Definition and Theorems (Chapter 7)

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Definition. If \alpha is any vector in V, the T-cyclic subspace generated by \alpha is the subspace Z(\alpha;V) of all vectors of the form g(T)\alpha,g\in F[x]. If Z(\alpha;T)=V, then \alpha is called a cyclic vector for T.

Definition. If \alpha is any vector in V, the T-annihilator of \alpha is the ideal M(\alpha;T) in F[x] consisting of all polynomials g over F such that g(T)\alpha=0. The unique monic polynomial p_{\alpha} which generates this ideal will also be called the T-annihilator of \alpha.

Theorem 1. Let \alpha be any non-zero vector in V and let p_{\alpha} be the T-annihilator of \alpha.
(i) The degree of p_{\alpha} is equal to the dimension of the cyclic subspace Z(\alpha;T).
(ii) If the degree of p_{\alpha} is k, then the vectors \alpha,T\alpha,T^2\alpha,\dots,T^{k-1}\alpha form a basis for Z(\alpha;T).
(iii) If U is the linear operator on Z(\alpha;T) induced by T, then the minimal polynomial for U is p_{\alpha}.

If U is an operator on W of dimension k which has a cyclic vector \alpha, then \alpha,\dots,U^{k-1}\alpha form a basis for W, and p_{\alpha} is the minimal polynomial for U. If we let \alpha_i=U^{i-1}\alpha,i=1,\cdots,k, and p_{\alpha}=c_0+c_1x+\cdots+c_{k-1}x^{k-1}+x^k, then

U\alpha_i=\alpha_{i+1},\quad i=1,\dots,k-1\\U\alpha_k=-c_0\alpha_1-c_1\alpha_2-\cdots-c_{k-1}\alpha_k

thus the matrix of U in the ordered basis \mathfrak B=\{\alpha_1,\dots,\alpha_k \} is called the companion matrix of the monic polynomial p_{\alpha}, which is

\displaystyle{\begin{bmatrix}0&0&0&\cdots&0&-c_0\\1&0&0&\cdots&0&-c_1\\0&1&0&\cdots&0&-c_2\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\0&0&0&\cdots&1&-c_{k-1}\end{bmatrix}.}

Theorem 2. If U is a linear operator on the finite-dimensional space W, then U has a cyclic vector if and only if there is some ordered basis for W in which U is represented by the companion matrix of the minimal polynomial for U.
Corollary. If A is the companion matrix of a monic polynomial p, then p is both the minimal and the characteristic polynomial of A.

Definition. Let T be a linear operator on a vector space V and let W be a subspace of V. We say that W is T-admissible if
(i) W is invariant under T.
(ii) if f(T)\beta is in W, there exists a vector \gamma in W such that f(T)\beta=f(T)\gamma.

Theorem 3 (Cyclic Decomposition Theorem). Let T be a linear operator on a finite-dimensional vector space V and let W_0 be a proper T-admissible subspace of V. There exist non-zero vectors \alpha_1,\dots,\alpha_r in V with respective T-annihilators p_1,\dots,p_r such that
(i) V=W_0\oplus Z(\alpha_1;T)\oplus\cdots\oplus Z(\alpha_r;T);
(ii) p_k divides p_{k-1},k=2,\dots,r.
Corollary. If T is a linear operator on a finite-dimensional vector space, then every T-admissible subspace has a complementary subspace which is also invariant under T.
Corollary. Let T be a linear operator on a finite-dimensional vector space V.
( a ) There exists a vector \alpha in V such that the T-annihilator of \alpha is the minimal polynomial for T.
( b ) T has a cyclic vector if and only if the characteristic and minimal polynomials for T are identical.

Theorem 4 (Generalized Cayley-Hamilton Theorem). Let T be a linear operator on a finite-dimensional vector space V. Let p and f be the minimal and characteristic polynomials for T, respectively.
(i) p divides f.
(ii) p and f have the same prime factors, except for multiplicities.
(iii) If p=f_1^{r_1}{\cdots}f_k^{r_k} is the prime factorization of p, then f=f_1^{d_1}{\cdots}f_k^{d_k} where d_i is the nullity of f_i(T)^n divided by the degree of f_i.
Corollary. If T is a nilpotent linear operator on a vector space of dimension n, then the characteristic polynomial for T is x^n.

An n\times n matrix A which is the direct sum of companion matrices of non-scalar monic polynomials p_1,\dots,p_r such that p_{i+1} divides p_i for i=1,\dots,r-1 will be said to be in rational form.

Theorem 5. Let F be a field and let B be an n\times n matrix over F. Then B is similar over the field F to one and only one matrix which is in rational form.
The polynomials p_1,\dots,p_r are called the invariant factors for the matrix B.

If M\in F[x]^{m\times n}, an elementary row operation on M is one of the following:

  1. multiplication of one row of M by a non-zero scalar in F.
  2. replacement of the rth row of M by row r plus f times row s, where f is any polynomial over F and r\neq s.
  3. interchange of two rows of M.

An elementary matrix in F[x]^{m\times n} is one which can be obtained from the m\times n identity matrix by means of a single elementary row operation.
Let M,N\in F[x]^{m\times n}, we say that N is row-equivalent to M if N can be obtained from M by a finite succession of elementary row operations: M=M_0{\rightarrow}M_1{\rightarrow}\cdots{\rightarrow}M_k=N.

Lemma. Let M be a matrix in F[x]^{m\times n} which has some non-zero entry in its first column, and let p be the greatest common divisor of the entries in column 1 of M. Then M is row-equivalent to a matrix N which has

\displaystyle{\begin{bmatrix}p\\0\\{\vdots}\\0\end{bmatrix}}

as its first column.

Theorem 6. Let P be an m\times m matrix with entries in the polynomial algebra F[x]. The following are equivalent.
(i) P is invertible.
(ii) The determinant of P is a non-zero scalar polynomial.
(iii) P is row-equivalent to the m\times m identity matrix.
(iv) P is a product of elementary matrices.
Corollary. Let M and N be m\times n matrices with entries in the polynomial algebra F[x]. Then N is row-equivalent to M if and only if
\displaystyle{N=PM}
where P is an invertible m\times m matrix with entries in F[x].

We define elementary column operations and column-equivalence in a manner analogous to row operations and row-equivalence.

Definition. The matrix N is equivalent to the matrix M if we can pass from M to N by means of a sequence of operations

\displaystyle{M=M_0{\rightarrow}M_1{\rightarrow}\cdots{\rightarrow}M_k=N}

each of which is an elementary row operation or an elementary column operation.

Theorem 7. Let M and N be m\times n matrices with entries in the polynomial algebra F[x]. Then N is equivalent to M if and only if

\displaystyle{N=PMQ}

where P is an invertible matrix in F[x]^{m\times n} and Q is an invertible matrix in F[x]^{m\times n}.

Theorem 8. Let A be an m\times n matrix with entries in the field F, and let p_1,\dots,p_r be the invariant factors for A. The matrix xI-A is equivalent to the n\times n diagonal matrix with diagonal entries p_1,\dots,p_r,1,1,\dots,1.

Definition. Let N be a matrix in F[x]^{m\times n}. We say that N is in (Smith) normal form if
( a ) every entry off the main diagonal of N is 0.
( b ) on the main diagonal of N there appear (in order) polynomials f_1,\dots,f_l such that f_k|f_{k+1},1\leq k\leq l-1, in which l=\min(m,n).

Theorem 9. Let M be an m\times n matrix with entries in the polynomial algebra F[x]. Then M is equivalent to a matrix N which is in normal form.

Definition. Let M be an m\times n matrix with entries in F[x]. If 1\leq k\leq \min(m,n), we define \delta_k(M) to be the greatest common divisor of the determinants of all k\times k submatrices of M.

Theorem 10. If M and N are equivalent m\times n matrices with entries in F[x], then

\displaystyle{\delta_k(M)=\delta_k(N),\qquad 1\leq k\leq \min(m,n)}

Corollary. Each matrix M in F[x]^{m\times n} is equivalent to precisely one matrix N which is in normal form. The polynomials f_1,\dots,f_l which occur on the main diagnal of N are

\displaystyle{f_k=\dfrac{\delta_k(M)}{\delta_{k-1}(M)},\qquad 1\leq k\leq \min(m,n)}

where, for convenience, we define \delta_0(M)=1.

Definition. Let V be a finite-dimensional vector space over the field F, and let T be a linear operator on V. We say that T is semi-simple if every T-invariant subspace has a complementary T-invariant subspace.

Lemma. Let T be a linear operator on the finite-dimensional vector space V, and let V=W_1\oplus\cdots\oplus W_k be the primary decomposition for T. In other words, if p is the minimal polynomial for T and p=p_1^{r_1}\cdots p_k^{r_k} is the prime factorization of p, then W_j is the null space of p_j(T)^{r_j}. Let W be any subspace of V which is invariant under T. Then

\displaystyle{W=(W\cap W_1)\oplus\cdots\oplus(W\cap W_k)}

Lemma. Let T be a linear operator on V, and suppose that the minimal polynomial for T is irreducible over the scalar field F. Then T is semi-simple.

Theorem 11. Let T be a linear operator on the finite-dimensional vector space V. A necessary and sufficient condition that T be semi-simple is that the minimal polynomial p for T be of the form p=p_1\cdots p_k, where p_1,\dots,p_k are distinct irreducible polynomials over the scalar field F.
Corollary. If T is a linear operator on a finite-dimensional vector space over an algebraically closed field, then T is semi-simple if and only if T is diagonalizable.

Lemma (Taylor’s Formula). Let F be a field of characteristic zero and let g and h be polynomials over F. If f is any polynomial over F with \deg f\leq n, then

\displaystyle{f(g)=f(h)+f^{(1)}(h)(g-h)+\frac{f^{(2)}(h)}{2!}(g-h)^2+\cdots+\frac{f^{(n)}(h)}{n!}(g-h)^n.}

Lemma. Let F be a subfield of the complex numbers, let f be a polynomial over F, and let f' be the derivative of f. The following are equivalent:
( a ) f is the product of distinct polynomials irreducible over F.
( b ) f and f' are relatively prime.
( c ) As a polynomial with complex coefficients, f has no repeated root.

Theorem 12. Let F be a subfield of the field of complex numbers, let V be a finite-dimensional vector space over F, and let T be a linear operator on V. Let \mathfrak B be an ordered basis for V and let A be the matrix of T in the ordered basis \mathfrak B. Then T is semi-simple if and only if the matrix A is similar over the field of complex numbers to a diagonal matrix.

Theorem 13. Let F be a subfield of the field of complex numbers, let V be a finite-dimensional vector space over F, and let T be a linear operator on V. There is a semi-simple operator S on V and a nilpotent operator N on V such that
(i) T=S+N;
(ii) SN=NS.
Furthermore, the semi-simple S and nilpotent N satisfying (i) and (ii) are unique, and each is a polynomial in T.

Linear Algebra (2ed) Hoffman & Kunze 7.5

christangdt留下评论

这一节是一个总结（summary）和一部分新的内容即semi-simple operator。
总结部分是对第6章和第7章的一个高度概括，叙述很简练有效，值得原样照搬：
We began to study T by means of characteristic values and characteristic vectors. We introduced diagonalizable operators, the operators which can be completely described in terms of characteristic values and vectors. We then observed that T might not have a single characteristic vector. Even in the case of an algebraically closed scalar field, when every linear operator does have at least one characteristic vector, we noted that the characteristic vectors of T need not span the space.
We then proved the cyclic decomposition theorem, expressing any linear operator as the direct sum of operators with a cyclic vector, with no assumption about the scalar field. If U is a linear operator with a cyclic vector, there is a basis \{\alpha_1,\dots,\alpha_n\} with

U\alpha_j=\alpha_{j+1},\quad j=1,\dots,n-1 \\ U\alpha_n=-c_0\alpha_1-c_1\alpha_2-\cdots -c_{n-1}\alpha_n.

The action of U on this basis is then to shift each \alpha_j to the next vector \alpha_{j+1}, except that U\alpha_n is some prescribed linear combination of the vectors in the basis. Since the general linear operator T is the direct sum of a finite number of such operators U, we obtained an explicit and reasonably elementary description of the action of T.
We next applied the cyclic decomposition theorem to nilpotent operators. For the case of an algebraically closed scalar field, we combined this with the primary decomposition theorem to obtain the Jordan form. The Jordan form gives a basis \{\alpha_1,\dots,\alpha_n\} for the space V such that, for each j, either T\alpha_j is a scalar multiple of \alpha_j or T\alpha_j=c\alpha_j+\alpha_{j+1}. Such a basis certainly describes the action of T in an explicit and elementary manner.
之后开始讲semi-simple operator， 其实质上是一个和diagonalizable几乎等价的概念，最终证明的结论是一个类似于上一章Theorem 13的结论，即每一个operator都可分解为semi-simple和nilpotent的算子之和。

Exercises

1.If N is a nilpotent linear operator on V, show that for any polynomial f the semi-simple part of f(N) is a scalar multiple of the identity operator (F a subfield of C).
Solution: Notice that N^r=0 for some r, thus if f is any polynomial, only the first r+1 items of f(N) is possibly different from zero, i.e., if we assume f=\sum_{i=0}a_ix^i, then we have

\displaystyle{f(N)=a_0I+a_1N+\cdots+a_{r-1}N^{r-1}}

notice that the item a_1N+\cdots+a_{r-1}N^{r-1} is nilpotent, and a_0I is semi-simple, since I commute with N, a_0I commutes with a_1N+\cdots+a_{r-1}N^{r-1}, thus a_0I is the semi-simple part of f(N).

2.Let F be a subfield of the complex numbers, V a finite-dimensional vector space over F, and T a semi-simple linear operator on V. If f is any polynomial over F, prove that f(T) is semi-simple.
Solution: We let \dim V=n. Since T is semi-simple, by Theorem 12, the matrix A=[T]_{\mathfrak B} is similar to a diagonal matrix D=\text{diag}(d_1,\dots,d_n), or D=P^{-1}AP for some invertible P. We have f(A)=[f(T)]_{\mathfrak B}, and P^{-1}f(A)P=f(P^{-1}AP)=f(D), notice that f(D)=\text{diag}(f(d_1),\dots,f(d_n)) is diagonal, f(T) is semi-simple again by Theorem 12.

3.Let T be a linear operator on a finite-dimensional space over a subfield of C. Prove that T is semi-simple if and only if the following is true: If f is a polynomial and f(T) is nilpotent, then f(T)=0.
Solution: If T is semi-simple, by Exercise 2 we know that f(T) is semi-simple, or [f(T)]_{\mathfrak B} is diagonal for some basis \mathfrak B of the finite-dimensional space. Since f(T) is nilpotent, all items on the diagonal of [f(T)]_{\mathfrak B} must be 0, which means [f(T)]_{\mathfrak B}=0, or f(T)=0.
Conversely, by Theorem 13 we shall have T=S+N, where S is semi-simple and N nilpotent, and SN=NS. If we use the notation in the proof of Theorem 13, let p_1^{r_1}\cdots p_k^{r_k} be the minimal polynomial for T, and f=p_1\cdots p_k, we shall have f(T) being nilpotent, also by the construction method we can see that N=\sum_{j=0}^{r-1}g_j(T)f(T)^j. From the condition given we know that f(T) is nilpotent means f(T)=0, this gives N=0 and so T=S is semi-simple.

Linear Algebra (2ed) Hoffman & Kunze 7.4

christangdt留下评论

这一节名义上是说计算，但是通篇都是在证明一种“算法”，而且还只是算法的原理，挺复杂。本节的目的是将一个多项式矩阵通过行列基本变换，得到一个等价的对角矩阵，其中该矩阵的invariant factors在对角线的下方出现。整个过程非常说理化，好在能看懂，最后的一些结论却是很有意思，比如n阶方阵A的最小多项式其实是特征多项式除以xI-A的所有n-1阶子矩阵行列式的最大公约数。

Exercises

1.True or false? Every matrix in F[x]^{n\times n} is row-equivalent to an upper-triangular matrix.
Solution: True. Since every matrix is row-equivalent to a matrix in normal form, which is upper-triangular.

2.Let T be a linear operator on a finite-dimensional vector space and let A be the matrix of T in some ordered basis. Then T has a cyclic vector if and only if the determinants of the (n-1)\times (n-1) submatrices of xI-A are relatively prime.
Solution: Let d be the g.c.d. of the determinants of the (n-1)\times (n-1) submatrices of xI-A, then we know that if p is the minimal polynomial for A and f the characteristic polynomial for A, then pd=f. Now by the corollary of the Cyclic Decomposition Theorem, T has a cyclic vector if and only if p=f, or if and only if d=1.

3.Let A be an n\times n matrix with entries in the field F and let f_1,\dots,f_n be the diagonal entries of the normal form of xI-A. For which matrices A is f_1\neq 1?
Solution: In order to make f_1\neq 1, we need \delta_1(xI-A)\neq 1, thus all entries of xI-A can be divided by some polynomial of degree at least 1, it follows that A shall be a scalar multiple of the identity matrix.

4.Construct a linear operator T with minimal polynomial x^2(x-1)^2 and characteristic polynomial x^3(x-1)^4. Describe the primary decomposition of the vector space under T and find the projections on the primary components. Find a basis in which the matrix of T is in Jordan form. Also find an explicit direct sum decomposition of the space into T-cyclic subspaces as in Theorem 3 and give the invariant factors.
Solution: We can construct T with the matrix of T under the standard basis \{\epsilon_1,\dots,\epsilon_7\} of R^7 is

\displaystyle{A=\begin{bmatrix}1\\1&1\\&&1\\&&1&1\\&&&&0\\&&&&1&0\\&&&&&&0\end{bmatrix}}

It is easy to verify the minimal polynomial is x^2(x-1)^2 and the characteristic polynomial is x^3(x-1)^4.
Under the primary decomposition, the vector spaces under T shall be W_1=\text{null }T^2 and W_2=\text{null }(T-I)^2, notice that

\displaystyle{A^2=\begin{bmatrix}1\\2&1\\&&1\\&&2&1\\&&&&0\\&&&&0&0\\&&&&&&0\end{bmatrix},(A-I)^2=\begin{bmatrix}0\\0&0\\&&0\\&&0&0\\&&&&1\\&&&&-2&1\\&&&&&&1\end{bmatrix}}

Thus W_1 consists of all vectors of the form (0,0,0,0,a,b,c), and W_2 consists of all vectors of the form (a,b,c,d,0,0,0). The projection for W_1 shall be

E_1(x_1,x_2,x_3,x_4,x_5,x_6,x_7)=(0,0,0,0,x_5,x_6,x_7) \\E_2(x_1,x_2,x_3,x_4,x_5,x_6,x_7)=(x_1,x_2,x_3,x_4,0,0,0)

Since A is already in Jordan form, the standard basis is what we need.
Since T\epsilon_1=\epsilon_1+\epsilon_2, we see Z(\epsilon_1;T) generate the subspace (a,b,0,0,0,0,0), by a similar argument, we can get a T-cyclic decomposition of R^7 to be

\displaystyle{R^7=Z(\epsilon_1;T){\oplus}Z(\epsilon_3;T){\oplus}Z(\epsilon_5;T){\oplus}Z(\epsilon_7;T)}

the invariant factors are (x-1)^2, (x-1)^2, x^2 and x.

5.Let T be the linear operator on R^3 which is represented in the standard basis by the matrix

\displaystyle{A=\begin{bmatrix}1&1&1&1&1&1&1&1\\0&0&0&0&0&0&0&1\\0&0&0&0&0&0&0&-1\\0&1&1&0&0&0&0&1\\0&0&0&1&1&0&0&0\\0&1&1&1&1&1&0&1\\0&-1&-1&-1&-1&0&1&-1\\0&0&0&0&0&0&0&0\end{bmatrix}.}

( a ) Find the characteristic polynomial and the invariant factors.
( b ) Find the primary decomposition of R^8 under T and the projections on the primary components. Find cyclic decompositions of each primary component as in Theorem 3.
( c ) Find the Jordan form of A.
( d ) Find a direct-sum decomposition of R^8 into T-cyclic subsapces as in Theorem 3.
Solution:
( a ) A computation shows the characteristic polynomial is

\displaystyle{\begin{aligned}f=\det(xI-A)&=\begin{vmatrix}x-1&-1&-1&-1&-1&-1&-1&-1\\0&x&0&0&0&0&0&-1\\0&0&x&0&0&0&0&1\\0&-1&-1&x&0&0&0&-1\\0&0&0&-1&x-1&0&0&0\\0&-1&-1&-1&-1&x-1&0&-1\\0&1&1&1&1&0&x-1&1\\0&0&0&0&0&0&0&x\end{vmatrix}\\&=(x-1)xx^3(x-1)^3=x^4(x-1)^4\end{aligned}}

also we can verify that A^2(A-I)^2=0, so the minimal polynomial for A is p=x^2(x-1)^2.
To compute the invariant factors for A, we proceed elementary operations on xI-A:

\begin{aligned}xI-A&=\begin{bmatrix}x-1&-1&-1&-1&-1&-1&-1&-1\\0&x&0&0&0&0&0&-1\\0&0&x&0&0&0&0&1\\0&-1&-1&x&0&0&0&-1\\0&0&0&-1&x-1&0&0&0\\0&-1&-1&-1&-1&x-1&0&-1\\0&1&1&1&1&0&x-1&1\\0&0&0&0&0&0&0&x\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&0&0&0&0&0&0\\0&x&0&0&0&0&0&-1\\0&0&x&0&0&0&0&1\\0&-1&-1&x&0&0&0&-1\\0&0&0&-1&x-1&0&0&0\\0&-x&-x&-x&-x&(1-x)^2&1-x&-x\\0&1&1&1&1&0&x-1&1\\0&0&0&0&0&0&0&x\end{bmatrix}=\begin{bmatrix}1&0\\0&S_1\end{bmatrix}\end{aligned}

thus we can proceed with the 7\times 7 matrix S_1.

\displaystyle{\begin{aligned}S_1&=\begin{bmatrix}x&0&0&0&0&0&-1\\0&x&0&0&0&0&1\\-1&-1&x&0&0&0&-1\\0&0&-1&x-1&0&0&0\\-x&-x&-x&-x&(1-x)^2&1-x&-x\\1&1&1&1&0&x-1&1\\0&0&0&0&0&0&x\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&0&0&0&0&0\\0&x&0&0&0&0&1\\0&x&-x^2&0&0&0&x-1\\0&0&-1&x-1&0&0&0\\0&-2x&x^2-x&-x&(1-x)^2&1-x&-2x\\0&2&1-x&1&0&x-1&2\\0&0&0&0&0&0&x\end{bmatrix}=\begin{bmatrix}1&0\\0&S_2\end{bmatrix}\end{aligned}}

proceed with the 6\times 6 matrix S_2:

\displaystyle{\begin{aligned}S_2&=\begin{bmatrix}x&&&&&1\\x&-x^2&&&&x-1\\&-1&x-1&\-2x&x^2-x&-x&(1-x)^2&1-x&-2x\\2&1-x&1&&x-1&2\\&&&&&x\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}1&&&&&x\\x-1&-x^2&&&&x\\&-1&x-1&\-2x&x^2-x&-x&(1-x)^2&1-x&-2x\\2&1-x&1&&x-1&2\\x&&&&&\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}1&&&&&\\&-x^2&&&&2x-x^2\\&-1&x-1&\\&x^2-x&-x&(1-x)^2&1-x&2x^2-2x\\&1-x&1&&x-1&2-2x\\&&&&&-x^2\end{bmatrix}=\begin{bmatrix}1&0\\0&S_3\end{bmatrix}\end{aligned}}

proceed with the 5\times 5 matrix S_3:

\displaystyle{\begin{aligned}S_3&=\begin{bmatrix}-x^2&&&&2x-x^2\\-1&x-1&\\x^2-x&-x&(1-x)^2&1-x&2x^2-2x\\1-x&1&&x-1&2-2x\\&&&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}1&1-x&\\-x^2&&&&2x-x^2\\x^2-x&-x&(1-x)^2&1-x&2x^2-2x\\1-x&1&&x-1&2-2x\\&&&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}1&&\\&x^2(1-x)&&&2x-x^2\\&x^2(x-2)&(1-x)^2&1-x&2x^2-2x\\&2x-x^2&&x-1&2-2x\\&&&&-x^2\end{bmatrix}=\begin{bmatrix}1&0\\0&S_4\end{bmatrix}\end{aligned}}

proceed with the 4\times 4 matrix S_4:

\displaystyle{\begin{aligned}S_4&=\begin{bmatrix}x^2(1-x)&&&2x-x^2\\x^2(x-2)&(1-x)^2&1-x&2x^2-2x\\2x-x^2&&x-1&2-2x\\&&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}&&x^2(1-x)&2x-x^2\\1-x&(1-x)^2&x^2(x-2)&2x^2-2x\\x-1&&2x-x^2&2-2x\\&&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}x-1&&2x-x^2&2-2x\\&(1-x)^2&(x^2-x)(x-2)&2(x-1)^2\\&&x^2(1-x)&2x-x^2\\&&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}x-1&&1&\\&(1-x)^2&(x^2-x)(x-2)&2(x-1)^2\\&&x^2(1-x)&2x-x^2\\&&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}1&&x-1&\\(x^2-x)(x-2)&(1-x)^2&&2(x-1)^2\\x^2(1-x)&&&2x-x^2\\&&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}1&&&\\&(1-x)^2&-x(x-1)^2(x-2)&2(x-1)^2\\&&x^2(x-1)^2&2x-x^2\\&&&-x^2\end{bmatrix}=\begin{bmatrix}1&0\\0&S_5\end{bmatrix}\end{aligned}}

proceed with the 3\times 3 matrix S_5:

\displaystyle{\begin{aligned}S_5&=\begin{bmatrix}(1-x)^2&-x(x-1)^2(x-2)&2(x-1)^2\\&x^2(x-1)^2&2x-x^2\\&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}(1-x)^2&-x(x-1)^2(x-2)&-4x+2\\&x^2(x-1)^2&2x\\&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}(1-x)^2&x(x-1)^2(x+2)&2\\&x^2(x-1)^2&2x\\&&-x^2\end{bmatrix}\\&{\rightarrow}\begin{bmatrix}1&&\\&-x^2(x-1)^2(x+1)&-x(x-1)^2\\&\frac{1}{2}x^3(x-1)^2(x+2)&\frac{1}{2}x^2(x-1)^2\end{bmatrix}=\begin{bmatrix}1&0\\0&S_6\end{bmatrix}\end{aligned}}

The matrix S_6 has \delta_2(S_6)=x(x-1)^2, so the invariant factors are x(x-1)^2,x^2(x-1)^2.

( b ) We can calculate

\displaystyle{A^2=\begin{bmatrix}1&2&2&2&2&2&2&2\\0&\\0&\\0&\\&1&1&1&1&&&1\\0&2&2&2&2&1&&2\\&-2&-2&-2&-2&&1&-2\\0\end{bmatrix}}

Thus W_1=\text{null }A^2 shall consists of the vectors of the form

\displaystyle{(0,x_2,x_3,x_4,x_5,0,0,-x_2-x_3-x_4-x_5)}

Also since

\displaystyle{(A-I)^2=\begin{bmatrix}0\\&1&0&0&0&0&0&-2\\&&1&&&&&2\\&-2&-2&1&&&&-2\\&1&1&-1&&&&1\\0\\0\\&&&&&&&1\end{bmatrix}}

thus W_2=\text{null }(A-I)^2 shall consists of the vectors of the form

\displaystyle{(x_1,0,0,0,x_5,x_6,x_7,0)}

And we can verify R^8=W_1\oplus W_2, since for any (a,b,c,d,e,f,g,h)\in R^8 we have

\displaystyle{\begin{aligned}&\quad(a,b,c,d,e,f,g,h)\\&=(0,b,c,d,-b-c-d-h,0,0,h)+(a,0,0,0,e+b+c+d+h,f,g,0)\end{aligned}}

The projections on the primary components are

E_1(a,b,c,d,e,f,g,h)=(0,b,c,d,-b-c-d-h,0,0,h) \\E_2(a,b,c,d,e,f,g,h)=(a,0,0,0,e+b+c+d+h,f,g,0)

The cyclic decomposition of W_1 is:
Let \alpha_1=(0,1,0,0,0,0,0,-1), then A\alpha_1=(0,-1,1,0,0,0,0,0). Let \alpha_2=(0,0,0,1,0,0,0,-1), then A\alpha_2=(0,-1,1,-1,1,0,0,0). So W_1=Z(\alpha_1;A)\oplus Z(\alpha_2;A).
The cyclic decomposition of W_2 is: W_2=Z(\epsilon_5,A)\oplus Z(\epsilon_6,A)

( c ) The Jordan form of A is

\displaystyle{\begin{bmatrix}0\\1&0\\&&0\\&&1&0\\&&&&1\\&&&&1&1\\&&&&&&1\\&&&&&&1&1\end{bmatrix}}

( d ) We let \alpha=\alpha_1+\epsilon_5 and \beta=\alpha_2+\epsilon_6, then R^8=Z(\alpha;A)\oplus Z(\beta;A).