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Section 30.2 Examples

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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 7×7 matrix

A=[341−10−4003−40400005−1−10000−22300006−1−20004−505−10431−10−3−1].

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

cA(λ)=(λ−3)4(λ+1)3.

Thus, the eigenvalues of A are λ1=3, with multiplicity m1=4, and λ2=−1, with multiplicity m2=3. 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 λ1=3:

3I−A=[0−4−110400040−40000−211000021−30000−615000−450−540−4−3−11034]row→reduce[100000−101000−100010−1000001−100000000000000000000000]⟹E3(A)=Span{[1000001],[0100010],[0011100]}.

Since the geometric multiplicity of the eigenvalue λ1=3 is not equal to the algebraic multiplicity, we need to continue with powers of (3I−A).

(3I−A)2=[0−1680−816000160−1600000000000160−160000−16016000−16240−24160−16−870−7816]row→reduce[100000−101000−100010−1000000000000000000000000000000]⟹E23(A)=Span{[1000001],[0100010],[0010100],[0001000]}

The dimension of this generalized eigensubspace is equal to the multiplicity of λ1=3, so we have G3(A)=E23(A) (Statement 5 of Theorem 29.6.1). Following Procedure 29.4.1, we extend our basis for E3(A) to one for E23(A) instead of just using the above basis.

G3(A)=Span{[1000001],[0100010],[0011100],[0001000]}.

Now we continue with eigenvalue λ2=−1:

(−1)I−A=[−4−4−110400−440−40000−61100002−3−30000−611000−450−500−4−3−11030]row→reduce[1000000010000000100000001000000010000000100000000]⟹E−1(A)=Span{[0000001]}.

We are not up to the algebraic multiplicity m2=3, so continue with powers of (−I−A):

(−I−A)2=[161616−8−8−160016−16016000032−8−80000088000032−8−800016−1601600161615−8−7−160]row→reduce[10000−10010000000100000001000000010000000000000000]⟹E2−1(A)=Span{[0000001],[1000010]}.

The dimension of E2−1(A) is still not equal to the algebraic multiplicity of λ2=−1, so continue:

(−I−A)3=[−64−64−9648486400−64640−640000−16048480000−32−16−160000−1604848000−64640−6400−64−64−964848640]row→reduce[1000−1−10010010000100000001100000000000000000000000]⟹E3−1(A)=Span{[0000001],[1000010],[1−10−1100]}.

We are now up to the algebraic multiplicity for λ2=−1, so G−1(A)=E3−1(A). 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 E3−1(A) above are already the same as our basis for E2−1(A), and the first vector in that basis is already the same as our basis vector for E−1(A). 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:

G−1(A)=Span{[0000001],[1000010],[1−10−1100]}.

Finally, let's take P to be the matrix whose columns are our basis vectors for G3(A) and G−1(A):

P=[1000011010000−10010000001100−1001000101000101000100].

If you compute the rank of P, you will find that it is 7. Since P is a 7×7 matrix, this tells us that the columns of P form a basis for R7 (Theorem 21.5.5). And since P was formed by combining the bases from two different subspaces of R7, 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 P−1AP anyway and see what happens. Remember that we can do this by row reducing [PAP]→[IP−1AP]:

[1000011341−10−40010000−103−404000010000005−1−100001100−100−223000010001006−1−200010001004−505−101000100431−10−3−1]row→reduce[1000000300−1000010000003000000010000003−10000001000000300000001000000−112000001000000−110000001000000−1].

It worked! Form matrix P−1AP 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:

P−1AP=[300−10300003−10003−1120−1100−1].

We have two blocks, one for each generalized eigenspace of A. 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.