Eigenvalues and eigenvectors

[arkturus]

1. The problem statement, all variables and given/known data

Given the matrix A = [1 0 0
-2 1 3
1 1 -1]

Find an invertable matrix X and a diagonal matrix D such that A = XDX^-1

2. Relevant equations
A = XDX^-1


3. The attempt at a solution
I've found that the eigenvalues are -2, 2, and 1, but I'm having issues finding the specific eigenvectors.

For example, with eigenvalue = -2 I get the matrix down to [3 0 0
-2 0 0
0 1 1]

Am I correct in saying that x1 = 0, x2 = 0, and x3 = t, thus the corresponding eigenvector is (0,0,1)^T

[Mark44]

1. The problem statement, all variables and given/known data

Given the matrix A = [1 0 0
-2 1 3
1 1 -1]

Find an invertable matrix X and a diagonal matrix D such that A = XDX^-1

2. Relevant equations
A = XDX^-1


3. The attempt at a solution
I've found that the eigenvalues are -2, 2, and 1, but I'm having issues finding the specific eigenvectors.

For example, with eigenvalue = -2 I get the matrix down to [3 0 0
-2 0 0
0 1 1]
This matrix is not completely reduced. The first row should be divided by 3, and then used to eliminate the second row. I would then swap the second and third rows.


Am I correct in saying that x1 = 0, x2 = 0, and x3 = t, thus the corresponding eigenvector is (0,0,1)^TNope, that's not it. As a check, your eigenvalue \lambda and eigenvector x should satisfy the equation Ax = \lambdax, or equivalently, (A - \lambdaI)x = 0.

[arkturus]

Thanks. I'm now down to:

1 0 0
0 1 1
0 0 0

It looks like x3 is a free variable, so x3 = t. In that case, x2 = -t and x1 = 0.

Good to go?

[HallsofIvy]

Personally, to find eigenvectors, I prefer to use the basic definition:
Saying that -2 is an eigenvalue means that there exist x, y, z, not all 0, such that
\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 3 \\ 1 & 1 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}-2x \\ -2y \\ -2z\end{bmatrix}
which is the same as the three equations
x= -2x
-2x+ y+ 3z= -2y
x+ y- z= -2z

The first equation, of course, just says that x= 0. The second and third then become
3y+ 3z= 0 and y+ z= 0, both of which reduce to z= -y. Yes, any vector of the form (0, y, -y)= y(0, 1, -1) is an eigenvector corresponding to eigenvalue -2.

Notice that your equations for eigenvalues 2 and 1 are almost the same:
x= 2x
-2x+ y+ 3z= 2y
x+ y- z= 2z
and
x= x
-2x+ y+ 3z= y
x+ y- z= z

[Mark44]

Thanks. I'm now down to:

1 0 0
0 1 1
0 0 0

It looks like x3 is a free variable, so x3 = t. In that case, x2 = -t and x1 = 0.

Good to go?See my previous post for how you can check whether this is correct.