Examples Jordan Normal Form UTK – M531 – Ordinary Differential Equations I Fall 2004, Jochen Denzler, TR 11:10–12:25, Ayres 309B

(These examples were generated with the help of symbolic algebra software, and also the calculations were done using such software.)

diag(λ_{1}, …, λ_{n}) denotes the diagonal matrix (size n×n) with the diagonal entries as specified.

I may write column vectors like [ \begin{matrix}1 \\ 2 \\ 3 \end{matrix} ] as transposed rows [1 , 2 , 3]^{T} for typographical

reasons.

Example 1:

The matrix

A = [ \begin{matrix}\frac{44}{3} & −\frac{82}{3} & −20 \\ \frac{19}{3} & −\frac{35}{3} & −10 \\ 4 & −8 & −4 \end{matrix} ]

has characteristic polynomial

det(A − λI) = −λ^{3} − λ^{2} + 10λ − 8 = −(λ − 2)(λ − 1)(λ + 4)

where the factorization is based on eyeballing (guesswork) to find λ = 1 as one root, and then the quadratic formula.

Since all eigenvalues are distinct, we can diagonalize the matrix A = SDS^{−1} with D = diag(1, 2, −4). (Any other order like eg. D̃ = diag(1, −4, 2) would do just as well, with a different S̃ ≠ S (similarly permuted. So we have chosen the numbering λ_{1} = 1, λ_{2} = 2, λ_{3} = −4. Let’s find corresponding eigenvectors v_{1}, v_{2}, v_{3}:

v_{1} is calculated as a solution to the linear system (A − λ_{1})v_{1} = 0:

\begin{matrix}\frac{41}{3}v_{1}^{(1)} + & \frac{82}{3}v_{1}^{(2)} + & (−20)v_{1}^{(3)} = 0 \\ \frac{19}{3}v_{1}^{(1)} + & \frac{−38}{3}v_{1}^{(2)} + & (−10)v_{1}^{(3)} = 0 \\ 4v_{1}^{(1)} + & (−8)v_{1}^{(2)} + & (−5)v_{1}^{(3)} = 0 \end{matrix}

(Upper indices denote components of the vector, lower indices identify vectors.) Gauss elimination from this system produces: v_{1}^{(1)} = 2v_{1}^{(2)}, v_{1}^{(3)} = 0. We may therefore choose v_{1} = [2 , 1 , 0]^{T}. Any nonzero multiple would have been just as legitimate a choice (and would have led to a somewhat different matrix S).

Likewise we obtain v_{2} = [1 , −1 , 2]^{T} and v_{3} = [4 , 2 , 1]^{T}, where again nonzero multiples would have been just as legitimate choices. We have therefore found (our choice of) the matrix S:

S = [ \begin{matrix}2 & 1 & 4 \\ 1 & −1 & 2 \\ 0 & 2 & 1 \end{matrix} ]

(Vertical lines inserted to make eigenvectors more clearly visible). It is another bunch of linear equations to find its inverse (if you need it):

S^{−1} = \frac{1}{3} [ \begin{matrix}5 & 7 & −6 \\ 1 & −2 & 0 \\ −2 & 4 & 3 \end{matrix} ]

You can now check explicitly that AS = SD, or A = SDS^{−1}.

Example 2:

This example must be higher dimensional, because its purpose is to display several bells and whistles at the same time. Took me quite a while to come up with an example with somewhat decent numbers. Let

A = [ \begin{matrix}28 & 4 & −1 & 1 & −9 & −15 & −8 \\ 90 & 4 & −2 & 2 & −36 & −45 & −27 \\ 93 & 0 & −2 & 0 & −40 & −45 & −28 \\ 93 & 0 & 0 & −2 & −40 & −45 & −28 \\ −560 & −12 & 3 & −3 & 235 & 275 & 169 \\ −554 & 4 & −1 & 1 & 243 & 267 & 168 \\ 1790 & 24 & −6 & 6 & −759 & −875 & −540 \end{matrix} ]

The characteristic polynomial is

det(A − λI) = −λ^{7} − 10λ^{6} − 41λ^{5} − 90λ^{4} − 120λ^{3} − 112λ^{2} − 80λ − 32 = −(λ + 2)^{5}(λ^{2} + 1)

where the factorization is a conspiratively designed lucky ‘coincidence’. So λ_{1} = −2 is an eigenvalue with algebraic multiplicity 5 and λ_{6,7} = ±i are single eigenvalues. To find the geometric multiplicity of λ_{1}, we must actually solve the linear system (A + 2I)v = 0. Let me do this ‘by hand’, with ‘unsystematic’ row transformations (i.e., smart pivoting) to preserve nice numbers:

\begin{matrix}30 & 4 & −1 & 1 & −9 & −15 & −8 \\ 90 & 6 & −2 & 2 & −36 & −45 & −27 \\ 93 & 0 & 0 & 0 & −40 & −45 & −28 \\ 93 & 0 & 0 & 0 & −40 & −45 & −28 \\ −560 & −12 & 3 & −3 & 237 & 275 & 169 \\ −554 & 4 & −1 & 1 & 243 & 269 & 168 \\ 1790 & 24 & −6 & 6 & −759 & −875 & −538 \end{matrix} step1_{⟶} \begin{matrix}30 & 4 & −1 & 1 & −9 & −15 & −8 \\ −3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 93 & 0 & 0 & 0 & −40 & −45 & −28 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ −560 & −12 & 3 & −3 & 237 & 275 & 169 \\ 6 & 16 & −4 & 4 & 6 & −6 & −1 \\ 110 & −12 & 3 & −3 & −48 & −50 & −31 \end{matrix}

\underline{step 1:} subtract 3rd row from 2nd and 4th; subtract 5th row from 6th and add three times to 7th. \underline{step 2:} add 5×last row to 5th, then subtract 3rd row from last, then use 2nd row to produce leading 0’s in 1st, 3rd, 6th row \underline{step 3:} move 4th row to bottom, 2nd row to top, and use it to create more zeros in the 1st column

\begin{matrix}0 & 64 & −21 & 21 & 31 & −15 & 2 \\ −3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 0 & 186 & −62 & 62 & 84 & −45 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ −10 & −72 & 18 & −18 & −3 & 25 & 14 \\ 0 & 28 & −8 & 8 & 14 & −6 & 1 \\ 17 & −12 & 3 & −3 & −8 & −5 & −3 \end{matrix} step3_{⟶} \begin{matrix}−3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 0 & 64 & −21 & 21 & 31 & −15 & 2 \\ 0 & 186 & −62 & 62 & 84 & −45 & 3 \\ 0 & −92 & \frac{74}{3} & \frac{−74}{3} & \frac{−49}{3} & 25 & \frac{32}{3} \\ 0 & 28 & −8 & 8 & 14 & −6 & 1 \\ 0 & 22 & \frac{−25}{3} & \frac{25}{3} & \frac{44}{3} & −5 & \frac{8}{3} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}

\underline{step 4:} subtract 2nd row 3 times from 3rd, add it to 4th, subtract 5th row from 6th \underline{step 5:} add 5th row to 4th, subtract it 5/2 times from 2nd; subtract 3rd row from 6th

\begin{matrix}−3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 0 & 64 & −21 & 21 & 31 & −15 & 2 \\ 0 & −6 & 1 & −1 & −9 & 0 & −3 \\ 0 & −28 & \frac{11}{3} & \frac{−11}{3} & \frac{44}{3} & 10 & \frac{38}{3} \\ 0 & 28 & −8 & 8 & 14 & −6 & 1 \\ 0 & −6 & \frac{−1}{3} & \frac{1}{3} & \frac{2}{3} & 1 & \frac{5}{3} \end{matrix} step5_{⟶} \begin{matrix}−3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 0 & −6 & −1 & 1 & −4 & 0 & \frac{−1}{2} \\ 0 & −6 & 1 & −1 & −9 & 0 & −3 \\ 0 & 0 & \frac{−13}{3} & \frac{13}{3} & \frac{86}{3} & 4 & \frac{41}{3} \\ 0 & 28 & −8 & 8 & 14 & −6 & 1 \\ 0 & 0 & \frac{−4}{3} & \frac{4}{3} & \frac{29}{3} & 1 & \frac{14}{3} \end{matrix}

\underline{step 6:} multiply 4th and 6th row by 3; use 3rd row to produce 0’s in column2 of 2nd and 5th row \underline{step 7:} exchange 3rd and 2nd row; use new 3rd (old 2nd) row to produce 0’s in 3rd column

\begin{matrix}−3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 0 & 0 & −2 & 2 & 5 & 0 & \frac{5}{2} \\ 0 & −6 & 1 & −1 & −9 & 0 & −3 \\ 0 & 0 & −13 & 13 & 86 & 12 & 41 \\ 0 & 0 & −\frac{10}{3} & \frac{10}{3} & −28 & −6 & −13 \\ 0 & 0 & −4 & 4 & 29 & 3 & 14 \end{matrix} step7_{⟶} \begin{matrix}−3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 0 & −6 & 1 & −1 & −9 & 0 & −3 \\ 0 & 0 & −2 & 2 & 5 & 0 & \frac{5}{2} \\ 0 & 0 & 0 & 0 & \frac{107}{2} & 12 & \frac{99}{4} \\ 0 & 0 & 0 & 0 & \frac{−109}{3} & −6 & \frac{−103}{6} \\ 0 & 0 & 0 & 0 & 19 & 3 & 9 \end{matrix}

\underline{step 8:} multiply row5 by 3/2 and add row 4 to it: it becomes (0, 0, 0, 0, −1, 3, −1); use it to produce 0’s in column 5. \underline{step 9:} the 4th row has become a multiple of the 6th and can be dropped; normalize the 6th row.

\begin{matrix}−3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 0 & −6 & 1 & −1 & −9 & 0 & −3 \\ 0 & 0 & −2 & 2 & 5 & 0 & \frac{5}{2} \\ 0 & 0 & 0 & 0 & 0 & \frac{345}{2} & \frac{−115}{4} \\ 0 & 0 & 0 & 0 & −1 & 3 & −1 \\ 0 & 0 & 0 & 0 & 0 & 60 & −10 \end{matrix} step9_{⟶} \begin{matrix}−3 & 6 & −2 & 2 & 4 & 0 & 1 \\ 0 & −6 & 1 & −1 & −9 & 0 & −3 \\ 0 & 0 & −2 & 2 & 5 & 0 & \frac{5}{2} \\ 0 & 0 & 0 & 0 & −1 & 3 & −1 \\ 0 & 0 & 0 & 0 & 0 & 6 & −1 \end{matrix}

We therefore find two eigenvectors: v^{(7)} can be chosen arbitrarily, but the 5th eqn requires 6v^{(6)} − v^{(7)} = 0. The 4th eqn then determines v^{(5)}, whereas v^{(4)} is again arbitrary. The other components are then determined again. The eigenspace is therefore 2-dimensional (geometric multiplicity 2), and we choose two linearly independent eigenvectors v and w by letting v^{(7)} = 2, v^{(4)} = 0, and w^{(7)} = 0, w^{(4)} = 1. Other choices would be equally legitimate and would simply produce a different matrix S.

We thus find the eigenvectors v = [\frac{1}{3} , \frac{1}{2} , 0 , 0 , −1 , \frac{1}{3} , 2]^{T} and w = [0 , 0 , 1 , 1 , 0 , 0 , 0]^{T}.

We try to find further vectors by solving (A + 2I)v′ = v and (A + 2I)w′ = w (if possible). And then, we try to solve (A + 2I)v′′ = v′ and (A + 2I)w′′ = w′ (if possible). We wouldn’t know yet if any particular among these equations has a solution, but the JNF theorem guarantees that altogether we find three solutions, i.e., among the tentative vectors {v′, v′′, v′′′, …, w′, w′′, w′′′, …} three will actually exist (such as to bring the total to 5 generalized eigenvectors, according to the algebraic multiplicity 5).

I skip the remaining linear systems (the row transformations could be reused) and merely give the results:

The system (A + 2I)v′ = v has the solutions v′ = [0 , \frac{1}{12} , 0 , 0 , 0 , 0 , 0]^{T} + c_{1}v + c_{2}w. The most convenient choice is of course c_{1} = c_{2} = 0, and we have v′ = [0 , \frac{1}{12} , 0 , 0 , 0 , 0 , 0]^{T}.

The system (A + 2I)w′ = w has the solutions w′ = [2 , \frac{3}{2} , 0 , 0 , −1 , 5 , 0]^{T} + c_{3}v + c_{4}w, and again we choose w′ = [2 , \frac{3}{2} , 0 , 0 , −1 , 5 , 0]^{T}.

The system (A + 2I)v′′ = v′ has the solutions v′′ = [\frac{1}{4} , \frac{1}{3} , \frac{−1}{6} , 0 , \frac{−3}{4} , \frac{1}{4} , \frac{3}{2}]^{T} + c_{5}v + c_{6}w. We choose c_{5} = c_{6} = 0.

The system (A + 2I)w′′ = w′ has NO solutions.

We have already 5 vectors, and indeed the system (A + 2I)v′′′ = v′′ (which we might yet consider trying) has no solutions. At this moment we know that the two Jordan blocks for λ = −2 have sizes 3 and 2 respectively, corresponding to the sets of vectors {v_{1} := v, v_{2} := v′, v_{3} := v′′} and {v_{4} := w, v_{5} := w′}. (Now that we know the sizes of the Jordan blocks, we can number them sensibly.)

We still need eigenvectors v_{6} and v_{7} for the eigenvalues i and −i respectively. Then we put them all as columns in our matrix S. We get AS = SJ with :

S = [ \begin{matrix}\frac{1}{3} & 0 & \frac{1}{4} & 0 & 2 & −1 & −1 \\ \frac{1}{2} & \frac{1}{12} & \frac{1}{3} & 0 & \frac{3}{2} & 0 & 0 \\ 0 & 0 & \frac{−1}{6} & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ −1 & 0 & \frac{−3}{4} & 0 & −1 & −1 + i & −1 − i \\ \frac{1}{3} & 0 & \frac{1}{4} & 0 & 5 & \frac{−9+8i}{5} & \frac{−9−8i}{5} \\ 2 & 0 & \frac{3}{2} & 0 & 0 & 1 − 4i & 1 + 4i \end{matrix} ] J = [ \begin{matrix}−2 & 1 & 0 \\ 0 & −2 & 1 \\ 0 & 0 & −2 & 0 \\ & & & −2 & 1 \\ & & & 0 & −2 \\ & & & & & i \\ & & & & & & −i \end{matrix} ]

S is invertible as predicted by the general theory. I won’t write down S^{−1}, but note that det S = i/540, therefore A = SJS^{−1}.

ker(A + 2I) is spanned by columns 1 and 4 of S, ker(A + 2I)^{2} is spanned by columns 1,2,4,5 of S. The complete eigenspace for eigenvalue −2 is ker(A+2I)^{3} = ker(A+2I)^{j} for any j ≥ 3, spanned by columns 1–5 of S. — Have a look at the (3,4) entry of J, which is 0. It is this 0 that distinguishes between the case of a 3-Jordan block and a 2-Jordan block, as opposed to a single 5-Jordan block.

The span of columns 6 and 7 of S can also be spanned by real vectors (which are then no longer eigenvectors), namely the real and imaginary parts of these two vectors, [−1 , 0 , 0 , 0 , −1 , −\frac{9}{5} , 1]^{T} and [0 , 0 , 0 , 0 , 1 , \frac{8}{5} , −4]^{T}.

Exponentials

The exponential of a Jordan block of size n can be calculated explicitly:

exp t [ \begin{matrix}λ & 1 \\ & λ & ... \\ & & ... & 1 \\ & & & λ \end{matrix} ] = exp (tλI + t [ \begin{matrix}0 & 1 \\ & 0 & ... \\ & & ... & 1 \\ & & & 0 \end{matrix} ]) = e^{tλ} [ \begin{matrix}1 & t & \frac{t^{2}}{2!} & ⋯ & \frac{t^{n−1}}{(n−1)!} \\ & 1 & t & ... & ⋮ \\ & & ... & ... & \frac{t^{2}}{2!} \\ & & & ... & t \\ & & & & t \end{matrix} ]

Therefore the JNF theorem implies that exp(tA) → 0 as t → ∞ if and only if all eigenvalues of A have Re λ < 0.

Moreover, exp(tA) remains bounded as t → ∞ if and only if all eigenvalues of A have Re λ ≤ 0 and all Jordan blocks for eigenvalues with Re λ = 0 have size 1.