MA22020: Exercise sheet 5

1. Let \(\lambda \in \F \) and define \(J(\lambda ,n)\in M_n(\F )\) by
\(\seteqnumber{0}{}{0}\)
\begin{equation*} J(\lambda ,n)= \begin{pmatrix} \lambda &1&0&\dots &0\\ &\ddots &\ddots &\ddots &\vdots \\ &&\ddots &\ddots &0\\ &&&\ddots &1\\ 0&&&&\lambda \end {pmatrix}. \end{equation*}

Set \(J_n:=J(0,n)\).

Prove:
- (a) \(\ker J_n^k=\Span {\lst {e}1k}\).
- (b) \(\im J_n^k=\Span {\lst {e}1{n-k}}\).
- (c) \(m_{J(\lambda ,n)}=\pm \Delta _{J(\lambda ,n)}=(x-\lambda )^n\).
- (d) \(\lambda \) is the only eigenvalue of \(J(\lambda ,n)\) and \(E_{J(\lambda ,n)}(\lambda )=\Span {e_1}\), \(G_{J(\lambda ,n)}(\lambda )=\F ^{n}\).
2. 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 \(\phi ^n(v_n)=0\).
3. 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
\(\seteqnumber{0}{}{0}\)
\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*}
4. Let \(\phi \in L(V)\) be a nilpotent linear operator on a finite-dimensional vector space \(V\) with Jordan normal form \(J_{n_1}\oplus \dots \oplus J_{n_k}\).

Use question 3 on sheet 4 to show that \(m_{\phi }=x^s\) where \(s=\max \set {\lst {n}1k}\).
5. 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 \(m_{\phi }=\prod _{i=1}^k(x-\lambda _i)\).
6. Let \(\phi =\phi _A\in L(\C ^3)\) where \(A\) is given by
\(\seteqnumber{0}{}{0}\)
\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 4.)
7. Let \(\phi =\phi _A\in L(\C ^3)\) where
\(\seteqnumber{0}{}{0}\)
\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 6 of sheet 4.)

MA22020: Exercise sheet 5—Solutions

1. Note that \(\phi _{J_n}(x)=(x_2,\dots ,x_n,0)\) so that \(\phi _{J_n}^k(x)=(x_{k+1},\dots ,x_{n},0,\dots ,0)\), \(k<n\) and \(\phi _{J_n}^{n}=0\).
- (a) It is clear from the above that \(\ker J_n^k=\set {x\in \F ^{n}\st x_{k+1}=\dots =x_n=0}=\Span {\lst {e}1k}\).
- (b) Similarly, \(\im J_n^k=\set {y\in \F ^n\st y_{n-k+1}=\dots =y_n=0}=\Span {\lst {e}1{n-k}}\).
- (c) \(J(\lambda ,n)\) is upper triangular so that \(\Delta _{J(\lambda ,n)}=(\lambda -x)^n\). Therefore \(m_{J(\lambda ,n)}=(x-\lambda )^{s}\), for some \(s\leq n\). However \((J(\lambda ,n)-\lambda I_n)^{k}=J_n^k\neq 0\), for \(k<n\), so that \(m_{J(\lambda ,n)}=(x-\lambda )^{n}\).
- (d) Finally, it is clear that \(\lambda \) is the only eigenvalue and the eigenspace is \(\ker (J(\lambda ,n)-\lambda I_{n})=\ker J_n=\Span {e_1}\) by part (a). Similarly, \(G_{J(\lambda ,n)}(\lambda )=\ker J_n^n=\F ^n\).
2. 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).
3. From lectures, we know that, for \(s\geq 1\),
\(\seteqnumber{0}{}{0}\)
\begin{equation*} \#\set {i\st n_i\geq s}=\dim \ker \phi ^s-\dim \ker \phi ^{s-1}. \end{equation*}

Now
\(\seteqnumber{0}{}{0}\)
\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*}
4. 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 question 1 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 \(s=\max \set {\lst {n}1k}\).
5. 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.
6. From sheet 4, 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 4 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
\(\seteqnumber{0}{}{0}\)
\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 \((0,1,0),(1/4,0,0),(0,0,1)\).
7. From question 6 of sheet 4, 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 4.

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 4 that we can take \(w=(1,0,2)\) and then
\(\seteqnumber{0}{}{0}\)
\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 \((0,1,1),(1,0,2),(0,1,0)\).