Is Lambda = 1 an Eigenvalue of the Given Matrix?

[matt_crouch]

Homework Statement

Show that lambda = 1 is an eigenvalue of the matrix

2,-1, 6
3,-3, 27
1,-1, 7

and find the eigenvalues and the corresponding eigenvectors

Homework Equations

The Attempt at a Solution

I don't understand how to actually get eigenvalues and eigenvectors
could someone help me out please
=]

 

[HallsofIvy]

Do you understand the definition of "eigenvalue"? "$\alpha$" is an eigenvaue of matrix A if and only if there is some non-zero vector, v, such that $Av= \alpha v$.

In particular, "1" is an eigenvalue of this matrix if and only if
$$\begin{bmatrix}2 &-1 & 6 \\ 3 & -3 & 27 \\ 1 & -1 & 7\end{bmatrix}\begin{bmatrix} x \\ y \\ z\end{bmatrix}= 1\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
for some not all of x, y, z zero.

Multiplying that left side gives
$$\begin{bmatrix}2x- y+ 6z \\3x- 3y+ 27z \\ x- y+ 7z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$

Now, solve that using whatever method (row reduction, determinants, etc.) you choose. The most basic is to write that as the system of equations
2x- y+ 6z= x, 3x- 3y+ 27z= y, and x- y+ 7z= z. Those are the same as x- y+ 6z= 0, 3x- 2y+ 27z= 0, and x- y+ 6z= 0.

Notice that the first and second equations are the same- we really have only two equations, not 3 and so there is no unique answer. That had to happen. Obviously, v= 0 is a solution to $Av= \alpha v$ for any A and $\alpha$. Saying there is a non-zero vector such that that is true means there is NO "unique" answer.

Getting back to our equations: if we multiply x- y+ 6z= 0 by 2 we get 2x- 2y+ 12z= 0 while the second equation above was 3x- 2y+ 27z= 0. Subtracting 2x- 2y+ 12z= 0 from that, x+ 15z= 0 so x= -15z. That's as far as we can go, we cannot solve for a specific 5value of x. But putting x= -15z into x- y+ 6z= 0 gives -15x- y+ 6z= 0 or y= -7z.

What does that mean? That means that any vector of the form v= <15z, -7z, z> satisfies Av= v, no matter what z is. Any such vector, for non-zero z, is an eigenvector corresponding to eigenvalue $\lambda$.

More generally, if $Av= \lambda v$, then $Av- \lambda v= (A- \lamba I)v= 0$. The point of writing it that way is that equation always has the solution v= 0 and will have a non-zero answer only if v= 0 is not unique. If $A- \lambda I$ had an inverse matrix, then we could multiply on both sides by that inverse and get the unique answer $v= (A- \lambda I)^{-1}0= 0$. That is, $Av= \lambda v$ has a non-zero solution if and only if $A- \lambda I$ is not invertible which is the same as saying it has determinant 0.

The "eigenvalue equation" or "characteristic equation" for matrix A is "$det(A- \lambda I)= 0$". For your example, that is
$$\left|\begin{array}{ccc}2-\lambda & -1 & 6 \\ 3 & -3-\lambda & 27 \\ 1 & -1 & 7-\lambda \end{array}\right|= 0$$
Working out that determinant on the left will give you a cubic polynomial equation (the "eigenvalue equation" or "characteristic equation" for an n by n matrix is typically an $n^{th}$ degree polynomial). Since you already know that $\lambda= 1$ is a solution, you can reduce it to a quadratic equation for the other two eigenvalues.

 

[matt_crouch]

ok i got to the cubic polynomial and just want to check that i have the right polynomial

im rubbish with latex so let x = lambda

x³+6x²+20x=0

i then simplified to get

-x(x²-6x-20)=0

 

[HallsofIvy]

ok i got to the cubic polynomial and just want to check that i have the right polynomial

im rubbish with latex so let x = lambda

x³+6x²+20x=0

i then simplified to get

-x(x²-6x-20)=0

Unfortunately, there is one serious problem with that! Remember that you were asked to show that "1" is an eigenvalue? (It is.) But $1^3+ 5(1)+ 10= 26$, not 0.