I have to find the Jordan Normal Form of the following matrix:

$$A=\left(\begin{matrix} 3 &-4 & 0 &2\\4 &-5 & -2 &4\\0 &0 &3 &-2\\0 &0 & 2 &-1\end{matrix}\right)$$

Finding the characteristic polynomial, I came across the following:


According to the solution I have, the professer made to following step, in order to find the geometric multiplicity:

$dim V_{\lambda=1}=dimKer(A-I)=$

dimKer$\left(\begin{matrix} 2 &-4 & 0 &2\\4 &-6 & -2 &4\\0 &0 &2 &-2\\0 &0 & 2 &-2\end{matrix}\right)=4-\rho\left(\begin{matrix} 2 &-4 & 0 &2\\4 &-6 & -2 &4\\0 &0 &2 &-2\\0 &0 & 2 &-2\end{matrix}\right)=1$

I have found out the the matrix $(A-I)$ has a rank of 3, so for some reason I don't get, the geometric multiplicity is 1.

I remember there's some theorem about the connection between dimension and rank of matrices, but I can't recall nor understand what's done here.

Can you please explain how to find the geometric multiplicity, and what the solution suggests?

After finding that the algebric and geometric multipicty don't match, it's easy to say that the Jordan Normal Form will be $$diag(J_2(-1),J_2(1))$$



  • $\begingroup$ Maybe it's related to the theorem $dim(KerT)+dim(ImT) =n$? $\endgroup$ – Alan Sep 28 '15 at 19:35
  • $\begingroup$ Yes, it is exactly this result. The rank is $3$ so the dimension of the kernel is $4-3 = 1$ which is the geometric mult, essentially by defintiiion. $\endgroup$ – quid Sep 28 '15 at 19:44

The geometric multiplicity of an eigen-value $\lambda$ of a matrix $A$ is usually defined as the dimension of the eigen-space associated to this eaigen-value. This space is the same as the kernel of $A - \lambda I$.

By the rank-nullity theorem one has that the dimension of the kernel of a $n \times n$ matrix $B$ is $n - \text{rank}(B)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.