M216: Exercise sheet 7

    Warmup questions

  • 1. Let \(\lst {v}1n\) be a basis for a vector space and \(\phi \in L(V)\). Show that the following are equivalent:

    • (1) \(\phi (v_1)=0\) and \(\phi (v_i)=v_{i-1}\), for \(\bw 2in\).

    • (2) \(v_i=\phi ^{n-i}(v_n)\) and \(\phi ^n(v_n)=0\).

  • 2. Let \(\phi \in L(V)\) be a nilpotent linear operator on a finite-dimensional vector space \(V\) with Jordan normal form \(J_{n_1}\oplus \dots \oplus J_{n_k}\).

    Show that

    \begin{equation*} \#\set {i\st n_i= s}=2\dim \ker \phi ^s-\dim \ker \phi ^{s-1}-\dim \ker \phi ^{s+1}. \end{equation*}

    Homework

  • 3. Let \(\phi \in L(V)\) be a linear operator on a finite-dimensional complex vector space \(V\) with distinct eigenvalues \(\lst {\lambda }1k\).

    Show that \(\phi \) is diagonalisable if and only if \(m_{\phi }=\prod _{i=1}^k(x-\lambda _i)\).

  • 4. Let \(\phi =\phi _A\in L(\C ^3)\) where

    \begin{equation*} \begin{pmatrix*}[r] 0&1&-1\\-10&-2&5\\-6&2&1 \end {pmatrix*}. \end{equation*}

    Find the JNF and a Jordan basis for \(\phi \).

    (You have studied \(\phi \) before in question 3 of sheet 6.)

    Extra questions

  • 5. Let \(\phi \in L(V)\) be a nilpotent linear operator on a finite-dimensional vector space \(V\) with Jordan normal form \(J_{n_1}\oplus \dots \oplus J_{n_k}\).

    Show that \(m_{\phi }=x^s\) where \(s=\max \set {\lst {n}1k}\).

    Hint: use question 6 on sheet 5.

  • 6. Let \(\phi =\phi _A\in L(\C ^3)\) where \(A\) is given by

    \begin{equation*} \begin{pmatrix*} 0&0&0\\4&0&0\\0&0&5 \end {pmatrix*}. \end{equation*}

    Find the JNF and a Jordan basis for \(\phi \).

    (You have studied \(\phi \) before in question 4 of sheet 6.)

Please hand in at 4W level 1 by NOON on Friday 24th November

M216: Exercise sheet 7—Solutions

  • 1. Assume (1). Then \(v_{n-1}=\phi (v_n)\) and induction gives \(v_i=\phi ^{n-i}(v_n)\). In particular, \(v_1=\phi ^{n-1}(v_n)\) so that \(\phi (v_1)\) gives \(\phi ^n(v_n)=0\). This establishes (2).

    Assume (2). Then \(0=\phi ^n(v_n)=\phi (\phi ^{n-1}(v_n))=\phi (v_1)\). Moreover \(\phi (v_i)=\phi (\phi ^{n-i}(v_n))=\phi ^{n-(i-1)}(v_n)=v_{i-1}\), for \(\bw 2in\). Thus we have (1).

  • 2. From lectures, we know that, for \(s\geq 1\),

    \begin{equation*} \#\set {i\st n_i\geq s}=\dim \ker \phi ^s-\dim \ker \phi ^{s-1}. \end{equation*}

    Now

    \begin{align*} \#\set {i\st n_i=s}&=\#\set {i\st n_i\geq s}-\#\set {i\st n_i\geq s+1}\\ &=\dim \ker \phi ^s-\dim \ker \phi ^{s-1}-(\dim \ker \phi ^{s+1}-\dim \ker \phi ^{s})\\ &= 2\dim \ker \phi ^s-\dim \ker \phi ^{s-1}-\dim \ker \phi ^{s+1}. \end{align*}

  • 3. Let \(p=\prod _{i=1}^k(x-\lambda _i)\in \C [x]\).

    If \(\phi \) is diagonalisable then \(p(\phi )=0\) since \(\phi -\lambda _i\id _{V}=0\) on \(E_{\phi }(\lambda _i)\) and \(V\) is the direct sum of these eigenspaces.

    Conversely, if \(m_{\phi }=\prod _{i=1}^k(x-\lambda _i)\), a result from lectures tells us that all Jordan blocks in the Jordan normal form of \(\phi \) have size \(1\). Otherwise said, the JNF is diagonal and the Jordan basis is an eigenbasis. So \(\phi \) is diagonalisable.

  • 4. From question 3 of sheet 6, we know that \(m_{\phi }=(x-3)(x+2)^2\) so that the JNF of \(\phi \) must be \(J(3,1)\oplus J(-2,2)\). A Jordan basis of \(G_{\phi }(3)=E_{\phi }(3)\) is an arbitrary basis and one is given by \((0,1,1)\) as we found out in question 3 of sheet 6.

    For \(G_{\phi }(-2)\), we want \((\phi +2\id _{\C ^3})(v),v\) with \((\phi +2\id _{\C ^3})^{2}(v)=0\) so work backwards from an eigenvector \(w\) with eigenvalue \(-2\) and then solve \((A+2I_3)\mathbf {v}=\mathbf {w}\) to get \(v\). We know from sheet 6 that we can take \(w=(1,0,2)\) and then

    \begin{equation*} (A+2I_{3})\mathbf {v}= \begin{pmatrix*}[r] 2&1&-1\\-10&0&5\\-6&2&3 \end {pmatrix*}\mathbf {v}= \begin{pmatrix} 1\\0\\2 \end {pmatrix} \end{equation*}

    is clearly solved by \((0,1,0)\). Our Jordan basis is therefore \((0,1,1),(1,0,2),(0,1,0)\).

  • 5. We follow the hint which tells us that when \(A=\oplst {A}1k\), \(m_{A}\) is the smallest monic polynomial divided by each \(m_{A_i}\).

    In the case at hand, with \(A=J_{n_1}\oplus \dots \oplus J_{n_k}\), we know from sheet 6 that \(m_{J_{n_i}}=x^{n_i}\). Thus \(m_A\) is the monic polynomial of smallest degree divided by each \(x^{n_i}\) which is \(x^s\), for \(s=\max \set {\lst {n}1k}\).

  • 6. From sheet 6, we have that \(m_{\phi }=x^2(x-5)\) so that the JNF is \(J(0,2)\oplus J(5,1)\). A Jordan basis for \(G_{\phi }(5)\) is any non-zero eigenvector with eigenvalue \(5\). We know from sheet 6 that \((0,0,1)\) is such a vector.

    For \(G_{\phi }(0)\), a Jordan basis is \(\phi (v),v\) with \(\phi ^2(v)=0\). As usual, work backwards from \(w\in \ker \phi \): take \(w=(0,1,0)\) and solve

    \begin{equation*} A\mathbf {v}=\begin{pmatrix*} 0&0&0\\4&0&0\\0&0&5 \end {pmatrix*}\mathbf {v}= \begin{pmatrix} 0\\1\\0 \end {pmatrix} \end{equation*}

    to get \(v=(1/4,0,0)\), for example.

    Thus, a Jordan basis is given by \((0,1,0),(1/4,0,0),(0,0,1)\).