## 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 such a matrix cannot be diagonalized.

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

(1)

Here 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 have different eigenvalues Then can be represented as a direct sum of invariant subspaces

(2)

The subspace consists of only root vectors belonging to the eigenvalue

**Proof**. By Exercise 2 we have where the subspace has two properties: it is invariant with respect to and the restriction of to does not have as an eigenvalue. consists of root vectors belonging to Applying Exercise 2 to we get where have similar properties. Applying Exercise 2

times we finish the proof.

Note that the restriction of onto 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

and

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 ( - any number).

**Exercise 5**. In addition to Exercise 4, show that in the matrix has only one eigenvector, up to a scaling factor. Hint: use the Jordan cell.

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