Discover Linear Algebra

Example 30.2.1. A transition matrix of generalized eigenvectors for a matrix with more than one eigenvalue.

Let's see what happens if we try to apply Procedure 29.4.1 to the \(7 \times 7\) matrix

If you compute the characteristic polynomial of \(A\) (maybe use a computer algebra system?), you will find

Thus, the eigenvalues of \(A\) are \(\lambda_1 = 3\text{,}\) with multiplicity \(m_1 = 4\text{,}\) and \(\lambda_2 = -1\text{,}\) with multiplicity \(m_2 = 3\text{.}\) Since \(A\) has two distinct eigenvalues, we will not be able to put \(A\) into scalar-triangular form. But let's compute the (generalized) eigenspaces of \(A\) anyway.

We begin with \(\lambda_1 = 3\text{:}\)

Since the geometric multiplicity of the eigenvalue \(\lambda_1 = 3\) is not equal to the algebraic multiplicity, we need to continue with powers of \((3 I - A)\text{.}\)

The dimension of this generalized eigensubspace is equal to the multiplicity of \(\lambda_1 = 3\text{,}\) so we have \(G_3(A) = E^2_3(A)\) (Statement 5 of Theorem 29.6.1). Following Procedure 29.4.1, we extend our basis for \(E_3(A)\) to one for \(E^2_3(A)\) instead of just using the above basis.

Now we continue with eigenvalue \(\lambda_2 = -1\text{:}\)

We are not up to the algebraic multiplicity \(m_2 = 3\text{,}\) so continue with powers of \((-I - A)\text{:}\)

The dimension of \(E^2_{-1}(A)\) is still not equal to the algebraic multiplicity of \(\lambda_2 = -1\text{,}\) so continue:

We are now up to the algebraic multiplicity for \(\lambda_2 = -1\text{,}\) so \(G_{-1}(A) = E^3_{-1}(A)\text{.}\) Again, remember that we need to build our basis for \(G_{-1}(A)\) one eigensubspace at a time. But notice that this time the first two vectors in our basis for \(E^3_{-1}(A)\) above are already the same as our basis for \(E^2_{-1}(A)\text{,}\) and the first vector in that basis is already the same as our basis vector for \(E_{-1}(A)\text{.}\) So we already have a basis for the generalized eigenspace \(G_{-1}(A)\) of the form required by the scalar-triangularization procedure, without any further adjustment:

Finally, let's take \(P\) to be the matrix whose columns are our basis vectors for \(G_3(A)\) and \(G_{-1}(A)\text{:}\)

If you compute the rank of \(P\text{,}\) you will find that it is \(7\text{.}\) Since \(P\) is a \(7 \times 7\) matrix, this tells us that the columns of \(P\) form a basis for \(\R^7\) (Theorem 21.5.5). And since \(P\) was formed by combining the bases from two different subspaces of \(\R^7\text{,}\) this calculation tells us that the generalized eigenspaces actually form a complete independent set of subspaces.

But are they a complete set of independent, \(A\)-invariant subspaces? We're not yet sure, but let's compute \(\inv{P}AP\) anyway and see what happens. Remember that we can do this by row reducing \(\left[\begin{array}{c|c} P \amp AP \end{array}\right] \to \left[\begin{array}{c|c} I \amp \inv{P}AP \end{array}\right] \text{:}\)

It worked! Form matrix \(\inv{P} A P\) is upper triangular. But it also has a block-diagonal form — to emphasize this, let's remove some of the zeros in the bottom left and top right:

We have two blocks, one for each generalized eigenspace of \(A\text{.}\) Each block is scalar-triangular, with the corresponding eigenvalue down the diagonal of the block, and of size equal to the algebraic multiplicity of the eigenvalue. This pattern is no coincidence.