Edu - "How to Find Eigenvectors of a Matrix

[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

 

[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).

 

[HallsofIvy]

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).

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.

 

[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.