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Let 3P (the vector space of real polynomials of degree less than 3) defined by
(())'()''()p x xp x p x σ=+.
(1) Find the matrix A representing σ with respect to the ordered basis [21,,x x ] for 3P .
(2) Find a basis for 3P such that with respect to this basis, the matrix B representing σ is diagonal.
(3) Find the kernel （核） and range （值域）of this transformation. Solution: (1)
221022x x x x σσσ===+（）（）（） 002010002A ⎛⎫
⎪
= ⎪ ⎪
⎝⎭
----------------------------------------------------------------------------------------------------------------- (2)
101010001T ⎛⎫ ⎪
= ⎪ ⎪⎝⎭
(The column vectors of T are the eigenvectors of A)
The corresponding eigenvectors in 3P are 1000010002T AT -⎛⎫
⎪
= ⎪ ⎪⎝⎭
(T diagonalizes A ) 22[1,,1][1,,]x x x x T += . With respect to this new basis 2
[1,,1]x x +, the representing
matrix of σis diagonal.
------------------------------------------------------------------------------------------------------------------- (3) The kernel is the subspace consisting of all constant polynomials.
The range is the subspace spanned by the vectors 2,1x x +
-----------------------------------------------------------------------------------------------------------------------
Let 020012A ⎛⎫
⎪
= ⎪ ⎪-⎝⎭
.
(1) Find all determinant divisors and elementary divisors of A .
(2) Find a Jordan canonical form of A .
(3) Compute At e . (Give the details of your computations.) Solution: (1)
1
100
20012I A λλλλ-⎛⎫ ⎪
-=- ⎪ ⎪-⎝⎭,(特征多项式 2()(1)(2)p λλλ=--. Eigenvalues are 1, 2, 2.)
Determinant divisor of order 1()1D λ=, 2()1D λ=, 23()()(1)(2)D p λλλλ==-- Elementary divisors are 2(1) and (2)λλ-- .
---------------------------------------------------------------------------------------------------------------------- (2) The Jordan canonical form is
100021002J ⎛⎫ ⎪
= ⎪ ⎪⎝⎭
--------------------------------------------------------------------------------------------------------------------------
(3) For eigenvalue 1, 0
100
1001
1I A ⎛⎫
⎪
-=- ⎪ ⎪-⎝⎭ ， An eigenvector is 1(1,0,0)T p = For eigenvalue 2, 11020
0001
0I A ⎛⎫
⎪
-= ⎪ ⎪⎝
⎭
, An eigenvector is 2(0,0,1)T p =
Solve 32(2)A I p p -=, 331100(2)00000101A I p p --⎛⎫⎛⎫
⎪ ⎪
-== ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭
we obtain that
3(1,1,0)T p =-
101001010P ⎛⎫ ⎪
=- ⎪ ⎪⎝⎭, 111000
1010P -⎛⎫
⎪
= ⎪ ⎪-⎝
⎭ 1At J e Pe P -=222101001100010
00101000010t
t t t e e te e ⎛⎫⎛⎫⎛⎫
⎪ ⎪ ⎪=- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝
⎭222200
00t t t t t t e e e e te
e ⎛⎫
-
⎪= ⎪ ⎪-⎝⎭ --------------------------------------------------------------------------------------------------------------------
Suppose that ∈R A and O I A A =--65.
(1) What are the possible minimal polynomials of A ? Explain.
(2) In each case of part (1), what are the possible characteristic polynomials of A ? Explain.
Solution:
(1) An annihilating polynomial of A is 2
56x x --.
The minimal polynomial of A divides any annihilating polynomial of A. The possible minimal polynomials are
6x -, 1x +, and 256x x --.
---------------------------------------------------------------------------------------------------------------
(2) The minimal polynomial of A divides the characteristic polynomial of A. Since A is a matrix of order 3, the characteristic polynomial of A is of degree 3. The minimal polynomial of A and the characteristic polynomial of A have the same linear factors. Case 6x -, the characteristic polynomial is 3(6)x - Case 1x +, the characteristic polynomial is 3(1)x + Case 256x x --, the characteristic polynomial is 2(1)(6)x x +- or 2(6)(1)x x -+
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-------------------------------
Let 120000A ⎛⎫=
⎪⎝⎭
. Find the Moore-Penrose inverse A +
of A .
