Chapter 6

Jordan Canonical Form

175 6.1 Primary Decomposition Theorem

For a general matrix, we shall now try to obtain the analogues of the ideas we have in the case of diagonalizable matrices.

Diagonalizable Case General Case

Minimal Polynomial: Minimal Polynomial:

k k Y rj Y rj mA(λ) = (λ − λj) mA(λ) = (λ − λj) j=1 j=1

where rj = 1 for all j where 1 ≤ rj ≤ aj for all j Eigenspaces: Eigenspaces:

Wj = Null Space of (A − λjI) Wj = Null Space of (A − λjI)

dim(Wj) = gj and gj = aj for all j dim(Wj) = gj and gj ≤ aj for all j Generalized Eigenspaces:

rj Vj = Null Space of (A − λjI)

dim(Vj) = aj Decomposition: Decomposition:

n n F = W1 ⊕ W2 ⊕ Wj ⊕ · · · ⊕ Wk W = W1⊕W2⊕Wj⊕· · ·⊕Wk is a subspace of F Primary Decomposition Theorem:

n F = V1 ⊕ V2 ⊕ Vj ⊕ Vk

n n x ∈ F =⇒ ∃ unique x ∈ F =⇒ ∃ unique xj ∈ Wj, 1 ≤ j ≤ k such that vj ∈ Vj, 1 ≤ j ≤ k such that

x = x1 + x2 + ··· + xj + ··· + xk x = v1 + v2 + ··· + vj + ··· + vk

176 6.2 Analogues of Lagrange Polynomial

Diagonalizable Case General Case

Define Define

k k Y Y ri fj(λ) = (λ − λi) Fj(λ) = (λ − λi) i=1 i=1 i6=j i6=j

This is a set of coprime polynomials This is a set of coprime polynomials and and hence there exist polynomials hence there exist polynomials Gj(λ), 1 ≤ gj(λ), 1 ≤ j ≤ k such that j ≤ k such that

k k X X fj(λ)gj(λ) = 1 Fj(λ)Gj(λ) = 1 j=1 j=1

The polynomials The polynomials

`j(λ) = fj(λ)gj(λ), 1 ≤ j ≤ k Lj(λ) = Fj(λ)Gj(λ), 1 ≤ j ≤ k

are the Lagrange Interpolation are the counterparts of the polynomials Lagrange Interpolation polynomials `1(λ) + `2(λ) + ··· + `k(λ) = 1 L1(λ) + L2(λ) + ··· + Lk(λ) = 1 λ1`1(λ)+λ2`2(λ)+···+λk`k(λ) = λ λ1L1(λ) + λ2L2(λ) + ··· + λkLk(λ) = d(λ) (a polynomial but it need not be = λ) For i 6= j the product `i(λ)`j(λ) has For i 6= j the product Li(λ)Lj(λ) has a

a factor mA (λ) factor mA (λ)

6.3 Matrix Decomposition

In the diagonalizable case, using the Lagrange polynomials we obtained a decomposition of the matrix A. We shall next look at the analogous situation for a general matrix.

177 Diagonalizable Case General Case

Define the matrices Analogously define the matrices

Aj = `j(A) for 1 ≤ j ≤ k Aj = Lj(A) for 1 ≤ j ≤ k

We have Analogously we have

A1 + A2 + ··· + Ak = I A1 + A2 + ··· + Ak = I since since

`1(λ) + `2(λ) + ··· + `k(λ) = 1 L1(λ) + L2(λ) + ··· + Lk(λ) = 1

AiAj = 0(n×n) if i 6= j AiAj = 0(n×n) since, then `i(λ)`j(λ) has the mini- since, then Li(λ)Lj(λ) has the minimal mal polynomial as a factor polynomial as a factor 2 2 Aj = Aj for 1 ≤ j ≤ k Aj = Aj for 1 ≤ j ≤ k Range of Aj is Wj and Ajx = x for Range of Aj is Vj and Ajx = x for all all x ∈ Wj x ∈ Vj We have We will not have

λ1A1 + λ2A2 + ··· + λkAk = A λ1A1 + λ2A2 + ··· + λkAk = A since since

λ1`1(λ)+λ2`2(λ)+···+λk`k(λ) = λ λ1L1(λ) + λ2L2(λ) + ··· + Lk(λ) ,

in general, may not be equal to λ. Let

D = λ1A1 + λ2A2 + ··· + λkAk

Then D is a diagonalizable matrix (by The- orem 5.7.1). 178 We then look at by how much A differs form D. For this purpose we define N = A − D so that A = D + N (In the case where A is a diagonalizable matrix we have N = 0(n×n) and A = D). Thus, in the general case, in addition to a diagonalizable matrix, we also have to analyse the part N = A − D. We shall do this in the next section.

6.4 Nilpotent Matrices

We want to analyse the difference N = A − D (6.4.1) where D is as defined in the previous section. Using the fact that

I = A1 + A2 + ··· + Ak we get

A = AIn×n

= AA1 + AA2 + ··· + AAk Combining this and the fact that

D = λ1A1 + λ2A2 + ··· + λkAk we get

A − D = (AA1 + AA2 + ··· + AAk) − (λ1A1 + λ2A2 + ··· + λkAk)

= (A − λ1In×n)A1 + (A − λ2In×n)A2 + ··· + (A − λkIn×n)Ak

All the Aj, being polynomials in A, commute with A, and furher, AiAj = 0(n×n). Using these facts we get

r r r r (A − D) = (A − λ1In×n) A1 + (A − λ2In×n) A2 + ··· + (A − λkIn×n) Ak (6.4.2)

r If r ≥ Max. {r1, r2, ··· , rj} then (A − λjIn×n) Aj has a factor mA (A), and hence

r (A − λjIn×n) Aj = 0n×n for all j if r ≥ Max.{r1, r2, ··· , rj} (6.4.3)

179 Hence we get from (6.4.2)

r r N = (A − D) = 0n×n for r ≥ Max.{r1, r2, ··· , rj} (6.4.4) Thus the difference between A and the diagonalizable matrix D is such that a power of it vanishes. This leads us to the notion of a nilpotent matrix. We have

n×n Definition 6.4.1 A matrix N ∈ C is said to be a “NILPOTENT” r matrix if there exists a positive integer r such that N = 0n×n. For r a nilpotent matrix the smallest power r for which N = 0n×n is called the “order of nilpotency” of N and is denoted by γ . N Thus we have that the matrix N = A − D is a nilpotent matrix the order of nilpotency being equal to Max. {r1, r2, ··· , rk}. We can summarise our analysis above as follows:

n×n Theorem 6.4.1 Every matrix A ∈ C can be expressed as the sum,

A = D + N, (6.4.5) of a diagonalizable matrix D and a nilpotent matrix N We have already analysed diagonalizable matrices in the last chapter. We shall now analyse a nilpotent matrix.

6.5 Nilpotent Matrices

Let N ∈ n×n be a nilpotent matrix with order of nilpotency γ . This means F N that

γ N N = 0n×n and (6.5.1) N r 6= 0 for any r < γ (6.5.2) n×n N From (6.5.1) we also get,

N r = 0 for all r ≥ γ (6.5.3) n×n N

γ From this it follows that the polynomial p(λ) = λ N is an annihilating poly- γ nomial for N. Hence the minimal polynomial must divide λ N . Thus the

180 minimal polynomial must be of the form m (λ) = λr where r ≤ γ . How- N N ever, from (6.5.2) we have that N r 6= 0 for any r < γ . Hence we get n×n N that the minimal polynomial of N is given by

γ N mN (λ) = (λ)

Thus we see that λ1 = 0 is the only eigenvalue of any nilpotent matrix N. Since the characteristic polynomial of N is a monic polynomial of degree n and its roots are the eigenvalues we get that the characteristic polynomial of N is given by

n cN (λ) = λ Hence we have,

n×n Theorem 6.5.1 If N ∈ F is a nilpotent matrix with order of nilpotency as γ then its characteristic and minimal polynomials N are given by

n cN (λ) = λ , and (6.5.4) γ N mN (λ) = λ (6.5.5) From the above simple properties we now obtain a useful result. Suppose A ∈ n×n is both nilpotent and diagonalizable. Since it is nilpotent matrix F γ A its minimal polynomial is of the form mA = λ , where γA is its order of nilpotency. Combining this with the fact that A is diagonalizable we get that its minimal polynomial must be mA (λ) = λ. Hence mA (A) = 0n×n gives us that A = 0n×n. Thus it follows that

n×n Theorem 6.5.2 The only matrix in F which is both diagonaliz- able and nilpotent is the zero matrix We shall now see an example to show that the sum and product of two nilpotent matrices need not be nilpotent.

n×n Example 6.5.1 . Consider the matrices N1,N2 ∈ C given below: 0 1 ! N = (6.5.6) 1 0 0 0 0 ! N = (6.5.7) 2 1 0

181 It is easy to see that

2 N1 = 02×2 2 N2 = 02×2 Thus both are nilpotent matrices both with order of nilpotency as 2. How- ever, we have,

0 1 ! N + N = 1 2 1 0

This matrix has characteristic polynomial given by

λ2 − 1 = 0

Since there are two eigenvalues λ1 = 1, λ2 = −1, both eigenvalues having algebraic and geometric multiplicity both as 1, the matrix is diagonalizable. Since this is not the zero matrix, by the Theorem 6.5.2, this matrix cannot be nilpotent. Thus we see that the sum N1 +N2 of the two nilpotent matrices N1 and N2 is not nilpotent. Similarly we have,

1 0 ! N N = 1 2 0 0 and this is a diagonal matrix. Again by Theorem 6.5.2 it follows that N1N2 is not nilpotent. Thus the product N1N2 of the two nilpotent matrices N1 and N2 is not nilpotent. However, we can show that the following is true: when two nilpotent matrices commute with each other, then their sum and product will also be nilpotent We shall now use this result to investigate the uniqueness of the decompo- sition of a matrix as the sum D + N of a diagonalizable matrix D and a nilpotent matrix N obtained in Section 6.3. The diagonalizable matrix D we got is a polynomial in A. Further the nilpotent matrix N we obtained was defined as N = A − D, and hence N is also a polynomial in A. Thus D, N and A all commute with each other. Thus we have the decomposition

A = D + N where (6.5.8) DN = ND (6.5.9)

182 Suppose now there exists another decomposition of A as

A = D1 + N1 (6.5.10) where D1 is diagonalizable and N1 is nilpotent and D1 and N1 commute with each other, that is,

D1N1 = N1D1 (6.5.11) Then we get from (6.5.13),

D1A = D1(D1 + N1) 2 = D1 + D1N1 2 = D1 + N1D1 by (6.5.14)

= (D1 + N1)D1

= AD1

Hence D1 commutes with A and hence with any polynomial in A. Thus D1 commutes with D and N. Similarly it can be shown that N1 commutes with D and N. Thus we have N − N1 is nilpotent. We can also show that the sum of two commuting diagonalizable matrices is diagonalizable and hence D − D1 is diagonalizable. Now we have from (6,2,11) and (6,2,13),

A − A = (D − N) − (D1 − N1) =⇒

0n×n = (D − D1) − (N − N1) =⇒

D − D1 = N − N1 The lhs above is a diagonalizable whereas the rhs is nilpotent, and since they are equal we must have both equal to 0n×n, since we have seen that the only matrix which is both diagonalizable and nilpotent is the zero matrix. Thus we have,

D = D1 (6.5.12)

N = N1 (6.5.13)

n×n Thus the decomposition of a matrix A ∈ F as the sum of a diagonalizable and nilpotent matrix is unique if these two matrices have to commute. Thus we have

183 n×n Theorem 6.5.3 Every matrix A ∈ F , can be decomposed as the sum,

A = D + N (6.5.14) of a diagonalizable matrix D and a nilpotent matrix N. Moreover such a decomposition is unique if

DN = ND (6.5.15)

We the call D as the “Diagonalizable Part” of A and N as the “Nilpotent Part” of A

Remark 6.5.1 The decomposition of A as the sum of a diagonalizable ma- trix and a nilpotent matrix is NOT unique if we do not insist that the di- agonalizable and Nilpotent matrices in the decomposition commute. For example for the matrix 1 1 ! A = 0 1 we have the following decompositions of the matrix:

A = D + N where 1 0 ! 0 1 ! D = andN = 0 1 0 0 and also the decomposition

A = D1 + N1 where 1 1 ! 0 0 ! D = and N = 1 1 1 −1 0

We have DN = ND but we do not have D1N1 = N1D1. Thus the second decomposition is a decomposition where the diagonalizable matrix D1 and the nilpotent matrix N1 in the decomposition do not commute.

184 6.6 Canonical Form of a Nilpotent Matrcix

We shall first look at simple examples of nilpotent matrices Example 6.6.1 The matrix 0 1 ! N = ∈ 2×2 0 0 C is a nilpotent matrix with order of nilpotency 2. The matrix  0 1 0    3×3 N =  0 0 1  ∈ C 0 0 0 is a nilpotent matrix with order of nilpotency 3. The matrix  0 0 1    3×3 N =  0 0 0  ∈ C 0 0 0 is a nilpotent matrix with order of nilpotency 2. The matrix  0 1 0    3×3 N =  0 0 0  ∈ C 0 0 0 is a nilpotent matrix with order of nilpotency 2. In general, the matrix  0 1 0 0 ······ 0     0 0 1 0 ······ 0     ......   ......  n×n N =   ∈ n  ......  C  ......       0 0 0 0 ······ 1  0 0 0 0 ······ 0 is a nilpotent matrix with order of nilpotency n. This is called the canonical (n × n) (nilpotent) matrix. (N is the n × n matrix whose (j, j + 1)th entry is 1 for j = 1, 2, ··· , (n − 1) and all other entries are zero). Thus we have  0 1 0  0 1 ! N = and N =  0 0 1  2 0 0 3   0 0 0

185 n×n Suppose N ∈ F is nilpotent. Then we know that its characteristic poly- nomial must be c (λ) = λn. The order of nilpotency γ satisfies, A N 1 ≤ γ ≤ n (6.6.1) N 1 When the order of nilpotency is 1 we have N = 0n×n and hence N is the zero matrix. Thus the only nilpotent matrix with order of nilpotency 1 is the zero matrix. We shall next look at nilpotent matrices whose order of nilpotency is n, that is γ = n. For such matrices we have N n N = 0n×n and (6.6.2) r N 6= 0n×n for 1 ≤ r ≤ n − 1 (6.6.3) (n−1) In particular we have N 6= 0n×n. Hence there exists a nonzero vector n (n−1) n x ∈ F such that N x 6= θn. Now we define n nonzero vectors {vj}j=1 as follows: (n−1)  v1 = N x   v = N (n−2)x  2   ·········  ((n−j))  vj = N x (6.6.4)  ·········    v(n−1) = Nx   vn = x  We now show the following: CLAIM: v1, v2, ··· , vn are linearly independent We shall prove this Claim later. But suppose this claim is true then we get n that v1, v2, ··· , vn form a basis for F . We then have, from (6.3.4),  Nv1 = θn   Nv2 = v1   ·········  (6.6.5) Nv = v j (j−1)   ·········   Nvn = v(n−1) 

If we now define the matrix P as the matrix whose jth column is vj, for j = 1, 2, ··· , n, that is,

P = [v1 v2 ··· vj ··· vn] (6.6.6)

186 then P is invertible, (since the columns are linearly independent), and,

NP = [Nv1 Nv2 ··· Nvj ··· Nvn]

= [θn v1 v2 ··· v(j−1) ··· v(n−1)]  0 1     0 1     .. ..  = P  . .     0 1    0 =⇒  0 1     0 1    −1  .. ..  P NP =  . .     0 1    0

= Nn

Thus we have,

Theorem 6.6.1 If N ∈ F n×n is nilpotent with order of nilpotency as n, then it is similar to the canonical n × n nilpotent matrix Nn, that is, there exists an invertible matrix P ∈ F n×n such that

−1 P NP = Nn (6.6.7)

We say that “canonical form of N is Nn and write Ncan = Nn”. We shall complete the proof of the above argument by proving the Claim. Proof of Claim: We have

α1v1 + α2v2 + ··· + αnvn = θn =⇒ (n−1) (n−2) α1N x + N x + ··· + α(n−1)Nx + αnx = θn =⇒ (n−1) (n−1) (n−2) N (α1N x + N x + ··· + α(n−1)Nx + αnx) = θn n (since all terms except the last have factor N which is 0n×n) =⇒ (n−1) αnN x = θn =⇒ (n−1) (since N x 6= θn) αn = 0 =⇒

187 (n−1) (n−2) α1N x + N x + ··· + α(n−1)Nx = θn =⇒ (n−2) (n−1) (n−2) N (α1N x + N x + ··· + α(n−1)Nx) = θn =⇒

(as above) α(n−1) = 0

Continuing this process step by step we get all αj as 0 and hence v1, v2, ··· , vn are linearly independent, thus proving the Claim. We next look at nilpotent matrices for which the order of nilpo- tency is between 1 and n, that is,

1 < γ < n (6.6.8) N We then have,

γ N N = 0n×n and (6.6.9) N r 6= 0 for 1 ≤ r ≤ γ − 1 (6.6.10) n×n N From this we get

γ n Null Space of N N = F (6.6.11) Null Space of N r 6= F n for 1 ≤ r ≤ γ − 1 (6.6.12) N We further have,

Null Space of N r is a proper subspace of ) (6.6.13) Null Space of N (r+1) for 1 ≤ r ≤ γ − 1 N Thus if we define,

V = Null Space of N j for 1 ≤ j ≤ γ (6.6.14) j N then the subspaces are strictly increasing, that is,

⊂ ⊂ ⊂ ⊂ ⊂ ⊂ ⊂ V1 6= V2 6= ··· 6= Vj 6= V(j+1) 6= ··· 6= Vγ −1 6= Vγ N N (6.6.15)

Let

dj = dimension of Vj (6.6.16)

188 Then by (6.6.15) we get

d1 < d2 < ··· < d(j−1) < dj < d(j+1) < ··· < d(γ −1) < dγ = n (6.6.17) N N

(Clearly d1 is the Nullity of the matrix N). We now look at the increase in dimension at each stage. We define

α1 = d1 (6.6.18) α = d − d for j = 2, 3, ··· , γ (6.6.19) j j (j−1) N (6.6.20)

Hence we can write

d1 = α1 (6.6.21)

d2 = α1 + α2 (6.6.22) d = α + α + ··· + α + α for 2 ≤ j ≤ γ (6.6.23) j 1 2 (j−1) j N We now state the following Lemma without proof:

Lemma 6.6.1 The αj are non increasing, that is,

α1 ≥ α2 ≥ · · · ≥ α(j−1) ≥ αj ≥ α(j+1) ≥ · · · ≥ αγ −1 ≥ αγ N N (6.6.24)

We now define

nj = αj − α(j+1) for 1 ≤ j ≤ γN − 1 (6.6.25) n = α (6.6.26) γN γN

By (6.6.24) we get nj ≥ 0 for each j. We now define a matrix Ncan with the following properties:

1. Ncan is an n × n matrix with diagonal blocks arranged such that the block sizes are non increasing as we go down the diagonal

2. The number of blocks is equal to α1 - the nullity of N 3. Each block is a canonical nilpotent matrix

4. The first block is of size γ × γ and hence all other blocks are of size N N less than or equal to γ × γ N N

189 5. The number of blocks of size j × j is equal to n , for j = 1, 2, ··· , γ j N

The matrix Ncan is called the “canonical form” of the matrix N. We shall now state the main theorem (without proof) for nilpotent matrices: n×n Theorem 6.6.2 Every nilpotent matrix N ∈ C is similar to its canonical form, that is, n×n n×n −1 N ∈ C =⇒ ∃ invertible P ∈ C such that P NP = Ncan 2×2 Example 6.6.2 Consider a nilpotent matrix N ∈ C . We have CASE 1: γ = 1 N In this case N has to be 02×2 CASE 2: γ = 2 N Since the leading block has to be of size 2 × 2 we get 0 1 ! N = N = , the canoncal 2 × 2 nilpotent matrix can 2 0 0

3×3 Example 6.6.3 Consider a nilpotent matrix N ∈ C . We have CASE 1: γ = 1 N In this case N has to be 03×3 CASE 2: γ = 2 N Since the leading block has to be of size 2 × 2 we get  0 1 0  N 0 ! N = 2 2×1 =  0 0 0  can 0 0   1×2 1×1 0 0 0

4×4 Example 6.6.4 Consider a nilpotent matrix N ∈ C . We have CASE 1: γ = 1 N In this case N has to be 04×4 CASE 2: γ = 2 N The leading block is 2 × 2. The remaining can be either one 2 × 2 block, (which will hold if the nullity is 2 because then there will be only two blocks), or two 1 × 1 blocks (which will hold when the nullity is 3 because then there will be three blocks). Hence we have CASE 2A: Nullity of N is 2  0 1 0 0  ! N 0  0 0 0 0  2 2×2   Ncan = =   02×2 N2  0 0 0 1  0 0 0 0

190 CASE 2B: Nullity of N is 3

 0 1 0 0   N 0 0  2 2×1 2×1  0 0 0 0  N =  0 0 0  =   can  1×2 1×1 1×1   0 0 0 0  0 0 0   1×2 1×1 1×1 0 0 0 0

CASE 3: γ = 3 N Since now the leading block has to be 3 × 3 the remaining has to be a 1 × 1 block. Hence we get

 0 1 0 0  ! N 0  0 0 1 0  3 3×1   Ncan = =   01×3 01×1  0 0 0 0  0 0 0 0

In this case the nullity has to be 2 CASE 4: γ = 4 N Since the leading block itself is 4 × 4 there will be only one block. We then have  0 1 0 0   0 0 1 0    Ncan = N4 =    0 0 0 1  0 0 0 0

6×6 Example 6.6.5 Consider a nilpotent matrix N ∈ C CASE 1: γ = 1 N N has to be 06×6 CASE 2: γ = 2 N All blocks have to be of size smaller than or equal to 2 × 2 and the leading block must be 2 × 2. Hence there must be at least 3 boocks. This means that nullity of N is at least 3 CASE 2A: Nullity is 3:

  N2   Ncan =  N2  N2

191 CASE 2B: Nullity is 4:

  N2    N2  N =   can  N   1  N1

CASE 2C: Nullity is 5

  N2    N1    N =  N  can  1     N1  N1

Nullity cannot be 6 when γ = 2(Why?) N CASE 3: γ = 3 N The leading block is 3 × 3. The remaining can be (i) one 3 × 3 block (when nullity is 2), or (ii) one 2 × 2 block and one 1 × 1 block (when nullity is 3), or iii) three 1 × 1 blocks (when nullity is 4) Nullity cannot be 1, 5 or 6 when γ = 3 (Why?) N CASE 3A: Nullity is 2

! N3 Ncan = N3

CASE 3B: Nullity is 3

  N3   Ncan =  N2  N1

CASE 3C: Nullity is 4

  N3    N1  N =   can  N   1  N1

192 CASE 4: γ = 4 N The leading block is 4 × 4 and the remaining can be, i) one 2 block (when nullity is 2), or ii) two 1 × 1 blocks (when nullity is 3 CASE 4A: Nullity is 2 ! N4 Ncan = N2

CASE 4B: Nullity is 3

  N4   Ncan =  N1  N1

CASE 5: γ = 5 N The leading block is 5 × 5 and hence the remaining is a 1 × 1 block. ! N5 Ncan = N1

In this case the nullity has to be 2. CASE 6: γ = 6 N Clearly there is only one 6 × 6 block and we have

Ncan = N6

12×12 Example 6.6.6 Consider a nilpotent matrix N ∈ C whoae canonical form is given by   N5    N3    N =  N  can  2     N1  N1 From the canonical form we can make the following conclusions:

1. Since there are 5 blocks we have Nullity of N must be 5. Thus we have

α1 = 5 (6.6.27)

193 2. Since the leading block is N5 it follows that the order of nilpotency of N is 5. Thus we have

γ = 5 (6.6.28) N

3. We have

n1 = α1 − α2 = 2 (since there are two 1 × 1 blocks) (6.6.29)

n2 = α2 − α3 = 1 (since there is one 2 × 2 block) (6.6.30)

n3 = α3 − α4 = 1 (since there is one 3 × 3 block) (6.6.31)

n4 = α4 − α5 = 0 (since there are no 4 × 4 blocks) (6.6.32)

n5 = α5 = 1 (since there is one 5 × 5 blocks) (6.6.33)

4. We have (6.6.27) and (6.6.29) =⇒

α2 = 3 (6.6.34)

Now (6.6.30) and (6.6.34) Longrightarrow

α3 = 2 (6.6.35)

Now (6.6.31) and (6.6.35) Longrightarrow

α4 = 1 (6.6.36)

Now (6.6.32) and (6.6.36) Longrightarrow

α5 = 1 (6.6.37)

j 5. We have Vj = Null Space of N for 1 ≤ j ≤ 5. From the above values for αj we get

dimension of V1 = α1 = 5 dimension of V2 = α1 + α2 = 5 + 3 = 8 dimension of V3 = α1 + α2 + α3 = 5 + 3 + 2 = 10 dimension of V4 = α1 + α2 + α3 + a4 = 5 + 3 + 2 + 1 = 11 dimension of V5 = α1 + α2 + α3 + a4 + α5 = 5 + 3 + 2 + 1 + 1 = 12

194 17×17 Example 6.6.7 Consider a nilpotent matrix N ∈ C of order of nilpo- j tency 5 and Vj = Null Space of N for 1 ≤ j ≤ 5. Suppose

dimension of V1 = 8 dimension of V2 = 12 dimension of V3 = 14 dimension of V4 = 16 dimension of V5 = 17

Then, the characteristic and minimal polynomials must be

17 cN (λ) = λ 5 mN (λ) = λ

Further,

α1 = d1 = 8

α2 = d2 − d1 = 12 − 8 = 4

α3 = d3 − d2 = 14 − 12 = 2

α4 = d4 − d3 = 16 − 14 = 2

α5 = d5 − d4 = 17 − 16 = 1

Hence we get

n1 = α1 − α2 = 8 − 4 = 4

n2 = α2 − α3 = 4 − 2 = 2

n3 = α3 − α4 = 2 − 2 = 0

n4 = α4 − α5 = 2 − 1 = 1

n5 = α5 = 1

Hence there is one 5 × 5 block, one 4 × 4 block, no 3 × 3 blocks, two 2 × 2 blocks, and four 1 × 1 blocks.

195 Hence the canonical form of N is given by

  N5  N   4     N2     N   2  Ncan =    N1     N   1     N1  N1

196