# Question on Eigenspaces

1. Jan 1, 2008

### smoothman

Hi there, I'm having a bit of a problem understanding eigenspaces, kernel and span. I've searched the net and wikipedia but there doesn't seem to be any clear examples.

I have an example in a book that says this:

Let,
A =
[ 2 2 2 ]
[ 0 2 2 ]
[ 0 0 2 ]

I can see the characteristic polynomial = $(X - 2)^3$ so 2 is the only eigenvalue.

It then calculates the generalised eigenspaces: $V_t(2)$
$V_1(2) =$
ker [ 0 2 2 ]
.....[ 0 0 2 ]
.....[ 0 0 0 ]

The kernel is calculated by row reducing the matrix:

$V_1(2) =$
ker [ 0 1 0 ] = span [1]
.....[ 0 0 1 ]...........[0]
.....[ 0 0 0 ]...........[0]

$V_2(2) =$
ker [ 0 0 1 ] = span [1] [0]
.....[ 0 0 0 ]...........[0] [1]
.....[ 0 0 0 ]...........[0] [0]

$V_3(2) =$
ker [ 0 0 0 ] = span [1] [0] [0]
.....[ 0 0 0 ]...........[0] [1] [0]
.....[ 0 0 0 ]...........[0] [0] [1]

that is the end of the example.

so now here are my questions:

QUESTION 1
What does it mean by : $V_1(2)$, $V_2(2)$, $V_3(2)$ etc.

QUESTION 2
What exactly is the kernal in this example and how is the span calculated from the kernal matrices....??

QUESTION 3
which part of this whole question/example is the eigenspace?

thankyou very much. if this could be explained then it would clear most of the confusion on this topic.

:)

2. Jan 1, 2008

### Hurkyl

Staff Emeritus
I'm confused -- is it just that you don't know the definitions, or is there more to your questions?

3. Jan 1, 2008

### HallsofIvy

Staff Emeritus
$V_1$ is X-2 (more correctly, X-2I), row reduced, $V_2= (X-2I)^2$, row reduced, and $V_3= (X-2I)^3$. Of course, $V_3= 0$ because X satisfies its "characteristic equation", (X- 2I)3= 0.

I don't know what you mean by "the kernel" or "the span". The kernel of any linear matrix, T, is the set of vectors, v, such that Tv= 0. If
$$V_1 x= \left[\begin{array}{ccc}0 & 1 & 0 \\0 & 0 & 1\\0 & 0 & 0\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\left]= \left[\begin{array}{c} y \\ z \\ 0\end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0\end{array}\right]$$
then we must have y= 0 and z= 0. x can be anything so the kernel consists of vectors of the form (x, 0, 0)= x(1, 0, 0). (1, 0, 0) spans that vector space. Similarly for the other two matrices. I would have expected you to have learned "span" and "kernel" long before you start working with eigenvectors.

Eigen space or "generalized eigenspaces"? The eigen space for A itself is the kernel of $V_1$ which is the set of all vectors (x, 0, 0), spanned by (1, 0, 0). The "generalized eigenspaces" include the kernel of $V_2$, all vectors of the form (x, y, 0) which is spanned by (1, 0, 0) and (0, 1, 0) and the kernel of $V_3$ which is all of R3, spanned, of course, by (1, 0, 0), (0, 1, 0), and (0, 0, 1).

4. Jan 1, 2008

### smoothman

i dont know the definitions.. i dont know how they got the kernels, the span etc etc
i also would appreciate what the difference between generalised eigenspace and normal eigenspace is? thanx

5. Jan 1, 2008

### smoothman

thnx. this was a brilliant explanation :) really helped me :) brilliant

6. Jan 1, 2008

### smoothman

oh just one question though.

for V_2(2)
$$V_2x= \left[\begin{array}{ccc}0 & 0 & 1 \\0 & 0 & 0\\0 & 0 & 0\end{array}\right]\left[\begin{array}{c} x \\ y \\ z\end{array}\left]= \left[\begin{array}{c} z \\ 0 \\ 0\end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0\end{array}\right]$$

here z must be 0... so x and y can be anything... the kernal therefore consists of the vectors of form: {1,0,0,} and {1,1,0}..
so how does that deduce the span as$$\left[\begin{array}{c} 1 \\ 0 \\ 0\end{array}\right]\left[\begin{array}{c} 0 \\ 1 \\ 0\end{array}\right]$$
the vector forms arent the same as the span for me...
please clear this final confusion :)

7. Jan 1, 2008

### Vid

x = t
y = s
z = 0

Therefore, (x,y,z) = (t,s,0) = (t,0,0) + (0,s,0) = t(1,0,0) + s(0,1,0) = span[(1,0,0), (0,1,0)]