Dimension of eigenspace, multiplicity of zero of char.pol.

[jostpuur]

Does the multiplicity of zero of characteristic polynomial restrict from above the possible dimension of the corresponding eigenspace?

For example if we have a 3x3 matrix A, and a characteristic polynomial

$$\textrm{det}(\lambda - A)=\lambda^2(\lambda - 1)$$

I can see that the eigenspace corresponding to the eigenvalue $$\lambda=0$$ can be either one or two dimensional. For example

$$A=\left(\begin{array}{ccc}<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 1 \\<br /> \end{array}\right)$$

or

$$A=\left(\begin{array}{ccc}<br /> 0 & 1 & 0 \\<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 1 \\<br /> \end{array}\right)$$

What about the eigenspace corresponding to the eigenvalue $$\lambda=1$$? Is it necessarily one dimensional, or could it be two dimensional?

I'm almost sure that it must be one dimensional, but its difficult to see certainly what's happening if the matrix is not diagonalizable.

 

[gel]

Yes.
If the lambda=1 eigenspace was 2d, then you could choose a basis for which
$$A=\left(\begin{array}{ccc}<br /> 1 & 0 & \cdot \\<br /> 0 & 1 & \cdot \\<br /> 0 & 0 & \cdot \\<br /> \end{array}\right)$$
- just take the first two vectors of the basis in the eigenspace.
Then, it should be clear that the determinant of
$$\lambda-A=\left(\begin{array}{ccc}<br /> \lambda-1 & 0 & \cdot \\<br /> 0 & \lambda-1 & \cdot \\<br /> 0 & 0 & \cdot \\<br /> \end{array}\right)$$
has a factor of $(\lambda-1)^2$, which would contradict your assumption.

 

[jostpuur]

I see.

 

[matt grime]

That is not correct: you cannot assume that the matrix is diagonalizable. What you get tells you about the dimension of the generalized eigenspaces. But really, you need the minimal polynomial, not just the characteristic polynomial, for full information.

 

[maze]

$$\frac{1}{2}\left(\begin{matrix}2 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1\end{matrix}\right)$$

nevermind this is a strange matrix, which doesn't quite work

Wow, i just made this way harder than it needs to be...

$$\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{matrix}\right)$$

should suffice

 

[matt grime]

Note, I misread Gel's post, but it's now too late to edit mine.