31
Jul 19

## Chipping off root subspaces

The basis in which a matrix is diagonal consists of eigenvectors. Therefore if the number of linearly independent eigenvectors is less than $n,$ such a matrix cannot be diagonalized.

The general result we are heading to is that any matrix in $C^{n}$ in an appropriately chosen basis can be written as a block-diagonal matrix

(1) $A=\left(\begin{array}{ccc}A_{1} & ... & ... \\ ... & ... & ... \\... & ... & A_{m}\end{array}\right) .$

Here $A_{i}$ are Jordan cells, with possibly different lambdas on the main diagonal and of possibly different sizes, and all off-diagonal blocks are zero matrices of compatible dimensions. The next exercise is an intermediate step towards that result.

Exercise 1. Let $A$ have $k$ different eigenvalues $\lambda _{1},...,\lambda_{k}.$ Then $C^{n}$ can be represented as a direct sum of $k$ invariant subspaces

(2) $C^{n}=N_{\lambda _{1}}^{(p_{1})} \dotplus ... \dotplus N_{\lambda_{k}}^{(p_{k})}.$

The subspace $N_{\lambda _{i}}^{(p_{i})}$ consists of only root vectors belonging to the eigenvalue $\lambda _{i}.$

Proof. By Exercise 2 we have $C^{n}=N_{\lambda _{1}}^{(p_{1})}\dotplus L_{1}$ where the subspace $L_{1}$ has two properties: it is invariant with respect to $A$ and the restriction of $A$ to $L_{1}$ does not have $\lambda _{1}$ as an eigenvalue. $N_{\lambda _{1}}^{(p_{1})}$ consists of root vectors belonging to $\lambda _{1}.$ Applying Exercise 2 to $L_{1}$ we get $L_{1}=N_{\lambda _{2}}^{(p_{2})}\dotplus L_{2}$ where $N_{\lambda_{2}}^{(p_{2})},L_{2}$ have similar properties. Applying Exercise 2 $k$
times we finish the proof.

Note that the restriction of $A$ onto $N_{\lambda _{i}}^{(p_{i})}$ may not be described by a single Jordan cell. A matrix may have more than one Jordan cell with the same eigenvalue. To make this point clearer, note that the matrices

$\left(\begin{array}{cccc}\lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\0 & 0 & \lambda & 1 \\0 & 0 & 0 & \lambda\end{array} \right)$   and   $\left(\begin{array}{cccc}\lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\0 & 0 & \lambda & 1 \\0 & 0 & 0 & \lambda\end{array}\right)$

are not the same (the first matrix has two Jordan cells on the main diagonal and the second one is itself a Jordan cell). It will take some effort to get from (2) to (1).

Exercise 4. Show that for the matrix from Exercise 3 $\det (A-\lambda _{1}I)=\left(\lambda -\lambda _{1}\right) ^p$ ($\lambda _{1}$ - any number).

Exercise 5. In addition to Exercise 4, show that in $N_{\lambda }^{(p)}$ the matrix $A$ has only one eigenvector, up to a scaling factor. Hint: use the Jordan cell.