist a non-trivial (i.e. non-zero) vector v such that Ax= \lambda x. Saying that -3 is an eigenvalue means that there is a non-zero vector
v= \left(\begin{array}{c}x \\ y\end{array}\right)
such that
\left(\begin{array}{cc}-2 & 1 \\ -1 & 4\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)= \left(\begin{array}{c}-2x+ y \\ -x+ 4y\end{array}\right)= \left(\begin{array}{c}-3x \\ -3y\end{array}\right).

The top row says -2x+ y= -3x and the second -x+ 4y= -3y. Normally, two equations in two variables would have a unique solution (and obviously x=0, y= 0 is a solution) but here the equations are not "independent" (precisely because -3 is an eigenvalue). It's easy to see that both equations reduce to y= -x. That is the same as x+y= 0 (your book uses e₁+e₂= 0 but it is the same thing). Any vector of the form (a, -a)= a(1,-1) is an eigenvector corresponding to eigenvalue -3.

 

[Quadruple Bypass]

thanks for the help, hopefully it will make sense with time