M216: Exercise sheet 4

    Warmup questions

  • 1. Let \(p,q\in \R [x]\) be given by \(p=x^2-2x-3\), \(q=x^3-2x^2+2x-5\).

    Let \(A\in M_2(\R )\) and \(B\in M_3(\R )\) be given by

    \begin{equation*} A= \begin{pmatrix*}[r] 1&2\\2&1 \end {pmatrix*}\qquad B= \begin{pmatrix*}[r] 1&2&1\\-2&0&1\\2&1&1 \end {pmatrix*}. \end{equation*}

    Compute \(p(A),p(B),q(A),q(B)\).

  • 2. Compute the characteristic polynomials of \(A\) and \(B\), from question 1.

    What do you notice?

  • 3. Let \(\F =\Z _2\), the field of two elements and let \(p=x^2+x\in \F [x]\).

    Show that \(p(t)=0\), for all \(t\in \F \).

    Homework questions

  • 4. Compute the minimum polynomial of \(A\in M_5(\R )\) given by

    \begin{equation*} \begin{pmatrix*}[r] 0&0&0&0&-3\\1&0&0&0&6\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0 \end {pmatrix*}. \end{equation*}

  • 5. Let \(\phi \in L(V)\) be an operator on a finite-dimensional vector space over \(\F \) and let \(p=m_{\phi }\in \F [x]\).

    Let \(\lambda \) be a root of \(p\).

    • (a) Show there is \(q\in \F [x]\) with \(\deg q<\deg p\) such that

      \begin{equation*} p=(x-\lambda )q. \end{equation*}

    • (b) Prove that \(q(\phi )\) is non-zero.

    • (c) Deduce that \(\lambda \) is an eigenvalue of \(\phi \).

      This shows that the roots of \(p\) are exactly the eigenvalues of \(\phi \) without recourse to the Cayley–Hamilton theorem.

    • (d) Deduce that \(\phi \) is invertible if and only if \(p\) has non-zero constant term.

    Extra questions

  • 6. Let \(\phi \in L(V)\) have minimal polynomial \(p=4+5x+6x^{2}-7x^3-8x^{4}+x^{5}\), so that \(\phi \) is invertible by question 5(d).

    Compute the minimal polynomial of \(\phi ^{-1}\).

    Hint: Think about multiplying \(a_0\id _V+\dots +\phi ^n\) by \(\phi ^{-n}\).

Please hand in at 4W level 1 by NOON on Friday 3rd November

M216: Exercise sheet 4—Solutions

  • 1. We just compute:

    \begin{equation*} A^{2}= \begin{pmatrix*}[r] 5&4\\4&5 \end {pmatrix*},\qquad A^3= \begin{pmatrix*}[r] 13&14\\14&13 \end {pmatrix*} \end{equation*}

    so that

    \begin{align*} p(A)&=A^2-2A-3I_2=\begin{pmatrix*}[r] 5&4\\4&5 \end {pmatrix*}-2\begin{pmatrix*}[r] 1&2\\2&1 \end {pmatrix*}-3 \begin{pmatrix*}[r] 1&0\\0&1 \end {pmatrix*}= \begin{pmatrix*}[r] 0&0\\0&0 \end {pmatrix*}\\ q(A)&=A^{3}-2A^2+2A-5I_{3}=\begin{pmatrix*}[r] 13&14\\14&13 \end {pmatrix*}-2\begin{pmatrix*}[r] 5&4\\4&5 \end {pmatrix*}+2\begin{pmatrix*}[r] 1&2\\2&1 \end {pmatrix*}-5\begin{pmatrix*}[r] 1&0\\0&1 \end {pmatrix*}= \begin{pmatrix*}[r] 0&10\\10&0 \end {pmatrix*}. \end{align*} Similarly,

    \begin{align*} p(B)&= \begin{pmatrix*}[r] -6&-1&2\\4&-6&-3\\-2&3&-1 \end {pmatrix*},\\ q(B)&= \begin{pmatrix*}[r] 0&0&0\\ 0&0&0\\ 0&0&0 \end {pmatrix*}. \end{align*}

  • 2. Again, we just compute:

    \begin{equation*} \Delta _A= \begin{vmatrix} 1-x&2\\2&1-x \end {vmatrix}=(1-x)^2-4=x^2-2x-3. \end{equation*}


    \begin{equation*} \begin{split} \Delta _B= \begin{vmatrix} 1-x&2&1\\-2&-x&1\\2&1&1-x \end {vmatrix}=(1-x)\bigl (x(x-1)-1\bigr ) -2\bigl (2(x-1)-2\bigr ) +(-2+2x)\\ =(-x^3+2x^2-1)-4x+8+2x-2=-x^3+2x^2-2x+5. \end {split} \end{equation*}

    We notice that, with \(p,q\) as in question 1, \(p=\Delta _A\) and \(q=-\Delta _B\) and so, again from question 1,

    \begin{equation*} \Delta _A(A)=\Delta _B(B)=0. \end{equation*}

    As we shall soon see, this is the Cayley–Hamilton theorem in action.

  • 3. We recall that \(\Z _2=\set {\bz ,\bo }\) with addition and multiplication given by

    \begin{align*} \bz &=\bz +\bz =\bo +\bo &\bo &=\bz +\bo =\bo +\bz \\ \bz &=\bz \bz =\bz \bo =\bo \bz &\bo &=\bo \bo . \end{align*} We immediately conclude that \(\bo ^{2}+\bo =\bz =\bz ^2+\bz \) so that \(p(t)=\bo \), for both \(t\in \F \).

  • 4. Let us compute the first few powers of \(A\):

    \begin{gather*} A^2= \begin{pmatrix*}[r] 0&0&0&-3&0\\0&0&0&6&-3\\1&0&0&0&6\\0&1&0&0&0\\0&0&1&0&0 \end {pmatrix*}\quad A^3= \begin{pmatrix*}[r] 0&0&-3&0&0\\0&0&6&-3&0\\0&0&0&6&-3\\1&0&0&0&6\\0&1&0&0&0 \end {pmatrix*}\quad A^4= \begin{pmatrix*}[r] 0&-3&0&0&0\\0&6&-3&0&0\\0&0&6&-3&0\\0&0&0&6&-3\\1&0&0&0&6 \end {pmatrix*}\\ A^{5}= \begin{pmatrix*}[r] -3&0&0&0&-18\\6&-3&0&0&36\\0&6&-3&0&0\\0&0&6&-3&0\\0&0&0&6&-3 \end {pmatrix*} \end{gather*} Stare at the top row to see that there can be no monic polynomial \(p=a_0\plus {}{x^{k}}\) with \(k\leq 4\) with \(p(A)=0\): the \(-3\) on the top row of the leading term would give \(a_0\plus 0{a_{k-1}}0-3=0\). On the other hand, we readily see that \(A^{5}-6A+3I_{5}=0\) so that \(m_A=x^5-6x+3\).

  • 5.

    • (a) The remainder theorem says we can write \(p=(x-\lambda )q+r\) with \(\deg r<\deg (x-\lambda )=1\) so that \(r\) is degree zero and so constant. Evaluating at \(\lambda \) gives \(0=p(\lambda )=0q+r=r\) and we are done.

    • (b) \(q(\phi )\) cannot be zero unless \(q=0\) since \(\deg q<\deg p\) and \(p\) is the minimal polynomial of \(\phi \). But \(q\) cannot be zero since \(p\) is non-zero.

    • (c) Since \(q(\phi )\) is non-zero, there is \(v\in V\) such that \(q(\phi )v\neq 0\). Now

      \begin{equation*} 0=p(\phi )(v)=(\phi -\lambda \id _V)(q(\phi )(v)) \end{equation*}

      so that \(q(\phi )v\) is an eigenvector with eigenvalue \(\lambda \).

    • (d) \(\phi \) is invertible if and only if \(\phi \) is injective if and only if zero is not an eigenvalue if and only if (thanks to the previous part) zero is not a root of \(p\) if and only if \(p\) has non-zero constant term.

  • 6. If \(a_0\id _V+a_1\plus \phi {\phi ^n}=0\) then, multiplying by \(\phi ^{-n}\) gives \(a_0\phi ^{-n}+a_1\plus {\phi ^{n-1}}{a_n} \id _V=0\). In the case at hand, this means that

    \begin{equation*} 4\phi ^{-5}+5\phi ^{-4}+6\phi ^{-3}-7\phi ^{-2}-8\phi ^{-1}+\id _V=0. \end{equation*}

    If there was a non-zero polynomial \(q=\sum _{k=1}^{4}b_kx^k\) of lower degree with \(q(\phi ^{-1})=0\) gives

    \begin{equation*} b_4\plus {\id _V}{b_0}\phi ^4=0, \end{equation*}

    contradicting the minimality of \(p\). Thus, dividing by \(4\) to get a monic polynomial, the minimum polynomial of \(\phi ^{-1}\) is \(1/4-2x-7/4x^2+3/2x^{3}+5/4x^4+x^5\).

    More generally, the same argument says that if \(\sum _{k=0}^na_kx^k\) is the minimal polynomial of invertible \(\phi \) with degree \(n\) then \(1/a_{0}\sum _{k=0}^{n}a_{n-k}x^k\) is the minimal polynomial of \(\phi ^{-1}\).