Geometric and Algebric multiplicity of a Matrix

Question

I'd like to proof that this matrix$$ A=\left(\begin{array}{ccccc} 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 3 & 0 & 0\\ 0 & 4 & 2 & 3 & 0\\ 0 & -3 & -2 & 1 & 3 \end{array}\right) $$ has this Jordan form $$ J=\left(\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 3 & 1 & 0\\ 0 & 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 0 & 3 \end{array}\right) $$ I have as characteristic polynomial $p_{A}(\lambda)=\lambda I-A=\left(\lambda+1\right)\left(\lambda-1\right)\left(\lambda-3\right)^{3}$ Can somebody help me to prove the geometric multiplicity for the eigenvalue =3 ?

Answer 1

i get the rank $4$ matrix $\pmatrix{1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&0}$ for the row reduced form of $A - 3I.$ that means the null space of $A-3I$ is of rank one. that indicates you have jordan block of size 3 of the form $\pmatrix{3&1&0\\0&3&1\\0&0&3}.$

edit: to find the eigenvector $u$ and the generalized eigenvectors $v, w$ you need to solve the system of equations $$(A-3I)u = 0, (A-3I)v = u, (A-3I)w = v$$ again by row reduction you will find that $u = e_5, v = e_4, \text{and } w = 0.5e_3 + e_4.$