Finding Eigenvectors

1. Dec 10, 2008

roam

1. Find the eigenvalues and the eigenvectors corresponding to eigenvalues of the matrix

A = $$\left[\begin{array}{ccccc} 1 & 3 \\ 4 & 2 \end{array}\right]$$

3. The attempt at a solution

$$(\lambda I - A)$$ = $$\lambda \left[\begin{array}{ccccc} 1 & 0 \\ 0 & 1 \end{array}\right] -$$ $$\left[\begin{array}{ccccc} 1 & 3 \\ 4 & 2 \end{array}\right]$$

$$\left[\begin{array}{ccccc} \lambda - 1 & -3 \\ -4 & \lambda - 2 \end{array}\right]$$ $$\left(\begin{array}{ccc}x\\y\end{ar ray}\right) =$$ $$\left(\begin{array}{ccc}0\\0\end{ar ray}\right)$$

det(λI - A) = 0
=> (λ-1)(λ-2)-12 = 0
λ2-3λ-10=0
(λ+2)(λ-5) = 0
λ = -2, 5

My problem is how to find the eigenspaces corresponding to these eigenvalues.

We have two cases, the first one is when $$\lambda = 5$$. In this case we have the following:

$$\left[\begin{array}{ccccc} 4 & -3 \\ -4 & 3 \end{array}\right]$$ $$\left(\begin{array}{ccc}x\\y\end{ar ray}\right) =$$ $$\left(\begin{array}{ccc}0\\0\end{ar ray}\right)$$

So to find the eigenvectors corresponding to $$\lambda = 5$$, I think I should solve the system

$$\left[\begin{array}{ccccc} 4 & -3 \\ -4 & 3 \end{array}\right]$$

4x-3y = 0 .....(1)
-4x+3y = 0 .....(2)

How can I solve this? I'm not sure how this is done (if I minus (1) from (2) to eliminate x then the y would be eliminated as well).

Thanks.

Roam

2. Dec 10, 2008

Avodyne

Notice that if (x,y) is an eigenvector with a certain eigenvalue, so is (cx,cy), where c is any constant. So you cannot expect to find unique values for x and y, but only their relative value. Either of your equations will do this for you (and, of course, give the same answer).

3. Dec 10, 2008

HallsofIvy

Staff Emeritus

Yes, that's exactly right! If you could solve the two equations for a single value of x and y, then there would be only one possible solution- and x= 0, y= 0 obviously satisfies it. The whole point of an "eigenvalue" is that there exist "non-trivial" (i.e. not x=0, y=0) solutions. And once that is true, there exist an infinite number of solutions: the set of all solutions to such an equation for an eigenvalue forms a subspace, the "eigen space".
What you can do is solve for one variable in term of the other. Here, for examply 4x= 3y so y= (4/3)x. Any vector of the form (x, (4/3)x) is in the eigenspace. You can also write that as x(1, 4/3) showing that any such vector is a multiple of (1, 4/3): the subspace is of dimension 1 and a basis for it is {(1, 4/3)}. Of course, if you don't like fractions, you could just factor a "3" out of that x and write (x/3)(3, 4). Since x could be any number, so can "x/3" and {(3, 4)} is a basis for the eigenspace corresponding to eigenvalue 5.

Similarly, for eigenvalue -2x you have
$$\left[\begin{array}{cc} 1 & 3 \\ 4 & 2 \end{array}\right]\left[\begin{array}{c} x \\ y \end{array}\right]= left[\begin{array}{c}-2x \\ -2y\end{array}\right]$$
That is the same as x+ 3y= -2x or 3x+ 3y= 0 and 4x+ 2y= -2y or 4x+ 4y= 0. What x and y satisfy that? Again, you cannot solve for specific x or y but you can solve for y in terms of x or viceversa.

4. Dec 11, 2008

roam

.....yes!

4x-3y = 0
-4x+3y = 0

x = 3/4y ( y is free)

x = $$\left(\begin{array}{ccc}x\\y\end{ar ray}\right) =$$ $$\left(\begin{array}{ccc}3/4 y\\y\end{ar ray}\right) =$$ $$y \left(\begin{array}{ccc}3/4\\1\end{ar ray}\right)$$

Ax = λx

Ax = $$\left[\begin{array}{ccccc} 1 & 3 \\ 4 & 2 \end{array}\right]$$ $$\left(\begin{array}{ccc}3/4 y\\y\end{ar ray}\right) =$$ $$5 \left(\begin{array}{ccc}3/4 y\\y\end{ar ray}\right) \Rightarrow$$ $$\left(\begin{array}{ccc}15/4 y\\5 y\end{ar ray}\right)$$

= 5x

(I know how to do the computations for the second case, $$\lambda = -2$$ )

I understand that system (λI-A) = 0 has non-trivial solutions.

Thanks very much for explaining, Best wishes.