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Homework 3

3-Homework-AlgebraB.pdf

  1. For the following pairs \(A, B\), decide if there exists an invertible matrix \(P\) such that both \(P^{-1} A P\) and \(P^{-1} B P\) are diagonal. If the answer is positive, find such \(P\).

    1. \(A = \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\)

      • For A, compute \(\chi_{A}(x)=\det \begin{pmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{pmatrix}=x^{3}-2x=x\left(x^{2}-2\right)\), then \(\lambda=0,\sqrt2,-\sqrt2\)​

      \(E_{0}=\ker \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}=\text{span}\{(1,0,-1)\}\)

      \(E_{\sqrt{2}}=\ker \begin{pmatrix} \sqrt{2} & 1 & 0 \\ 1 & \sqrt{2} & 1 \\ 0 & 1 & \sqrt{2} \end{pmatrix}=\ker \begin{pmatrix} \sqrt{2} & 1 & 0 \\ 0 & 1 & \sqrt{2} \\ 0 & 0 & 0 \end{pmatrix}=\text{span}\{(\sqrt{2},-2,\sqrt{2})\}\)

      \(E_{-\sqrt{2}}=\ker \begin{pmatrix} -\sqrt{2} & 1 & 0 \\ 1 & -\sqrt{2} & 1 \\ 0 & 1 & -\sqrt{2} \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & -\sqrt{2} \end{pmatrix}=\text{span}\{(1,\sqrt{2},1)\}\)

      Thus take \(\mathcal{B}=\{(1,0,-1),(\sqrt2,-2,\sqrt2),(1,\sqrt2, 1)\}\), \(A\) is diagonalizable - For B, compute \(\chi_{B}(x)=\det \begin{pmatrix} x-1 & 0 & -1 \\ 0 & x+1 & 0 \\ -1 & 0 & x-1 \end{pmatrix}=\left(x-1\right)^{2}\left(x+1\right)-\left(x+1\right)=\left(x+1\right )\left(\left(x-1\right)^{2}-1\right)=0\), then \(\lambda=0,2,-1\)

      \(E_{0}=\ker \begin{pmatrix} -1 & 0 & -1 \\ 0 & 1 & 0 \\ -1 & 0 & -1 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}=\text{span}\{(1,0,-1)\}\)

      \(E_{2}=\ker \begin{pmatrix} 1 & 0 & -1 \\ 0 & 3 & 0 \\ -1 & 0 & 1 \end{pmatrix}=\ker \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}=\text{span}\{(1,0,1)\}\)

      \(E_{-1}=\ker \begin{pmatrix} -2 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & -2 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}=\text{span}\{(0,1,0)\}\)

      Thus take \(\mathcal{B}=\{(1,0,-1),(1,0,1),(0,1,0)\}\), \(B\) is diagonalizable - Then we need to check whether they are commutative.

      \(AB= \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix}= \begin{pmatrix} 0 & 1 & 0 \\ -2 & 0 & -2 \\ 0 & 1 & 0 \end{pmatrix}\)​

      \(BA= \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix}= \begin{pmatrix} 0 & -2 & 0 \\ 1 & 0 & 1 \\ 0 & -2 & 0 \end{pmatrix}\)

      Since \(AB\neq BA\), then they are not commutative, thus we cannot find such \(P\) 2. \(A = \begin{pmatrix} 3 & 2 & 2 \\ -2 & -1 & -2 \\ 2 & 2 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & -1 & -1 \\ 1 & 3 & 1 \\ -1 & -1 & 1 \end{pmatrix}\)

      • Firstly, we check commutativity

      \(AB= \begin{pmatrix} 3 & 2 & 2 \\ -2 & -1 & -2 \\ 2 & 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & -1 & -1 \\ 1 & 3 & 1 \\ -1 & -1 & 1 \end{pmatrix}= \begin{pmatrix} 3 & 1 & 1 \\ -1 & 1 & -1 \\ 1 & 1 & 3 \end{pmatrix}\)

      \(BA= \begin{pmatrix} 1 & -1 & -1 \\ 1 & 3 & 1 \\ -1 & -1 & 1 \end{pmatrix} \begin{pmatrix} 3 & 2 & 2 \\ -2 & -1 & -2 \\ 2 & 2 & 3 \end{pmatrix}= \begin{pmatrix} 3 & 1 & 1 \\ -1 & 1 & -1 \\ 1 & 1 & 3 \end{pmatrix}\)

      Since \(AB=BA\), then they are commutative - For A, compute \(\chi_{A}(x)=\det \begin{pmatrix} x-3 & -2 & -2 \\ 2 & x+1 & 2 \\ -2 & -2 & x-3 \end{pmatrix}=\left(x-3\right)^{2}\left(x+1\right)+16-4\left(x+1\right)+8\left(x- 3\right)=\left(x-1\right)^{2}\left(x-3\right)=0\), then \(\lambda=1,3\)

      \(E_{1}=\ker \begin{pmatrix} -2 & -2 & -2 \\ 2 & 2 & 2 \\ -2 & -2 & -2 \end{pmatrix}=\ker \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}=\text{span}\{(1,0,-1),\left(0,1,-1\right)\}\)

      \(E_{3}=\ker \begin{pmatrix} 0 & -2 & -2 \\ 2 & 4 & 2 \\ -2 & -2 & 0 \end{pmatrix}=\ker \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}=\text{span}\{\left(1,-1,1\right)\}\)​

      Thus take \(\mathcal{B}=\{(1,0,-1),(0,1,-1),(1,-1,1)\}\), \(A\) is diagonalizable - For B, compute \(\chi_{B}(x)=\det \begin{pmatrix} x-1 & 1 & 1 \\ -1 & x-3 & -1 \\ 1 & 1 & x-1 \end{pmatrix}=\left(x-1\right)^{2}\left(x-3\right)-2-\left(x-3\right)+2\left(x-1\right )=\left(x-1\right)\left(x-2\right)^{2}=0\), then \(\lambda=1,2\)

      \(E_{1}=\ker \begin{pmatrix} 0 & 1 & 1 \\ -1 & -2 & -1 \\ 1 & 1 & 0 \end{pmatrix}=\ker \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & -1 \end{pmatrix}=\text{span}\{\left(1,-1,1\right)\}\)

      \(E_{2}=\ker \begin{pmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 1 & 1 & 1 \end{pmatrix}=\ker \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}=\text{span}\{\left(1,0,-1\right),\left(1,-1,0\right)\}\)

      Thus take \(\mathcal{B}=\{(1,-1,1),(1,0,-1),(1,-1,0)\}\), \(B\) is diagonalizable

      Then by theorem: \(\exists\) a basis \(\mathcal{B}'\) of \(V\) such that \([A]_{\mathcal{B}^{\prime}},[B]_{\mathcal{B}^{\prime}}\) is diagonal

      Since \((1,0,-1)-(0,1,-1)=(1,-1,0)\), then the common basis is \(\{(1,-1,1),(1,0,-1),(1,-1,0)\}\)​

      Thus \(P= \begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix}\)​

  2. Let \(A, A' \in K^{5 \times 5}\) be the following matrices \(A = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}, \quad A' = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}\)

    1. Prove that both matrices are nilpotent.

      \(A^{2}= \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}\)

      \(A^{3}= \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}\)

      \(A^{4}= \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}\)

      \(A^{5}= \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}\)​

      Thus \(A^{5}=0\), thus \(A\)​ is nilpotent

      \(A^{\prime2}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{pmatrix}\)

      \(A^{\prime3}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{pmatrix}\)

      \(A^{\prime4}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{pmatrix}\)

      \(A^{\prime5}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}\)​

      Thus \(A^{\prime 5}=0\), then \(A'\) is nilpotent 2. Find bases \(B\) and \(B'\) for \(R_4[x]\) so that the matrix representation of the derivative operator with respect to \(B\) is \(A\), and with respect to \(B'\) is \(A'\).

      We know \(A=[f]_{\mathcal{B}}= \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}\), let \(\mathcal{B}=\{v_1,...,v_5\}\)

      Then \(f(v_1)=0,f(v_2)=v_1,...,f(v_4)=v_3,f(v_5)=v_4\) and since \(f\) is a derivative operator.

      Then \(v_{1}=1,v_{2}=x,v_{3}=\frac{x^{2}}{2},v_{4}=\frac{x^{3}}{6},v_{5}=\frac{x^{4}}{24}\), thus \(\mathcal{B}=\left\lbrace1,x,\frac{x^{2}}{2},\frac{x^{3}}{6},\frac{x^{4}}{24}\right \rbrace\)

      We know \(A^{\prime}=[f]_{\mathcal{B}^{\prime}}= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}\), let \(\mathcal{B'}=\{v_{1},...,v_{5}\}\)

      Then \(f(v_{1})=v_{2},f(v_{2})=v_{3},...,f(v_{4})=v_{5},f(v_{5})=0\) and since \(f\) is a derivative operator.

      Then \(v_{5}=1,v_{4}=x,v_{3}=\frac{x^{2}}{2},v_{2}=\frac{x^{3}}{6},v_{1}=\frac{x^{4}}{24}\), thus \(\mathcal{B}^{\prime}=\left\lbrace\frac{x^{4}}{24},\frac{x^{3}}{6},\frac{x^{2}}{2} ,x,1\right\rbrace\) 3. Show that \(A\) is similar to \(A'\).

      From above we know \(\mathcal{B}=\left\lbrace1,x,\frac{x^{2}}{2},\frac{x^{3}}{6},\frac{x^{4}}{24}\right \rbrace\) and \(\mathcal{B}^{\prime}=\left\lbrace\frac{x^{4}}{24},\frac{x^{3}}{6},\frac{x^{2}}{2} ,x,1\right\rbrace\)​

      Then there exists a change matrix \(C\left(B,B^{\prime}\right)= \begin{pmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{pmatrix}\) such that \(A=[f]_{\mathcal{B}}=C(B,B')^{-1}[f]_{\mathcal{B'}}C(B,B')\) where \([f]_\mathcal{B'}=A'\)​

      Thus \(A\) is similar to \(A'\)

  3. Consider \(R^4\) as a vector space over \(R\) and let \(W_{1}= \text{span}\{ e_{1}- e_{3}, e_{1}- e_{2}\}, W_{2}= \text{span}\{ e_{3}+ e _{4}\}, W_{3}= \text{span}\{ -e_{1}+ e_{2}- e_{3}\},\) where \(\mathcal{E} = \{ e_{1}, e_{2}, e_{3}, e_{4} \}\) is the standard basis for \(R^{4}\).

    1. Show that \(R^4 = W_1 \oplus W_2 \oplus W_3\).

      Consider \((0,0,0,0)=a_{1}(e_{1}-e_{3})+a_{2}(e_{1}-e_{2})+a_{3}(e_{3}+e_{4})+a_{4}(-e_{1}+ e_{2}-e_{3})\)​

      \(=\left(a_{1}+a_{2}-a_{4}\right)e_{1}+\left(a_{4}-a_{2}\right)e_{2}+\left(a_{3}-a _{1}-a_{4}\right)e_{3}+\left(a_{3}\right)e_{4}\)​

      Since \(e_1,e_2,e_3,e_4\neq 0\), then \(\begin{cases} a_{1}+a_{2}-a_{4}=0 \\ a_{4}-a_{2}=0 \\ a_{3}-a_{1}-a_{4}=0 \\ a_{3}=0 \end{cases}\Rightarrow \begin{cases} a_{1}=0 \\ a_{2}=0 \\ a_{3}=0 \\ a_{4}=0 \end{cases}\)

      Thus we have the trivial way to express \((0,0,0,0)\). Then \(R^4 = W_1 \oplus W_2 \oplus W_3\). 2. Find projections \(F_i\) on \(T_{i}= \bigoplus\limits_{1 \leq j \leq 3\atop j \neq i}W_{j}\) along \(W_i\) for \(i = 1, 2, 3\)​

      We know \(T_{1}=W_{1}\oplus W_{2}\oplus W_{3},\) where \(W_1\) is kernel of \(F_1\)​

      Proposition: \(V\) is a vector space over \(\mathbb{K}\) and \(S, T \leq V\) such that \(V = S \oplus T\)​ . Then there exists a unique projection \(f: V \to V\) such that \(S = \text{Ker} f\) and \(T = \text{Im} f\)​ .

      Then exists \(F_1:T_1\to T_1\) such that \(W_{1}=\ker F_{1},W_{2}\oplus W_{3}=\text{Im}F_{1}\) and basis of \(T_1\) is \(\mathcal{B}_{1}=\{e_{1}-e_{3},e_{1}-e_{2},e_{3}+e_{4},-e_{1}+e_{2}-e_{3}\}\), that is \(\mathcal{B}_{1}=\{\left(1,0,-1,0\right),\left(1,-1,0,0\right),\left(0,0,1,1\right ),\left(-1,1,-1,0\right)\}\)​

      Then \(F_{1}(e_{1}-e_{3})=F_{1}(1,0,-1,0)=(0,0,0,0),F_{1}(e_{1}-e_{2})=F_{1}(1,-1,0,0)= (0,0,0,0)\)​

      \(F_{1}(e_{3}+e_{4})=F_{1}(0,0,1,1)=(0,0,1,1)\), \(F_{1}\left(-e_{1}+e_{2}-e_{3})=F_{1}\left(-1,1,-1,0\right)=(-1,1,-1,0\right)\)

      Thus \([F_{1}]_{\mathcal{B_1}}= \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\), then we know \([F_{1}]_{e}=C(B_{1},E)[F_{1}]_{B_1}C(E,B_{1})^{-1}\) and \(C(B,E)= \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\)​

      Then \([F_{1}]_{e}= \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 & 0 \\ -1 & -2 & -1 & 1 \\ 0 & 0 & 0 & 1 \\ -1 & -1 & -1 & 1 \end{pmatrix}= \begin{pmatrix} 1 & 1 & 1 & -1 \\ -1 & -1 & -1 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\)

      Thus \(F_{1}(e_{1})=e_{1}-e_{2}+e_{3},F_{1}(e_{2})=e_{1}-e_{2}+e_{3},F_{1}(e_{3})=e_{1} -e_{2}+e_{3},F_{1}(e_{4})=-e_{1}+e_{2}+e_{4}\)​

      Thus \(F_{1}(x,y,z,w)=(x+y+z-w,-x-y-z+w,x+y+z,w)\)​

      Then exists \(F_{2}:T_{2}\to T_{2}\) such that \(W_{2}=\ker F_{2},W_{1}\oplus W_{3}=\text{Im}F_{2}\) and basis of \(T_{2}\) is \(\mathcal{B}_{2}=\{e_{1}-e_{3},e_{1}-e_{2},e_{3}+e_{4},-e_{1}+e_{2}-e_{3}\}\), that is \(\mathcal{B}_{2}=\{\left(1,0,-1,0\right),\left(1,-1,0,0\right),\left(0,0,1,1\right ),\left(-1,1,-1,0\right)\}\)

      Then \(F_{2}(e_{1}-e_{3})=F_{2}(1,0,-1,0)=(1,0,-1,0),F_{2}(e_{1}-e_{2})=F_{2}(1,-1,0,0) =(1,-1,0,0)\)

      \(F_{2}(e_{3}+e_{4})=F_{2}(0,0,1,1)=(0,0,0,0)\), \(F_{2}\left(-e_{1}+e_{2}-e_{3})=F_{2}\left(-1,1,-1,0\right)=(-1,1,-1,0\right)\)

      Thus \([F_{2}]_{\mathcal{B_2}}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\), then we know \([F_{2}]_{e}=C(B_{2},E)[F_{2}]_{B_2}C(E,B_{2})^{-1}\) and \(C(B_{2},E)= \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\)

      Then \([F_{2}]_{e}= \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 & 0 \\ -1 & -2 & -1 & 1 \\ 0 & 0 & 0 & 1 \\ -1 & -1 & -1 & 1 \end{pmatrix}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}\)

      Thus \(F_{2}(e_{1})=e_{1},F_{2}(e_{2})=e_{2},F_{2}(e_{3})=e_{3},F_{2}(e_{4})=-e_{3}\)

      Thus \(F_{2}(x,y,z,w)=(x,y,z-w,0)\)

      Then exists \(F_{3}:T_{3}\to T_{3}\) such that \(W_{3}=\ker F_{3},W_{1}\oplus W_{2}=\text{Im}F_{3}\) and basis of \(T_{3}\) is \(\mathcal{B}_{3}=\{e_{1}-e_{3},e_{1}-e_{2},e_{3}+e_{4},-e_{1}+e_{2}-e_{3}\}\), that is \(\mathcal{B}_{3}=\{\left(1,0,-1,0\right),\left(1,-1,0,0\right),\left(0,0,1,1\right ),\left(-1,1,-1,0\right)\}\)

      Then \(F_{3}(e_{1}-e_{3})=F_{3}(1,0,-1,0)=(1,0,-1,0),F_{3}(e_{1}-e_{2})=F_{3}(1,-1,0,0) =(1,-1,0,0)\)

      \(F_{3}(e_{3}+e_{4})=F_{3}(0,0,1,1)=(0,0,1,1)\), \(F_{3}\left(-e_{1}+e_{2}-e_{3})=F_{3}\left(-1,1,-1,0\right)=(0,0,0,0\right)\)

      Thus \([F_{3}]_{\mathcal{B_3}}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\), then we know \([F_{3}]_{e}=C(B_{3},E)[F_{3}]_{B_3}C(E,B_{3})^{-1}\) and \(C(B,E)= \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\)

      Then \([F_{3}]_{e}= \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 & 0 \\ -1 & -2 & -1 & 1 \\ 0 & 0 & 0 & 1 \\ -1 & -1 & -1 & 1 \end{pmatrix}= \begin{pmatrix} 0 & -1 & -1 & 1 \\ 1 & 2 & 1 & -1 \\ -1 & -1 & 0 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}\)

      Thus \(F_{3}(e_{1})=e_{2}-e_{3},F_{3}(e_{2})=-e_{1}+2e_{2}-e_{3},F_{3}(e_{3})=-e_{1}+e_{2} ,F_{3}(e_{4})=e_{1}-e_{2}+e_{3}+e_{4}\)

      Thus \(F_{3}(x,y,z,w)=(-y-z+w,x+2y+z-w,-x-y+w,w)\)

      Thus \(F_{1}(x,y,z,w)=(x+y+z-w,-x-y-z+w,x+y+z,w)\)

      \(F_{2}(x,y,z,w)=(x,y,z-w,0)\)

      \(F_{3}(x,y,z,w)=(-y-z+w,x+2y+z-w,-x-y+w,w)\)

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