- #1
jostpuur
- 2,116
- 19
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
[tex]
\textrm{det}(\lambda - A)=\lambda^2(\lambda - 1)
[/tex]
I can see that the eigenspace corresponding to the eigenvalue [tex]\lambda=0[/tex] can be either one or two dimensional. For example
[tex]
A=\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}\right)
[/tex]
or
[tex]
A=\left(\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}\right)
[/tex]
What about the eigenspace corresponding to the eigenvalue [tex]\lambda=1[/tex]? 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.
For example if we have a 3x3 matrix A, and a characteristic polynomial
[tex]
\textrm{det}(\lambda - A)=\lambda^2(\lambda - 1)
[/tex]
I can see that the eigenspace corresponding to the eigenvalue [tex]\lambda=0[/tex] can be either one or two dimensional. For example
[tex]
A=\left(\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}\right)
[/tex]
or
[tex]
A=\left(\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{array}\right)
[/tex]
What about the eigenspace corresponding to the eigenvalue [tex]\lambda=1[/tex]? 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.