1. eigenvalue and eigenspaces

So I've to find the eigenvalues and a basis for each corresponding eigenspaces.

Matrix A=[(2,0,0), (1,-1,0), (-1,3,3)]
After finding the determinant, I got(λ-2) (λ+1) (λ-3)=0. So λ = 2, -1, 3 are the eigenvalues.

[(λ-2, 0,0),(-1, λ+1, 0),(1,-3,λ-3)]

For λ=2; -v_2+v_3=0
3v_2-3v_3=0
-v_3=0

Let v_3 = t, then v_2=-t, but what is v_1? Is it 0? So the span is [0, -1, 1]?

For λ=-1; -3v_1-v_2+v_3=0
-3v_3=0
-4v_3=0

Let v_3=t, then what is v_2 and v_1? Are they both 0? So the span is [0,0,1]?

For λ=3; v_1-v_2+v_3=0
4v_2-3v_3=0

Let v_3=t, then v_2=(3/4)t and v_1=(-1/4)t. So the span is [(-1/4), (3/4), 1]? I think I got this right.

2. Originally Posted by Taurus3
For λ=2; -v_2+v_3=0
3v_2-3v_3=0
-v_3=0

I don't know what are you doing. For $\displaystyle \lambda=2$ the corresponding eigenspace is:

$\displaystyle \ker (A-2I) \equiv\begin{Bmatrix}0=0\\x_1-3x_2=0\\-x_1+3x_2+x_3=0\end{matrix}$

and a basis is:

$\displaystyle B_2=\{(3,1,0)\}$

Fernando Revilla

3. are you sure?

4. Originally Posted by Taurus3
are you sure?
More than that. Check:

$\displaystyle A \begin{bmatrix}{3}\\{1}\\{0}\end{bmatrix}=2\begin{ bmatrix}{3}\\{1}\\{0}\end{bmatrix}$

Fernando Revilla

5. I think you mistook the format of my matrix. My bad. (2,0,0) is one column, (1,-1,0) is the second one, and (-1,3,3) the third one. Now can you check?

6. Originally Posted by Taurus3
So I've to find the eigenvalues and a basis for each corresponding eigenspaces.

Matrix A=[(2,0,0), (1,-1,0), (-1,3,3)]
After finding the determinant, I got(λ-2) (λ+1) (λ-3)=0. So λ = 2, -1, 3 are the eigenvalues.
Yes, for any upper or lower triangular matrices, the eigenvalues are the number on the main diagonal.

[(λ-2, 0,0),(-1, λ+1, 0),(1,-3,λ-3)]
Okay. I tend to think of eigenvector $\lambda$ and eigenvector v as satifying $\displaystyle Av= \lambda v$. You are writing it is $\displaystyle (\lambda I- A)v= 0$ which is the same thing.

For λ=2; -v_2+v_3=0
3v_2-3v_3=0
-v_3=0[/tex]
What happened to $\displaystyle v_1$? The equations, with $\displaystyle \lambda= 2$ are
$\displaystyle (0)v_1= 0$
$\displaystyle -v_1+ 3v_2= 0$
$\displaystyle v_1- 3v_2- v_3$

The first equation does NOT say "$\displaystyle v_1= 0$", it is true for any $\displaystyle v_1$.
The second equation says $\displaystyle v_1= 3v_2$ and the third that $\displaystyle v_3= v_1- 3v_2= 3v_2- 3v_2= 0$
Taking $\displaystyle v_2= 1$, an eigenvector, corresponding to eivenvalue 2, is [3, 1, 0] and the eigenspace is spanned by [3, 1, 0].

(Added: If those were to be columns rather then rows, then your equations are
$\displaystyle v_2- v_3= 0$
$\displaystyle -3v_2+ 3v_3= 0$
$\displaystyle 3v_3= 2v_3= 0$
The third equation gives $\displaystyle v_3= 0$, both first and second equations give $\displaystyle v_2= 0$ but there is no $\displaystyle v_1$ in any of the equations so it can be anything: [t, 0, 0]= t[1, 0, 0]. The eigenspace is spanned by [1, 0, 0])
Of course, it is easy to check whether or not a given vector is an eigenvector. Did you do that?

Let v_3 = t, then v_2=-t, but what is v_1? Is it 0? So the span is [0, -1, 1]?

For λ=-1; -3v_1-v_2+v_3=0
-3v_3=0
-4v_3=0

Let v_3=t, then what is v_2 and v_1? Are they both 0? So the span is [0,0,1]?

For λ=3; v_1-v_2+v_3=0
4v_2-3v_3=0

Let v_3=t, then v_2=(3/4)t and v_1=(-1/4)t. So the span is [(-1/4), (3/4), 1]? I think I got this right.

7. Originally Posted by Taurus3
I think you mistook the format of my matrix. My bad. (2,0,0) is one column, (1,-1,0) is the second one, and (-1,3,3) the third one. Now can you check?
To avoid cofusion, I suggest you next time to write in $\displaystyle \LaTeX$ code. Alternatively, use (a,b,...)^T for columns.

Fernando Revilla

8. for λ= -1, would the span be [0, 0, 0]?

-3v_1-v_2+v_3=0
-3v_3=0
-4v_3=0

Thanks.

9. Originally Posted by Taurus3
for λ= -1, would the span be [0, 0, 0]?

Impossible. If $\displaystyle E$ is an eigenspace then $\displaystyle E\neq \{0\}$ .

-3v_1-v_2+v_3=0
-3v_3=0
-4v_3=0

$\displaystyle (v_1,v_2,v_3)^T=(\alpha,-3\alpha ,0)^T=\alpha(1.-3,0)^T\quad (\alpha\in \mathbb{R})$

Fernando Revilla