Section 11.5

Jordan Normal Form of Linear Operators

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Problem 1187

(Root subspace). Let А be a linear operator on a space VV. Prove that the set Rλ(A)⊂VR_{\lambda }(\mathrm{A}) \subset V, consisting of all root vectors of the operator A\mathrm{A} corresponding to a given eigenvalue λ\lambda, together with the zero vector, is a vector subspace of VV.

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Problem 1188

Prove that if the operators А and В commute, then every root subspace RλR_{\lambda } (A) of the operator А is invariant under В.

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Problem 1189

Find the eigenvalues and root subspaces of the linear operators given by the following matrices:

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(1)

(26−1511−512−6)\left(\begin{array}{rrc}2 & 6 & -15 \\ 1 & 1 & -5 \\ 1 & 2 & -6\end{array}\right);

(2)

(0−23211−1−100201−101)\left(\begin{array}{rrrr}0 & -2 & 3 & 2 \\ 1 & 1 & -1 & -1 \\ 0 & 0 & 2 & 0 \\ 1 & -1 & 0 & 1\end{array}\right).

Problem 1190

Find the eigenvalues and root subspaces of the differentiation operator d/dxd / d x, acting on the space of real infinitely differentiable functions on the line R\mathbb {R}.

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Problem 1191

(Root decomposition). Prove that for any operator А on a finite-dimensional space VV over an algebraically closed field the following decomposition holds

V=⨁λRλ(A) V = \bigoplus _{\lambda } R_{\lambda }(\mathbf{A})

where the sum is taken over all eigenvalues λ\lambda of the operator A\mathbf{A}.

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Problem 1192

Prove that the dimension of the root subspace Rλ(A)R_{\lambda }(\mathrm{A}) equals the multiplicity of the root λ\lambda in the characteristic polynomial χA(λ)\chi_{\mathrm{A}}(\lambda ).

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Problem 1193

Let А be a linear operator on a space VV over an algebraically closed field. For an eigenvalue λ\lambda of the operator А, denote by πλ\pi_{\lambda } the projection operator onto the root subspace Rλ(A)R_{\lambda }(\mathbf{A}) along the direct sum of the remaining root subspaces. Prove that the formulas

S=∑λλπλ,N=A−S \mathbf{S} = \sum _{\lambda } \lambda \pi _{\lambda }, \quad \mathbf{N} = \mathbf{A}-\mathbf{S}

define the semisimple and nilpotent operators from the Jordan decomposition (see problem 1148).

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Problem 1194

Prove that the Jordan decomposition A=S+N\mathrm{A} = \mathrm{S}+\mathrm{N} for the operator А on a finite-dimensional vector space over R,C\mathbb {R}, \mathbb {C} (over an arbitrary field of characteristic zero) is unique.

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Problem 1195

Let А be a nilpotent operator on a finite-dimensional vector space VV, i.e., Ak=0\mathrm{A}^{k} = 0 for some kk. Prove that VV can be decomposed into a direct sum of invariant subspaces Vi,i=1,…,mV_{i}, i = 1, \ldots , m, in such a way that the restriction of the operator A\mathbf{A} to ViV_{i} has, in some basis, the matrix

(01…00001…0………00…0100…00) \left(\begin{smallmatrix} 0 & 1 & \ldots & 0 & 0 \\ 0 & 0 & 1 & \ldots & 0 \\ \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & 0 & 1 \\ 0 & 0 & \ldots & 0 & 0 \end{smallmatrix}\right)
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Problem 1196

Under the conditions of the previous problem, prove that m=dim⁡(Ker⁡A)m = \operatorname {dim}(\operatorname {Ker} A), and thus the number of blocks of the indicated form does not depend on the way the decomposition is carried out.

Problem 1195: А is a nilpotent operator on a finite-dimensional vector space VV, i.e., Ak=0\mathrm{A}^{k} = 0 for some kk; VV is decomposed into a direct sum of invariant subspaces Vi,i=1,…,mV_{i}, i = 1, \ldots , m, such that the restriction of the operator A\mathbf{A} to ViV_{i} has, in some basis, a Jordan block with zeros on the diagonal.

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Problem 1197

Denote by rjr_{j} the number of Jordan blocks of size jj in the decomposition of a nilpotent operator A\mathbf{A} for which Ak=0\mathbf{A}^{k} = 0. Prove the equalities:

rk=dim⁡(Ker⁡Ak)−dim⁡(Ker⁡Ak−1)rk−1+rk=dim⁡(Ker⁡Ak−1)−dim⁡(Ker⁡Ak−2)rk−2+rk−1+rk=dim⁡(Ker⁡Ak−2)−dim⁡(Ker⁡Ak−3)……………………………………………………………………r1+r2+…+rk=dim⁡(Ker⁡A) \begin{aligned} & r_{k} = \operatorname {dim}\left(\operatorname {Ker} \mathbf{A}^{k}\right)-\operatorname {dim}\left(\operatorname {Ker} \mathbf{A}^{k-1}\right) \\ & r_{k-1}+r_{k} = \operatorname {dim}\left(\operatorname {Ker} \mathbf{A}^{k-1}\right)-\operatorname {dim}\left(\operatorname {Ker} \mathbf{A}^{k-2}\right) \\ & r_{k-2}+r_{k-1}+r_{k} = \operatorname {dim}\left(\operatorname {Ker} \mathbf{A}^{k-2}\right)-\operatorname {dim}\left(\operatorname {Ker} \mathbf{A}^{k-3}\right) \\ & \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ & r_{1}+r_{2}+\ldots +r_{k} = \operatorname {dim}(\operatorname {Ker} \mathbf{A}) \end{aligned}

Using these relations, find expressions for rjr_{j} in terms of dim⁡(Ker⁡Am)\operatorname {dim}\left(\operatorname {Ker} \mathbf{A}^{m}\right), where m=1,…,km = 1, \ldots , k.

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Problem 1198

Find the Jordan form of the double-differentiation operator on the space C10[t]\mathbb {C}_{10}[t] of polynomials of degree at most 10.

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Problem 1199

Find the Jordan form JJ and a Jordan basis (ei)(e_{i}) for the following operators:

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(1)

(−20−1−1−1−1100)\left(\begin{array}{rrr}-2 & 0 & -1 \\ -1 & -1 & -1 \\ 1 & 0 & 0\end{array}\right);

(2)

(−2−3−4−100111)\left(\begin{array}{rrr}-2 & -3 & -4 \\ -1 & 0 & 0 \\ 1 & 1 & 1\end{array}\right)

(3)

(−2−4−5−1−3−3144)\left(\begin{array}{rrr}-2 & -4 & -5 \\ -1 & -3 & -3 \\ 1 & 4 & 4\end{array}\right);

(4)

(10151327−20−2−10)\left(\begin{array}{rrr}10 & 1 & 5 \\ 13 & 2 & 7 \\ -20 & -2 & -10\end{array}\right);

(5)

(2−3−4−1551−4−4)\left(\begin{array}{rrr}2 & -3 & -4 \\ -1 & 5 & 5 \\ 1 & -4 & -4\end{array}\right);

(6)

(0−2−3−1241−3−5)\left(\begin{array}{rrr}0 & -2 & -3 \\ -1 & 2 & 4 \\ 1 & -3 & -5\end{array}\right);

(7)

(1−4−31−5−4−154)\left(\begin{array}{rrr}1 & -4 & -3 \\ 1 & -5 & -4 \\ -1 & 5 & 4\end{array}\right)

(8)

(0−1−1−1−1−2011)\left(\begin{array}{rrr}0 & -1 & -1 \\ -1 & -1 & -2 \\ 0 & 1 & 1\end{array}\right)

(9)

(a00b0a000ba0000a)\left(\begin{array}{llll}a & 0 & 0 & b \\ 0 & a & 0 & 0 \\ 0 & b & a & 0 \\ 0 & 0 & 0 & a\end{array}\right);

(10)

(210001100011−1−1−10)\left(\begin{array}{rrrr}2 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ -1 & -1 & -1 & 0\end{array}\right);

(11)

(1012301012001010001000001)\left(\begin{array}{lllll}1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1\end{array}\right);

(12)

(01000001000001000001132−2−3)\left(\begin{array}{rrrrr}0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 3 & 2 & -2 & -3\end{array}\right);

(13)

(201100111010−100001000201000111000−100)\left(\begin{array}{rrrrrr}2 & 0 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & -1 & 0 & 0\end{array}\right);

(14)

(301111030000003011000300000030000003)\left(\begin{array}{llllll}3 & 0 & 1 & 1 & 1 & 1 \\ 0 & 3 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 1 & 1 \\ 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3\end{array}\right);

(15)

(−3−525−201263−4−8−6−337)\left(\begin{array}{rrrr}-3 & -5 & 2 & 5 \\ -2 & 0 & 1 & 2 \\ 6 & 3 & -4 & -8 \\ -6 & -3 & 3 & 7\end{array}\right)

(16)

(34−1−347−1−4−12−203121220−4−13)\left(\begin{array}{rrrr}3 & 4 & -1 & -3 \\ 4 & 7 & -1 & -4 \\ -12 & -20 & 3 & 12 \\ 12 & 20 & -4 & -13\end{array}\right);

(17)

(−1111−11002−5−10−250−1)\left(\begin{array}{rrrr}-1 & 1 & 1 & 1 \\ -1 & 1 & 0 & 0 \\ 2 & -5 & -1 & 0 \\ -2 & 5 & 0 & -1\end{array}\right);

(18)

(−5−927−3−21399−5−12−9−9310)\left(\begin{array}{rrrr}-5 & -9 & 2 & 7 \\ -3 & -2 & 1 & 3 \\ 9 & 9 & -5 & -12 \\ -9 & -9 & 3 & 10\end{array}\right).

Problem 1200

Find the Jordan form JJ, the transition matrix CC to the Jordan basis, the projector matrices onto the root subspaces, and the Jordan decomposition for the following operators:

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(1)

(7−2−1410−104−56−2−13103−1−64)\left(\begin{array}{rrcc}7 & -2 & -14 & 10 \\ -1 & 0 & 4 & -5 \\ 6 & -2 & -13 & 10 \\ 3 & -1 & -6 & 4\end{array}\right)

(2)

(325−8−3−3−12194616−22237−9)\left(\begin{array}{rccc}3 & 2 & 5 & -8 \\ -3 & -3 & -12 & 19 \\ 4 & 6 & 16 & -22 \\ 2 & 3 & 7 & -9\end{array}\right);

(3)

(1−2−810−104−52−2−9101−1−44)\left(\begin{array}{rrrc}1 & -2 & -8 & 10 \\ -1 & 0 & 4 & -5 \\ 2 & -2 & -9 & 10 \\ 1 & -1 & -4 & 4\end{array}\right);

(4)

(3−1133111330056400−3−4−400123)\left(\begin{array}{rrrrr}3 & -1 & 1 & 3 & 3 \\ 1 & 1 & 1 & 3 & 3 \\ 0 & 0 & 5 & 6 & 4 \\ 0 & 0 & -3 & -4 & -4 \\ 0 & 0 & 1 & 2 & 3\end{array}\right)

(5)

(21−2−11−102210034400−2−3−400123)\left(\begin{array}{rrrrr}2 & 1 & -2 & -1 & 1 \\ -1 & 0 & 2 & 2 & 1 \\ 0 & 0 & 3 & 4 & 4 \\ 0 & 0 & -2 & -3 & -4 \\ 0 & 0 & 1 & 2 & 3\end{array}\right)

(6)

(−4−142540−8−4−1200−4−400022000000)\left(\begin{array}{rrrrr}-4 & -1 & 4 & 2 & 5 \\ 4 & 0 & -8 & -4 & -12 \\ 0 & 0 & -4 & -4 & 0 \\ 0 & 0 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right);

(7)

(−796−41−45302001−6−4001−4−200−131)\left(\begin{array}{rrrrr}-7 & 9 & 6 & -4 & 1 \\ -4 & 5 & 3 & 0 & 2 \\ 0 & 0 & 1 & -6 & -4 \\ 0 & 0 & 1 & -4 & -2 \\ 0 & 0 & -1 & 3 & 1\end{array}\right);

(8)

(−7−4−468956−9−12005−9−3004−7−20000−1)\left(\begin{array}{rrrrr}-7 & -4 & -4 & 6 & 8 \\ 9 & 5 & 6 & -9 & -12 \\ 0 & 0 & 5 & -9 & -3 \\ 0 & 0 & 4 & -7 & -2 \\ 0 & 0 & 0 & 0 & -1\end{array}\right);

(9)

(−1−1−2−1−4438−5800−410800−497003−6−5)\left(\begin{array}{rrrrr}-1 & -1 & -2 & -1 & -4 \\ 4 & 3 & 8 & -5 & 8 \\ 0 & 0 & -4 & 10 & 8 \\ 0 & 0 & -4 & 9 & 7 \\ 0 & 0 & 3 & -6 & -5\end{array}\right).