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Ordinary Differential Equations 25 | Example for Non-Diagonalizable Matrix thumbnail

Ordinary Differential Equations 25 | Example for Non-Diagonalizable Matrix

4 min read

Based on The Bright Side of Mathematics's video on YouTube. If you like this content, support the original creators by watching, liking and subscribing to their content.

TL;DR

For ẋ = Ax with x(0)=x0, the unique solution is x(t)=e^{tA}x0, so computing e^{tA} is the central task.

Briefing

A system of linear differential equations with a non-diagonalizable matrix still has a closed-form solution once the matrix exponential is computed—by converting the matrix to Jordan normal form and exploiting the special structure of each Jordan block. With ẋ = Ax and an initial condition x(0)=x0, the unique solution is x(t)=e^{tA}x0, so the real task is calculating e^{tA} when A cannot be diagonalized.

The approach hinges on Jordan normal form: A is written as a block diagonal matrix whose blocks correspond to eigenvalues, with ones on the superdiagonal inside each block and zeros elsewhere. In the worked example, A is already in Jordan form with two Jordan blocks: one for eigenvalue 2 and one for eigenvalue 3. Even when blocks share the same size, their internal “box” structure differs, and that changes the polynomial factors that appear in e^{tA}.

For each Jordan block, the matrix exponential is computed by splitting the block into a diagonal part plus a nilpotent part. Concretely, for the Jordan block associated with eigenvalue 3 (a 3×3 block in the example), the block is decomposed as D+N, where D is diagonal (all 3’s on the diagonal) and N contains only the ones above the diagonal. The nilpotent matrix N satisfies N^2≠0 but N^3=0, so the exponential’s infinite power series collapses to a finite sum: e^{tN}=I+tN+(t^2/2)N^2. A key technical step is that D and N commute, which allows the factorization e^{t(D+N)}=e^{tD}e^{tN}. Since e^{tD} is just the exponential applied to each diagonal entry, the result becomes a triangular matrix with e^{3t} on the diagonal and additional terms like t e^{3t} and (t^2/2)e^{3t} in the superdiagonal positions.

The same method applies to the Jordan block for eigenvalue 2 (a smaller block in the example). Again, the block splits into a diagonal component and a nilpotent component. Here the nilpotent part squares to zero, so e^{tN} truncates after the linear term: e^{tN}=I+tN. Multiplying by e^{tD} produces another triangular matrix with e^{2t} on the diagonal and a t e^{2t} term in the appropriate superdiagonal entry.

Finally, the full matrix exponential e^{tA} is assembled by combining the exponentials of all Jordan blocks in the correct order. The takeaway is practical: even without diagonalizability, Jordan form turns e^{tA} into manageable algebra—triangular matrices whose entries are products of exponentials e^{λt} and low-degree polynomials in t determined by the size of each Jordan block.

Cornell Notes

For ẋ = Ax with x(0)=x0, the solution is x(t)=e^{tA}x0, so the main challenge is computing e^{tA} when A is not diagonalizable. Putting A into Jordan normal form breaks it into Jordan blocks, and each block can be written as D+N where D is diagonal (eigenvalue on the diagonal) and N is nilpotent (ones on the superdiagonal). Because D and N commute, e^{t(D+N)}=e^{tD}e^{tN}. Nilpotency makes e^{tN} a finite sum: for a block where N^3=0, e^{tN}=I+tN+(t^2/2)N^2. The resulting e^{tA} is block-diagonal (in Jordan form) with e^{λt} times polynomials in t, and assembling all blocks yields the full solution.

Why does the solution of ẋ = Ax reduce to computing a matrix exponential?

For a linear system ẋ = Ax with initial condition x(0)=x0, there is a unique solution given by x(t)=e^{tA}x0. The entire dynamics are encoded in e^{tA}, so once that matrix exponential is known, the state vector at time t follows immediately by multiplying by x0.

How does Jordan normal form make e^{tA} computable when A is not diagonalizable?

Jordan normal form expresses A as a block diagonal matrix of Jordan blocks. Each Jordan block has a single eigenvalue λ on the diagonal and ones on the superdiagonal. Because the exponential of a block diagonal matrix is block diagonal, e^{tA} can be built by computing e^{tJ} for each Jordan block J separately.

What is the D+N split inside a Jordan block, and why does it matter?

Inside a Jordan block, write J = D + N, where D is diagonal (λ on the diagonal) and N contains only the superdiagonal ones. D is easy because e^{tD} is just exponentials of the diagonal entries. N is nilpotent, so its exponential truncates to a finite polynomial in t. The commuting property D N = N D enables the factorization e^{t(D+N)}=e^{tD}e^{tN}.

How does nilpotency turn the infinite exponential series into a finite sum?

Nilpotency means N^k=0 for some k. Then e^{tN} = I + tN + (t^2/2!)N^2 + … stops at the highest nonzero power. In the example for the eigenvalue-3 Jordan block, N^3=0, so e^{tN}=I+tN+(t^2/2)N^2, producing polynomial factors like t and t^2 in the superdiagonal entries.

What new terms appear in e^{tJ} for a Jordan block compared with a diagonalizable case?

Diagonalizable matrices yield e^{λt} times constants along eigenvector directions. Jordan blocks add polynomial-in-t factors multiplying e^{λt}. For the 3×3 Jordan block example, the superdiagonal entries include terms proportional to t e^{3t} and (t^2/2)e^{3t}, reflecting the block’s size and the powers of the nilpotent part.

Review Questions

  1. Given a Jordan block J = D+N with N^3=0, write the truncated form of e^{tN} and explain what determines the truncation point.
  2. How does the commuting condition between D and N enable the factorization e^{t(D+N)} = e^{tD}e^{tN}?
  3. Why do polynomial factors in t appear in e^{tA} for non-diagonalizable matrices, and how are they linked to Jordan block size?

Key Points

  1. 1

    For ẋ = Ax with x(0)=x0, the unique solution is x(t)=e^{tA}x0, so computing e^{tA} is the central task.

  2. 2

    Converting A to Jordan normal form reduces the problem to computing exponentials of individual Jordan blocks.

  3. 3

    Each Jordan block can be written as J = D + N, where D is diagonal (eigenvalue on the diagonal) and N is nilpotent (ones on the superdiagonal).

  4. 4

    Because D and N commute, e^{tJ} factors as e^{tD}e^{tN}, turning matrix exponential computation into two simpler exponentials.

  5. 5

    Nilpotency forces e^{tN} to truncate to a finite polynomial in t, with the highest power determined by the smallest k such that N^k=0.

  6. 6

    Assembling the block exponentials in the correct Jordan order yields the full e^{tA}, and the final solution follows by multiplying by x0.

Highlights

Non-diagonalizable systems still have exact solutions: x(t)=e^{tA}x0, provided e^{tA} is computed correctly.
Jordan blocks turn e^{tA} into triangular matrices whose entries combine e^{λt} with low-degree polynomials in t.
Nilpotent parts make the exponential series finite: if N^3=0, then e^{tN}=I+tN+(t^2/2)N^2.
The commuting property D N = N D is what allows e^{t(D+N)} to split into e^{tD}e^{tN}.
The size of each Jordan block determines the highest power of t that appears in e^{tA}.

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