Finding nonzero entries in Jordan canonical form

Question

Let $A$ be a $7×7$ matrix such that $2A^2-A^4=I$,where $I$ is identity matrix.If $A$ has Two distinct eigenvalues and each of geometric multiplicity $3$,then what would be possible Jordan Canonical form of $A$?

Answer 1

Let $p(t)=t^4-2t^2+1 = (t-1)^2(t+1)^2.$ We know $p(A)=0,$ which means that $1$ and $-1$ can be the only eigenvalues (we know that the minimal polynomial of $A$ has each eigenvalue of $A$ as its root, and we know that the minimal polynomial divides $p,$ because it divides each polynomial $q$ with $q(A)=0.$) The geometric multiplicity of an eigenvalue is the number of Jordan blocks for that eigenvalue. The sizes of the Jordan blocks must sum up to 7. Putting all those facts together, the following two possibilities are left: Either the three Jordan blocks of the eigenvalue $1$ have sizes $1$, $1$ and $2$ and the three Jordan blocks of the eigenvalue $-1$ all have size $1$, or the three Jordan blocks of the eigenvalue $-1$ have sizes $1$, $1$ and $2$ and the three Jordan blocks of the eigenvalue $1$ all have size $1.$ So either $$ J = \begin{pmatrix} -1 & & & & & & \\ & -1 & & & & & \\ & & -1 & & & & \\ & & & \phantom{-}1 & & & \\ & & & & \phantom{-}1 & & \\ & & & & & \phantom{-}1 & \phantom{-}1\\ & & & & & & \phantom{-}1 \end{pmatrix} $$ or $$ J = \begin{pmatrix} -1 & & & & & & \\ & -1 & & & & & \\ & & -1 & \phantom{-}1 & & & \\ & & & -1 & & & \\ & & & & \phantom{-}1 & & \\ & & & & & \phantom{-}1 & \\ & & & & & & \phantom{-}1 \end{pmatrix} $$ It can now easily be verified that $p(J) =0$ and hence indeed $p(A)=0.$ This can also be concluded from the fact that the size of each Jordan block is smaller than or equal to $2,$ and that each linear factor in $p$ has at least multiplicity of $2.$