Homework 5
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Let \(f\) be a linear operator on \(K^2\). Prove that any non-zero vector which is not a characteristic vector for \(f\) is a cyclic vector for \(f\). Hence, prove that either \(f\) has a cyclic vector or \(f\) is a scalar multiple of the identity operator.
Proof
Suppose \(f\neq \lambda Id\) and \(B=\{e_{1},e_{2}\}\) be basis of \(K^2\). NTP: \(f\) has a cyclic vector
By theorem, it's equivalent to prove \(m_f(x)=\chi_f(x)\)
Consider \(\chi_{f}(x)=\det(xId-[f]_{B})\) and \([f]_{B}= \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\) where \(f\neq \lambda Id\)
Then \(\chi_{f}(x)=(x-a_{11})(x-a_{22})+a_{12}a_{21}=x^{2}-(a_{11}+a_{22})x+a_{11}a_{22} +a_{12}a_{21}\)
If \(\chi_f(x)\) is irreducible, then by theorem \(m_f(x)=\chi_f(x)\)
If \(\chi_f(x)\) is reducible, then \(\chi_f(x)=(x-c)(x-d)\).
If \(c\neq d\), then we have two eigenvalues and two corresponding eigenvectors. Then \(m_f(x)=\chi_f(x)\)
If \(c=d\), then \(\chi_f(x)=(x-c)^2\). NTP: \(m_{f}(x)=(x-c)^{2}\)
Suppose \(m_f(x)=(x-c)\), then \(m_f(f)=0=f-cId\), then \(f=cId\) contradicts to \(f\neq \lambda Id\)
Thus \(m_f(x)=(x-c)^2\), then \(m_f(x)=\chi_f(x)\)
Thus in all cases, \(m_f(x)=\chi_f(x)\), then \(f\) has a cyclic vector.
Then any non-zero vector which is not a characteristic vector for \(f\) is a cyclic vector for \(f\).
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Define the linear operator \(f\) on \(\mathbb{R}^7\) by the matrix \(\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & -1 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ -1 & 0 & 0 & -1 & 0 & 1 & 0 \end{pmatrix}\)
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Show that \(f\) does not admit a cyclic vector.
Let's compute \(\chi_{f}(x)=\det \begin{pmatrix} x & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & x & 0 & 0 & 0 & 0 & 0 \\ -2 & -1 & x & -1 & 0 & 1 & 0 \\ -1 & 0 & 0 & x-1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & x & 0 & 1 \\ -1 & 0 & 0 & -1 & 0 & x+1 & 0 \\ 1 & 0 & 0 & 1 & 0 & -1 & x \end{pmatrix}=x^{7}\)
Then \(E_{0}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & -1 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ -1 & 0 & 0 & -1 & 0 & 1 & 0 \end{pmatrix}=\ker \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}=\text{span}\{\left(0,0,1,0,0,0,0\right),\left(0,0,0,1,0,1,-1\right) ,\left(0,0,0,0,1,0,0\right)\}\)
Since \(\dim E_0<7\), then we need to compute \(E_{0}^{2}\)
\(E_{0}^{2}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & -1 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ -1 & 0 & 0 & -1 & 0 & 1 & 0 \end{pmatrix}^{2}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}=\text{span}\{\left(0,1,0,0,0,0,0),\left(0,0,1,0,0,0,0\right),\left( 0,0,0,1,0,0,0\right),(0,0,0,0,1,0,0),(0,0,0,0,0,1,0),(0,0,0,0,0,0,1)\right\}\)
Since \(\dim E_{0}^2<7\), then we need to compute \(E_{0}^{3}\)
\(E_{0}^{3}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & -1 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ -1 & 0 & 0 & -1 & 0 & 1 & 0 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}=\text{span}\{\left(1,0,0,0,0,0,0,0),(0,1,0,0,0,0,0),\left(0,0,1,0,0,0,0\right),\left( 0,0,0,1,0,0,0\right),(0,0,0,0,1,0,0),(0,0,0,0,0,1,0),(0,0,0,0,0,0,1)\right\}\)
Since \(\dim E_{0}^{3}=7\), then \(m_f(x)=x^3\)
Since \(m_f(x)\neq \chi_f(x)\), then by theorem: \(f\) does not admit a cyclic vector. 2. Show that \(W = \text{span}\{e_{3}, e_{4}+ e_{6}, e_{5}\}\) is not an \(f\)-admissible subspace.
Let \(w\in W:w=(0,0,x,y,z,y,0)\) where \(x,y,z\in \mathbb{R}\), then \(\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 1 & 0 & -1 & 0 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & -1 \\ 1 & 0 & 0 & 1 & 0 & -1 & 0 \\ -1 & 0 & 0 & -1 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ x \\ y \\ z \\ y \\ 0 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ -y \\ 0 \\ 0 \end{pmatrix}\in W\)
Thus \(W\) is \(f\)-invariant.
Then let \(p(x)=x\), then \(p(f)(e_2)=f(e_2)=e_3\in W\)
If \(\exists w\in W:p(f)(w)=e_{3}\), then \(w=a_{1}e_{3}+a_{2}(e_{4}+e_{6})+a_{3}e_{5}\), then \(p(f)(a_{1}e_{3}+a_{2}\left(e_{4}+e_{6})+a_{3}e_{5}\right)=a_{1}f\left(e_{3}\right )+a_{2}f\left(e_{4}+e_{6})+a_{3}f\left(e_{5}\right)\right.=-a_{2}e_{5}\neq e_3\)
Thus there doesn't exist such \(w\), thus \(W\) is not \(f\)-admissible
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Let \(f\) be a linear operator on \(\mathbb{R}^{9}\) whose matrix representation in the standard basis is \(\begin{pmatrix} 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -3 & 3 & 2 & 0 & 0 & 0 & -2 & 0 & 3 \\ -2 & 2 & 0 & 2 & 0 & 0 & -2 & 0 & 2 \\ 1 & -1 & 0 & -1 & 2 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}\)
Define \(W = \text{span}\{e_{6}, e_{1} + e_{9}\}\).
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Show that \(W\) is an \(f\)-admissible subspace of \(\mathbb{R}^9\).
Let \(w\in W:w=(x,0,0,0,0,y,0,0,x)\), then call the matrix \(A\), then \(Aw=\left(x,0,0,0,2y,0,0,0,0\right)\in W\)
Thus \(W\) is \(f\)-invariant
Then we are going to use this proposition \(W\) has an \(f\)-invariant complement\(\iff\)\(W\) is \(f\)-admissible
Assume \(W'=\text{span}\{e_{1},e_{2},e_{3},e_{4},e_{5},e_{7},e_{8}\}\), then \(v=(x_{1},...,x_{9})=\underbrace{(x_{9},0,0,0,0,x_{6},0,0,x_{9})}_{\in W}+\underbrace{(x_{1}-x_{9},x_{2},x_{3},x_{4} ,x_{5},0,x_{7},x_{8},0)}_{\in W'}\) \(\forall v\in V\)
Thus \(V=W+W'\). Also \(W\cap W'=\{0\}\). Then NTP: \(W'\) is \(f\)-invariant
Let \(w'\in W':w'=(x_1,x_2,x_3,x_4,x_5,0,x_7,x_8,0)\), then $Aw^{\prime}=\left(2x_{1},x_{1}+2x_{2},-3x_{1}+3x_{2}+2x_{3}-2x_{7},-2x_{1}+2x_{2} +2x_{4}-2x_{7},x_{1}-x_{2}-x_{4}+2x_{5}+x_{7},0,-x_{1}+x_{2},0,0\right)\in W' $
Thus \(W'\) is \(f\)-invariant, thus \(W\) is an \(f\)-admissible subspace of \(\mathbb{R}^9\). 2. Using \(W\), determine two distinct cyclic decompositions of \(\mathbb{R}^9\) associated with the linear operator \(f\).
\(\chi_{f}(x) = \det(xI - A) = \det \begin{pmatrix} x-2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ -1 & x-2 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 3 & -3 & x-2 & 0 & 0 & 0 & 2 & 0 & -3 \\ 2 & -2 & 0 & x-2 & 0 & 0 & 2 & 0 & -2 \\ -1 & 1 & 0 & 1 & x-2 & 0 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & x-2 & 0 & 0 & 0 \\ 1 & -2 & 0 & 0 & 0 & 0 & x & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & x & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & x-1 \end{pmatrix}\)
\(\chi_f(x) = x (x-2) (x-1) [x(x-2)^5] = x^2 (x-1) (x-2)^6\)
Thus the eigenvalues are \(0,1,2\)
\(E_{0}^{1}=\ker(A)=\ker \begin{pmatrix} 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ -3 & 3 & 2 & 0 & 0 & 0 & -2 & 0 & 3 \\ -2 & 2 & 0 & 2 & 0 & 0 & -2 & 0 & 2 \\ 1 & -1 & 0 & -1 & 2 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}=\text{span}\{e_{3}+e_{4}+e_{7},e_{8}\}\)
\(E_{1}^{1}=\ker(Id-A)=\ker \begin{pmatrix} -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ -1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 3 & -3 & -1 & 0 & 0 & 0 & 2 & 0 & -3 \\ 2 & -2 & 0 & -1 & 0 & 0 & 2 & 0 & -2 \\ -1 & 1 & 0 & 1 & -1 & 0 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 1 & -2 & 0 & 0 & 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}=\text{span}\{e_{1}+e_{9}\}\)
\(E_{2}^{1}=\ker(Id-A)=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 3 & -3 & 0 & 0 & 0 & 0 & 2 & 0 & -3 \\ 2 & -2 & 0 & 0 & 0 & 0 & 2 & 0 & -2 \\ -1 & 1 & 0 & 1 & 0 & 0 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 0 & 0 & 0 & 0 & 2 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}=\text{span}\{e_{3},e_{5},e_{6}\}\)
Since \(\dim(E_2)<6\), thus we compute \(E_2^2\)
\(E_{2}^{2}=\ker(Id-A)^{2}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 3 & -3 & 0 & 0 & 0 & 0 & 2 & 0 & -3 \\ 2 & -2 & 0 & 0 & 0 & 0 & 2 & 0 & -2 \\ -1 & 1 & 0 & 1 & 0 & 0 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 0 & 0 & 0 & 0 & 2 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}^{2}=\text{span}\{e_{2}+e_{7},e_{3},e_{4},e_{5},e_{6}\}\)
Since \(\dim(E_{2}^{2})<6\), thus we compute \(E_{2}^{3}\)
\(E_{2}^{3}=\ker(2Id-A)^{3}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 3 & -3 & 0 & 0 & 0 & 0 & 2 & 0 & -3 \\ 2 & -2 & 0 & 0 & 0 & 0 & 2 & 0 & -2 \\ -1 & 1 & 0 & 1 & 0 & 0 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 0 & 0 & 0 & 0 & 2 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}^{2}=\text{span}\{e_{1}+e_{2},e_{3},e_{4},e_{5},e_{6},-e_{1}+e_{7}\}\)
Since \(\dim(E_{2}^{3})=6\), therefore \(m_f(x) = x(x-1)(x-2)^3\)
Thus \(\ker(f)\supsetneq \{0\}\), \(\ker(f-Id)\supsetneq\{0\}\), \(\ker(f-2Id)^{3}\supsetneq\ker(f-2Id)^{2}\supsetneq\ker(f-2Id)\supsetneq\{0\}\)
Then \(e_{3}+e_{4}+e_{7}\overset{f}{\mapsto}0,e_{8}\overset{f}{\mapsto}0\)
\(e_{1}+e_{9}\overset{f-Id}{\longmapsto}0\)
\(e_{1}+e_{2}\overset{f-2Id}{\longmapsto}e_{2}+e_{7}\overset{f-2Id}{\longmapsto}e_{3} \overset{f-2Id}{\longmapsto}0\)
\(e_{4}\overset{f-2Id}{\longmapsto}-e_{5}\overset{f-2Id}{\longmapsto}0\)
\(e_{6}\overset{f-2Id}{\longmapsto}0\)
\(\mathbb{R}^{9}=W\oplus\langle e_{1}+e_{2}\rangle_{f}\oplus\langle e_{4}\rangle_{f} \oplus\langle e_{3}+e_{4}+e_{7}\rangle_{f}\oplus\langle e_{8}\rangle_{f}\)
\(\mathbb{R}^{9}=W\oplus\langle e_{1}+e_{2}+e_{4}\rangle_{f}\oplus\langle e_{4}\rangle _{f}\oplus\langle e_{3}+e_{4}+e_{7}\rangle_{f}\oplus\langle e_{8}\rangle_{f}\)
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