Vectors and Vector Operations

10.2 Generalized Eigenvectors

In the previous section we looked at the case where each eigenvalue of a square matrix A has as many linearly independent eigenvectors as its multiplicity. In that case we could diagonalize A and use this to compute its powers as in chapter six. Now we turn to situations where some eigenvalues do not have as many linearly independent eigenvectors as their multiplicities. In such cases we shall consider what are called generalized eigenvectors. These turn out to be a substitute for regular eigenvectors. In particular, they can be used to express A as A = TJT-1 where J, called the Jordan canonical form of A, is "almost" diagonal. This is a generalization of diagonalization and can be used to compute the powers in a fashion similar to chapter six. Let's look at an example.

Example 1. Let A = EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3)). We find the eigenvalues.

0 = EQ \b\lc\|(\a(1 - (,-1)) EQ \b\rc\|(\a(1,3 - ()) = (1 - ()(3 - () + 1 = (2 - 4( + 4 = (( - 2)2The only eigenvalue is ( = 2 which is of multiplicity two. An eigenvector v = EQ \B(\A\al( x , y )) satisfies

EQ \B(\A\al( 0 , 0 )) = (A - 2I)v = EQ \b\lc\((\a(-1,-1)) EQ \b\rc\)(\a(1,1)) EQ \B(\A\al( x , y ))

Both equations are - x + y = 0 or y = x so the eigenvectors are multiples of v1 = EQ \B(\A\al( 1 , 1 )). So even though the eigenvalue ( = 2 is of multiplicity two it has only one linearly independent eigenvector. In situations like this we turn to generalized eigenvectors as a substitute for eigenvectors. In order to appreciate the definition of a generalized eigenvector, note that an eigenvector v for an eigenvalue ( satisfies (A - (I)v = 0. If we replace A - (I by (A - (I)m we get the definition of a generalized eigenvector.

Definition 1. Let A be a square matrix and ( be an eigenvalue of A. A vector v is a generalized eigenvector for ( if

1.v ( 0

2.(A - (I)mv = 0 for some positive integer mThe smallest positive integer m such that (A - (I)mv = 0 is called the degree of the generalized eigenvector.

Note that any eigenvector of A is a generalized eigenvector of degree one since it satisfies (A - (I)mv = 0 with m = 1. In fact an eigenvector v of A satisfies (A - (I)mv for any positive integer m since (A - (I)mv = (A - (I)m-1(A - (I)v = (A - (I)m-10 = 0.

Example 2. Find the generalized eigenvectors of A = EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3)).

In Example 1 we saw that (= 2 was the only eigenvalue and the only eigenvectors were multiples of v1 = EQ \B(\A\al( 1 , 1 )). These are the generalized eigenvectors of degree one. The generalized eigenvectors of degree two are the solutions of (A 2I)2v= 0. One has A 2I= EQ \b\lc\((\a(-1,-1)) EQ \b\rc\)(\a(1,1)) so (A 2I)2 = EQ \b\lc\((\a(-1,-1)) EQ \b\rc\)(\a(1,1)) EQ \b\lc\((\a(-1,-1)) EQ \b\rc\)(\a(1,1)) = EQ \b\lc\((\a(0,0)) EQ \b\rc\)(\a(0,0)) = 0. So the equation (A 2I)2v= 0 is 0v= 0. Every vector v = EQ \B(\A\al( x , y )) satisfies this equation, so every non-zero vector v is a generalized eigenvector of A. The ones that do not lie on the line through v1 = EQ \B(\A\al( 1 , 1 )) are of degree two.

Now we look at how the generalized eigenvectors can be used to create a substitute for the diagonalization of the matrix. For simplicity we consider a 2(2 matrix A with a single eigenvalue ( that has only one linearly independent eigenvector v1 as in Example 1. It turns out that A must have a generalized eigenvector v of degree two. So (A - (I)v is a regular eigenvector. So (A - (I)v = cv1. Let v2 = v/c. Then (A - (I)v2 = v1 or Av2=v1+(v2. Let

T = matrix whose columns are v1 and v2

J = EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())Since Av1 = (v1 and Av2=v1+(v2 the matrix AT has columns equal to (v1 and v1+(v2. Consider TJ. Recall that the kth column of TJ is a linear combination of the columns of T using the entries in the kth column of J as coeffients. So TJ has columns equal to (v1 and v1+(v2. So AT = TJ or

(1)

A = TJT-1 = T EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,()) T-1

J = EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,()) is called the Jordan canonical form of A and we shall call formula (1) the Jordanization of A. It turns out to be a convenient substitute for the diagonalization of A when it comes to computing An and other computations with A.

Example 3. Let A = EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3)). Find the Jordanization (1) of A.

In Example 1 we saw that (= 2 was the only eigenvalue and the only eigenvectors are multiples of v1 = EQ \B(\A\al( 1 , 1 )). We need to find v = EQ \B(\A\al( x , y )) such that (A - (I)v2 = v1 = EQ \B(\A\al( 1 , 1 )). Since A 2I = EQ \b\lc\((\a(-1,-1)) EQ \b\rc\)(\a(1,1)) so EQ \b\lc\((\a(-1,-1)) EQ \b\rc\)(\a(1,1)) EQ \B(\A\al( x , y )) = EQ \B(\A\al( 1 , 1 )). Both equations are x + y = 1 or y = 1 + x so v= EQ \B(\A\al( x , y ))= EQ \B(\A( x , 1 + x ))= EQ \B(\A\al( 0 , 1 ))+ x EQ \B(\A\al( 1 , 1 )) where x can be any number. Let's take x = 0 so v2= EQ \B(\A\al( 0 , 1 )). Then (1) becomes

EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3)) = EQ \b\lc\((\a(1,1)) EQ \b\rc\)(\a(0,1)) EQ \b\lc\((\a(2,0)) EQ \b\rc\)(\a(1,2)) EQ \b\lc\((\a(1,1)) EQ \b\rc\)(\a(0,1))-1Now let's consider how to compute the powers of a matrix A using its Jordanization (1). As in chapter six one has

An = TJnT-1 = T EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())n T-1

It turns out that EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())n is quite simple. One has

EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())2 = EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,()) EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,()) = EQ \b\lc\((\a((2,0)) EQ \b\rc\)(\a(2(,(2))

EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())3 = EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())2 EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,()) = EQ \b\lc\((\a((2,0)) EQ \b\rc\)(\a(2(,(2)) EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,()) = EQ \b\lc\((\a((3,0)) EQ \b\rc\)(\a(3(2,(3))

EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())4 = EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())3 EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,()) = EQ \b\lc\((\a((3,0)) EQ \b\rc\)(\a(3(2,(3))

EQ \b\rc\)(\a(2(,(2)) EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,()) = EQ \b\lc\((\a((4,0)) EQ \b\rc\)(\a(4(3,(4))and in general

EQ \b\lc\((\a((,0)) EQ \b\rc\)(\a(1,())n = EQ \b\lc\((\a((n,0)) EQ \b\rc\)(\a(n(n-1,(n))So

An = T EQ \b\lc\((\a((n,0)) EQ \b\rc\)(\a(n(n-1,(n)) T-1Example 4. Let A = EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3)). Find An and use it to solve the difference equations

xn+1 = xn + yn

x0 = 4

xn+1 = - xn + 3yn

y0 = 5From Example 3 one has EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3)) = EQ \b\lc\((\a(1,1)) EQ \b\rc\)(\a(0,1)) EQ \b\lc\((\a(2,0)) EQ \b\rc\)(\a(1,2)) EQ \b\lc\((\a(1,1)) EQ \b\rc\)(\a(0,1))-1, so

EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3))n = EQ \b\lc\((\a(1,1)) EQ \b\rc\)(\a(0,1)) EQ \b\lc\((\a(2n,0)) EQ \b\rc\)(\a(n2n-1,2n)) EQ \b\lc\((\a(1,1)) EQ \b\rc\)(\a(0,1))-1 = 2n EQ \b\lc\((\a(1,1)) EQ \b\rc\)(\a(n/2,n/2 + 1)) EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(0,1))

= 2n EQ \b\lc\((\a(1 - n/2,- n/2)) EQ \b\rc\)(\a(n/2,n/2 + 1))The difference equations can be written as

EQ \B(\A\al( xn , yn )) = EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3)) EQ \B(\A\al( xn , yn ))

EQ \B(\A\al( x0 , y0 )) = EQ \B(\A\al( 4 , 5 ))The solution is

EQ \B(\A\al( xn , yn )) = EQ \b\lc\((\a(1,-1)) EQ \b\rc\)(\a(1,3))n EQ \B(\A\al( x0 , y0 )) = 2n EQ \b\lc\((\a(1 - n/2,- n/2)) EQ \b\rc\)(\a(n/2,n/2 + 1)) EQ \B(\A\al( 4 , 5 )) = 2n EQ \b(\a(4 + n/2,5 + n/2))So

xn = (4 + n/2) 2n

yn = (5 + n/2) 2nHere are some elementary properties of generalized eigenvectors.

Definition 2. Let A be a square matrix and ( be an eigenvalue of A. Let N(,m={v:(A (I)mv = 0} be the set of generalized eigenvectors for ( of degree at most m along with the zero vector. N(,m is called the generalized eigenspace of degree m for (.

The generalized eigenspaces are subspaces since they are the null spaces of the matrices (A - (I)m.

Proposition 1. Let A be a square matrix and ( be an eigenvalue of A.

(a)If (A - (I)mv = 0 for some positive integer m then (A - (I)m+kv = 0 for any positive integer k, i.e. (A - (I)pv = 0 for some positive integer p that is larger than m.

(b)N(,1 ( N(,2 ( ( N(,m ( N(,m+1 ( (

(c)If N(,m = N(,m+1 for some m then N(,m = N(,m+k for all positive integers k, i.e. N(,m = N(,p for all positive integers p larger than m.

(d)v is a generalized eigenvector of degree m if and only if (A - (I)v is a generalized eigenvector of degree m 1.

(e)v is a generalized eigenvector of degree m if and only if Av = (v + w where w is a generalized eigenvector of degree m 1.

Proof. (a) If (A - (I)mv = 0 then (A - (I)m+kv = (A - (I)k(A - (I)mv = (A - (I)k0 = 0. (b) follows from (a). (c) We shall show that N(,m = N(,m+1 implies N(,m+1 = N(,m+2. Applying this over and over again will show N(,m = N(,m+k for all positive integers k. Note that N(,m= N(,m+1 is equivalent to (A - (I)m+1v = 0 ( (A - (I)mv = 0. Suppose (A - (I)m+2v = 0. We write (A - (I)m+2v = (A - (I)m+1(A - (I)v. So (A - (I)m+1(A - (I)v = 0. So (A (I)m(A (I)v = 0. So (A - (I)m+1v = 0. Thus (A - (I)m+2v = 0 ( (A - (I)m+1v = 0. So N(,m+1 = N(,m+2. (d) This follows from the fact that (A - (I)mv = (A - (I)m-1(A - (I)v. (e) follows from (d). /

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