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Linear Algebra II (pdf, 500 kB)

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14<br />

5.6. Application. One important application of the Jordan Normal Form Theorem<br />

is to the explicit solution of systems of linear first-order differential equations<br />

with constant coefficients. Such a system can be written<br />

d<br />

y(t) = A · y(t) ,<br />

dt<br />

where y is a vector-valued function and A is a matrix. One can then show (Exercise)<br />

that there is a unique solution with y(0) = y0 for any specified initial value y0,<br />

and it is given by<br />

y(t) = exp(tA) · y0<br />

with the matrix exponential<br />

exp(tA) =<br />

∞<br />

n=0<br />

t n<br />

n! An .<br />

If A is in Jordan Normal Form, the exponental can be easily determined. In<br />

general, A can be transformed into Jordan Normal Form, the exponential can be<br />

evaluated for the transformed matrix, then we can transform it back — note that<br />

exp(tP −1 AP ) = P −1 exp(tA)P .<br />

5.7. Remark. What can we do when the characteristic polynomial does not split<br />

into linear factors (which is possible when the field F is not algebraically closed)?<br />

In this case, we have to use a weaker notion than that of diagonalizability. Define<br />

the endomorphism f : V → V to be semi-simple if every f-invariant subspace<br />

U ⊂ V has an f-invariant complementary subspace in V . One can show (Exercise)<br />

that if the characteristic polynomial of f splits into linear factors, then f is semisimple<br />

if and only if it is diagonalizable. The general version of the Jordan Normal<br />

Form Theorem then is as follows.<br />

Let V be a finite-dimensional vector space, f : V → V an endomorphism. Then<br />

f = s + n with endomorphisms s and n of V such that s is semi-simple, n is<br />

nilpotent, and s ◦ n = n ◦ s.<br />

Unfortunately, we do not have the means and time to prove this result here.<br />

However, we can state the result we get over F = R.<br />

5.8. Theorem (Real Jordan Normal Form). Let V be a finite-dimensional<br />

real vector space, f : V → V an endomorphism. Then there is a basis of V such<br />

that the matrix representing f with respect to this basis is a block diagonal matrix<br />

with blocks of the form B(λ, m) and of the form (with µ > 0)<br />

B ′ ⎛<br />

λ −µ 1 0 · · · 0 0 0<br />

⎞<br />

0<br />

⎜<br />

µ<br />

⎜0<br />

⎜<br />

⎜0<br />

⎜<br />

(λ, µ, m) = ⎜ .<br />

⎜<br />

⎜0<br />

⎜<br />

⎜0<br />

⎝0<br />

λ<br />

0<br />

0<br />

.<br />

0<br />

0<br />

0<br />

0<br />

λ<br />

µ<br />

.<br />

0<br />

0<br />

0<br />

1<br />

−µ<br />

λ<br />

.<br />

0<br />

0<br />

0<br />

· · ·<br />

· · ·<br />

· · ·<br />

...<br />

· · ·<br />

· · ·<br />

· · ·<br />

0<br />

0<br />

0<br />

.<br />

λ<br />

µ<br />

0<br />

0<br />

0<br />

0<br />

.<br />

−µ<br />

λ<br />

0<br />

0<br />

0<br />

0<br />

.<br />

1<br />

0<br />

λ<br />

0 ⎟<br />

0 ⎟<br />

0 ⎟<br />

.<br />

⎟ ∈ Mat(R, 2m) .<br />

0<br />

⎟<br />

1 ⎟<br />

−µ ⎠<br />

0 0 0 0 · · · 0 0 µ λ<br />

Blocks B(λ, m) occur for eigenvalues λ of f; blocks B ′ (λ, µ, m) occur if Pf(x) is<br />

divisible by x 2 − 2λx + λ 2 + µ 2 .

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