Find the eigenvalues and a basis of each eigenspace

[Melawrghk]

Homework Statement

Matrix A is
-4 4 4
-4 4 4
4 -4 -4

It has two real eigenvalues, one of multiplicity 1 and one of multiplicity 2. Find the eigenvalues and a basis of each eigenspace.

The Attempt at a Solution

I got the eigenvalues:
the one of multiplicity 1 is -4
the one of multiplicity 2 is 0

I can also get the eigenvectors for both:
for -4 : [1 1 -1]^T
for 0: [2 1 1]^T

But... I don't know where to go from here. At all. Any help would be greatly appreciated.

 

[gabbagabbahey]

Remember, when you diagonalize a matrix you get $D=P^{-1}MP \Rightarrow M=PDP^{-1}$ not just $M = D$. This means that

$$M^n=(PDP^{-1})^n=(PDP^{-1}) \cdot (PDP^{-1}) \cdot (PDP^{-1}) \ldots (PDP^{-1})=PD(P^{-1}P)D(P^{-1}P)\ldots DP^{-1}=PD^nP^{-1}$$

So, while you have correctly calculated $D^n$, you forgot to multiply it by $P$ and $P^{-1}$ to get $M^n$.

Edit-this was in reply to the original question ;0)

 

[gabbagabbahey]

I can also get the eigenvectors for both:
for -4 : [1 1 -1]^T
for 0: [2 1 1]^T

Your second eigenvector is incorrect. You should also be looking for two (generalized) eigenvectors for $\lambda=0$ since that eigenvalue has multiplicity of 2.

Your first eigenvector $x$ will be given by $(A-\lambda I)x=0$ but(!) your second one will not. Your matrix A is degenerate, and so you need to look for generalized eigenvectors. In this case, since $\lambda=0$ has multiplicity two, its eigenvectors are given by $(A-\lambda I)x=Ax=0$ and $(A-\lambda I)^2x=A^2x=0$.

 

[Melawrghk]

edit: oh okay, so the eigenvectors for 0 are:
(1, 1, 0)
(1, 0, 1) ?

 

[gabbagabbahey]

Yup!

 

[Melawrghk]

Sweet, how do I get eigenspaces though? To be honest, I'm not even sure what they are...

 

[gabbagabbahey]

An eigenspace of a given transformation for a particular eigenvalue is the set (linear span) of the eigenvectors associated to this eigenvalue, together with the zero vector (which has no direction).

So, what is the eigenspace of the $\lambda=-4$ eigenvalue? How about the $\lambda=0$ eigenvalue.

 

[HallsofIvy]

You already have two eigenvectors for the space of eigenvectors corresponding to eigenvalue 0, <1,1,0> and <1, 0,1>. Since those are not multiples of one another, they are independent. Two independent vectors in a space of dimension 2? What does that tell you about a basis?

 

[Melawrghk]

So... For eigenvalue -4, the eigenspace would just be the eigenvector? Like [-1, 1, 1]?

And for the 0, it'd be:
[1 1]
[1 0]
[0 1]?

 

[gabbagabbahey]

For the -4, the eigenspace is just <1,1,-1>.

And for the eigenvalue zero, the eigenspace is {<1,1,0>,<1,0,1>} (that is the set of the two vectors)

Clearly, the eigenspace of the -4 eigenvalue is 1D, while the eigenspace of the eigenvalue zero is 2D

 

[HallsofIvy]

No, no, no! An eigenspace is never a single vector. The eigenspace corresponding to eigenvalue -4 is the subspace spanned by {<1, 1, -1>} or having that set as basis.

Similarly, the eigenspace corresponding to eigenvalue 0 is NOT "{<1, 1, 0>, <1, 0, 1>}, it is the subspace spanned by that set or having that set as basis.