Homework 2

2-Homework-AlgebraB.pdf

1. Let \(A \in \mathbb{R}^{3\times3}\) be a matrix such that \(\chi_{A}(x) = (x^{2} - 2)(x+ 1)\). Consider \(B = -A^{4}+ 2A^{2}- I\).

   1. (10 p.) Is \(B\) diagonalizable?

      Since \(\chi_A(x)=0\), then \(x=\pm\sqrt{2},-1\), then we have three different eigenvectors and matrix is rank 3

      Thus \(A\) is diagonalizable, then \(\exists P\in\mathbb{R}^{3\times3}:D=P^{-1}AP\)

      Then \(B=-A^{4}+2A^{2}-I=-PD^{4}P^{-1}+2PD^{2}P^{-1}-PIP^{-1}=-P\left(D^{4}-2D^{2}+I\right )P^{-1}\)

      And we know \(D\) is diagonal matrix, thus \(D^{4}-2D^{2}+I=D'\) is also diagonal. Thus we find the diagonal matrix of \(B\) is \(D'\) and the change of basis matrix is \(P\)​

      Thus \(B\) is diagonalizable 2. (5 p.) What are the eigenvalues of \(B\)?

      Since \(B = -A^{4}+ 2A^{2}- I\), then \(Bv=-\left(Av\right)^{4}+2\left(Av\right)^{2}-\left(Iv\right)=\lambda^{\prime}v=- \lambda^{4}v+2\lambda^{2}v-v\).

      Thus we have \(\lambda^{\prime}=-\lambda^{4}+2\lambda^{2}-1\), then we know \(\lambda=\pm \sqrt2,-1\), thus \(\lambda'=-1,-1,0\)​ 3. (5 p.) What is \(\det(B)\)?

      First, we compute \(D^{4}-2D^{2}+I= \begin{pmatrix} 4 & \placeholder{} & \placeholder{} \\ \placeholder{} & 4 & \placeholder{} \\ \placeholder{} & \placeholder{} & 1 \end{pmatrix}- \begin{pmatrix} 4 & \placeholder{} & \placeholder{} \\ \placeholder{} & 4 & \placeholder{} \\ \placeholder{} & \placeholder{} & 2 \end{pmatrix}+ \begin{pmatrix} 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & 1 & \placeholder{} \\ \placeholder{} & \placeholder{} & 1 \end{pmatrix}= \begin{pmatrix} 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & 1 & \placeholder{} \\ \placeholder{} & \placeholder{} & 0 \end{pmatrix}\)

      Then \(\det(B)=\det\left\lbrack-P\left(D^{4}-2D^{2}+I\right)P^{-1}\right\rbrack=-\det\left (D^{4}-2D^{2}+I\right)=0\) 4. (5 p.) What is \(\operatorname{tr}(B)\)?

      We know \(B=-P\left(D^{4}-2D^{2}+I\right)P^{-1}\), then \(\text{tr}(B)=\text{tr}(-P\left(D^{4}-2D^{2}+I\right)P^{-1})=\text{tr}(-PP^{-1}\left (D^{4}-2D^{2}+I\right))=-\text{tr}(D^{4}-2D^{2}+I)\)​

      Then \(D^{4}-2D^{2}+I= \begin{pmatrix} 4 & \placeholder{} & \placeholder{} \\ \placeholder{} & 4 & \placeholder{} \\ \placeholder{} & \placeholder{} & 1 \end{pmatrix}- \begin{pmatrix} 4 & \placeholder{} & \placeholder{} \\ \placeholder{} & 4 & \placeholder{} \\ \placeholder{} & \placeholder{} & 2 \end{pmatrix}+ \begin{pmatrix} 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & 1 & \placeholder{} \\ \placeholder{} & \placeholder{} & 1 \end{pmatrix}= \begin{pmatrix} 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & 1 & \placeholder{} \\ \placeholder{} & \placeholder{} & 0 \end{pmatrix}\)

      Thus \(\text{tr}(B)=-2\)​

2. Consider the matrix \(A \in \mathbb{R}^{4\times4}\) given by \(A = \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}.\)

   1. (12 p.) For each integer \(d\) from 1 to 4, find \(d\)-dimensional subspaces that are invariant under the linear transformation \(f\) associated with \(A\).

      1. \(1\)-dimensional subspaces

         Take canonical basis, then \(A(e_{1})=2e_{1},A(e_{2})=e_{1}+2e_2,A(e_{3})=e_{2}+2e_3,A(e_{4})=2e_{4}\),

         Thus 1-dimensional subspaces is \(\lang e_1\rang\) or \(\lang e_4\rang\)​ 2. \(2\)-dimensional subspaces

         \(\lang e_1,e_2\rang\) or \(\lang e_1,e_4\rang\)​ 3. \(3\)-dimensional subspaces

         \(\lang e_1,e_2,e_3\rang\) or \(\lang e_1,e_2,e_4\rang\)​ 4. \(4\)-dimensional subspaces

         \(\R^4\)​ 2. (13 p.) Find an infinite collection of 1-dimensional \(f\)-invariant subspaces.

      We know \(A(e_{1})=2e_{1},A(e_{2})=e_{1}+2e_2,A(e_{3})=e_{2}+2e_3,A(e_{4})=2e_{4}\)​

      Thus we can consider 1-dimensional subspaces \(S=\langle ae_{1}+be_{2}+ce_{3}+de_{4}:a,b,c,d\in\mathbb{R}\rangle\)

      Then take \(v=ae_{1}+be_{2}+ce_{3}+de_{4}\), then \(f(v)=f(ae_{1}+be_{2}+ce_{3}+de_{4})=af\left(e_{1}\right)+bf\left(e_{2}\right)+cf \left(e_{3}\right)+df\left(e_{4}\right)\)​

      \(=2ae_{1}+be_{1}+2be_{2}+ce_{2}+2ce_{3}+2de_{4}=(2a+b)e_{1}+(2b+c)e_{2}+2ce_{3}+2 de_{4}\)​

      By def of invariant subspaces, this should be in \(S\)​

      Thus we have \(\begin{cases} 2a+b=\lambda a \\ 2b+c=\lambda b \\ 2c=\lambda c \\ 2d=\lambda d \end{cases}\Rightarrow \begin{pmatrix} 2-\lambda & 1 & 0 & 0 \\ 0 & 2-\lambda & 1 & 0 \\ 0 & 0 & 2-\lambda & 0 \\ 0 & 0 & 0 & 2-\lambda \end{pmatrix}\Rightarrow\lambda=2,b=c=0,a,d\in\mathbb{R}\)

      Thus \(S=\langle e_{1}+de_{4}:d\in\mathbb{R}\rangle\)

3. (20 p.) Let \(f\) be a linear operator on \(V\). Prove that every subspace of \(V\) is \(f\)-invariant if and only if \(f\) is a scalar multiple of the identity.

   \(\Rightarrow\))

   When \(\dim W=1\), then \(W=\langle w_{1}\rangle\), then take \(w=\lambda_{1}w_{1}\in W\), \(f(w)=f(\lambda_{1}w_{1})=u_{1}w_{1}\) since \(W\) is \(f-\text{invariant}\)​

   Since \(f\) is linear operator, then \(\lambda_{1}f\left(w_{1}\right)=u_{1}w_{1}\Rightarrow f\left(w_{1}\right)=\frac{u_{1}}{\lambda_{1}} w_{1}=kw_{1}\)

   Suppose \(f(v)=kv\) for \(v=w_1,...,w_{m-1}\) and \(W=\langle w_{1},...,w_{m-1}\rangle\) which means \(\dim W=1,2,...,m-1\)​

   When \(\dim W=m\), then \(W=\langle w_{1},w_{2},\ldots,w_{m}\rangle\)​, then take \(v=w_{1}+w_{2}+\cdots+w_{m}\in W\), \(f(v)=tv\) since \(W\) is \(f-\text{invariant}\)

   Then \(f(w_{1}+w_{2}+\cdots+w_{m})=t(w_{1}+w_{2}+\cdots+w_{m})\Rightarrow f(w_{1})+f(w_{2} )+\cdots+f\left(w_{m}\right)=tw_{1}+tw_{2}+\cdots+tw_{m}\) since \(f\) is linear

   Then \(kw_1+...+f(w_m)=tw_1+...+tw_m\) by hypothesis

   Again, \(f\) is invariant, then assume \(f(w_m)=pw_m\), then \(kw_{1}+...+pw_{m}=tw_{1}+...+tw_{m}\)​

   Thus \((k-t)w_1+(k-t)w_2+...+(p-t)w_m=0\), since \(w_1,...,w_m\) is basis, then \(k=p=t\)​

   Thus \(f(w_m)=kw_m\), then for any \(v\in W:v=\lambda_{1}w_{1}+....+\lambda_{m}w_{m}\), \(f(v)=f\left(\lambda_{1}w_{1}+....+\lambda_{m}w_{m})=\lambda_{1}f\left(w_{1}\right )+\cdots+\lambda_{m}f\left(w_{m}\right)=\lambda_{1}tw_{1}+\cdots+\lambda_{m}tw_{m} =t(\lambda_{1}w_{1}+...+\lambda_{m}w_{m}\right)\)​

   \(\Leftarrow\)) We know \(f\) is linear and \(f(v)=kv\)​

   Then take \(W=\langle w_{1},...,w_{m}\rangle\subseteq V\) for \(m\in \N\), then \(w=\lambda_{1}w_{1}+\ldots+\lambda_{m}w_{m}\)​

   Then \(f(w)=k(\lambda_{1}w_{1}+\ldots+\lambda_{m}w_{m})\in W\), thus every subspace of \(V\) is \(f\)-invariant

4. Indicate whether the following statements are true or false. Justify your response.

   1. (10 p.) If the minimal polynomial of the matrix \(A\) is a divisor of \(p(x) = (x^2 - 1)(x^2 + 2x)\), then \(A\) is diagonalizable.

      True

      Since \(p(x)=(x+1)(x-1)(x)(x+2)\) and \(m_{A}(x)\mid p(x)\), then \(m_{A}(x)\) is a product of linear factors, then by theorem, \(A\) is diagonalizable 2. (10 p.) If \(A \in \mathbb{R}^{3\times3}\) is diagonalizable and \((A - 3I)^{2}(A + 2I) = 0\), then the characteristic polynomial is \(\chi_A(x) = (x- 3)^2(x+ 2)\).

      False

      \((A - 3I)^{2}(A + 2I) (v)= 0\Rightarrow (Av - 3v)^{2}(Av + 2v) = 0\Rightarrow (\lambda v-3v)^2(\lambda v+2v)=0\)​

      Then \((\lambda -3)^2(\lambda +2)=0\), thus \(\lambda=3,-2\) are eigenvalues

      Thus \(3,-2\) are roots of \(\chi_{A}(x)\), then \(\chi_{A}(x)=(x-3)^{a}(x+2)^{b}\), but we cannot ensure the multiplicity 3. (10 p.) If \(u\) and \(v\) are eigenvectors of a linear operator \(f\) on a finite-dimensional vector space \(V\), then \(\operatorname{span}\{u+ v\}\) is an \(f\)-invariant subspace of \(V\).

      False

      Consider \(w=a\left(u+v\right)\in\text{span}\{u+v\}\), then \(f(w)=f(au+av)=af(u)+af(v)=a\lambda_{1}u+a\lambda_{2}v\), but we don't know if \(u,v\) belong to the same eigenvalue

      Thus \(\lambda_1\) may not equal to \(\lambda_2\), then \(f(w)\) may not in \(\text{span}\{u+v\}\), thus it may not be \(f\)-invariant subspace of \(V\)​