31
Jul 19

Chipping off root subspaces

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.

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