Raising the bar

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 $latex n,$ such a matrix cannot be diagonalized.

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

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

Here $latex A_{i}$ 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 $latex A$ have $latex k$ different eigenvalues $latex \lambda_1,...,\lambda_k.$ Then $latex C^n$ can be represented as a direct sum of $latex k$ invariant subspaces

(2) $latex C^n=N_{\lambda_1}^{(p_1)} \dotplus ... \dotplus N_{\lambda_k}^{(p_k)}.$

The subspace $latex N_{\lambda_i}^{(p_i)}$ consists of only root vectors belonging to the eigenvalue $latex \lambda_i.$

Proof. By Exercise 2 we have $latex C^n=N_{\lambda_1}^{(p_1)}\dotplus L_1$ where the subspace $latex L_1$ has two properties: it is invariant with respect to $latex A$ and the restriction of $latex A$ to $latex L_1$ does not have $latex \lambda_1$ as an eigenvalue. $latex N_{\lambda_1}^{(p_1)}$ consists of root vectors belonging to $latex \lambda_1.$ Applying Exercise 2 to $latex L_1$ we get $latex L_1=N_{\lambda_2}^{(p_2)}\dotplus L_2$ where $latex N_{\lambda_2}^{(p_2)},L_{2}$ have similar properties. Applying Exercise 2 $latex k$ times we finish the proof.

Note that the restriction of $latex A$ onto $latex N_{\lambda_i}^{(p_i)}$ 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

$latex \left(\begin{array}{cccc}\lambda & 1 & 0 & 0 \\
0 & \lambda & 0 & 0 \\0 & 0 & \lambda & 1 \\0 & 0 & 0 & \lambda\end{array}
\right) $   and   $latex \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 $latex \det (A-\lambda _{1}I)=\left(\lambda -\lambda _{1}\right) ^p$ ($latex \lambda _{1}$ - any number).

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

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