MA22020: Exercise sheet 3

    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 \(t\in \F \).

  • 4. Let \(\phi \in L(V)\) be an operator on a finite-dimensional vector space over \(\F \). Show that \(\phi \) is invertible if and only if \(m_{\phi }\) has non-zero constant term.

    Homework questions

  • 5. 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*}

  • 6. Compute the characteristic and minimum polynomials of

    \begin{equation*} A= \begin{pmatrix*}[r] 1&-5&-7\\1&4&2\\0&1&4 \end {pmatrix*}. \end{equation*}

Please hand in at 4W level 1 by NOON on Thursday 14th November

MA22020: Exercise sheet 3—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*}

    Similarly,

    \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*}

    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 \(p(t)=\bz \), for both \(t\in \F \).

  • 4. \(\phi \) 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 corollary to the Cayley–Hamilton theorem) zero is not a root of \(m_{\phi }\) if and only if \(m_{\phi }\) has non-zero constant term.

  • 5. 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 \(m_A=x^5-6x+3\).

  • 6. We compute the characteristic polynomial of \(A\) to be

    \begin{equation*} \Delta _A=-x^3+9x^2-27x+27=-(x-3)^3. \end{equation*}

    We learn from the Cayley–Hamilton theorem that \(m_A=(x-3)^k\), for some \(k\) with \(\bw {k}13\). Clearly \(k=1\) is out, since \(A\) is not diagonal, so we try \(k=2\):

    \begin{equation*} (A-3I)^2= \begin{pmatrix*}[r] -2&-5&-7\\1&1&2\\0&1&1 \end {pmatrix*} \begin{pmatrix*}[r] -2&-5&-7\\1&1&2\\0&1&1 \end {pmatrix*} = \begin{pmatrix*}[r] -1&-2&-3\\ -1&-2&-3\\ 1&2&3 \end {pmatrix*}, \end{equation*}

    which is non-zero. This means we must have \(m_A=(x-3)^3\).