Notes - Linear Algebra MT23, Jordan normal form
Flashcards
What does it mean for a linear transformation $T$ to be nilpotent?
For some $n > 0$, $T^n = \pmb 0$.
What can be said about the minimal polynomial of $T$ if $T$ is nilpotent?
Suppose $T$ is some nilpotent linear map. Then what can be said about the matrix of $T$ with respect to a certain fun basis?
$T$ is the zero matrix, but with $0$ or $1$ entries on the superdiagonal.
Define $J _ i(\lambda)$, a Jordan block of size $i$.
State the “Jordan normal form” theorem, describing a linear transformation in terms of its Jordan blocks.
- $T : V \to V$
- $m _ T(x) = (x-\lambda _ 1)^{q _ 1} \cdots (x-\lambda _ r)^{q _ r}$
Then there exists a basis $\mathcal B$ of $V$ such that $T$ is block diagonal and each diagonal block is of the form $J _ i(\lambda _ j)$ for some $1 \le i \le q _ j$ and $1 \le j \le r$.
Suppose
- $T : V \to V$
- $m _ T(x) = (x-\lambda _ 1)^{q _ 1} \cdots (x-\lambda _ r)^{q _ r}$
then what are the high-level ingredients to show that $T$ can be put into Jordan normal form?
- Primary decomposition theorem: $V$ can be split into $T$-invariant subspaces, so there is a block diagonal matrix consisting of $T$ restricted to these subspaces
- $T$ restricted to each of these subspaces has minimal polynomial $(x - \lambda _ i)^{q _ i}$, so we can consider it as the sum of $\lambda _ i I + N$ where $N$ is a nilpotent matrix
- Nilpotent matrices have bases in which they are zero apart from the $0$s and $1$s on the superdiagonal
- Breaking these up into sections with contiguous ones gives Jordan blocks
It is not true every matrix has a Jordan normal form. What do we need to assume about the matrix?
It is over $\mathbb C$.
How do the algebraic and geometric multiplicities, and the multiplicity in the minimal polynomial of an eigenvalue $\lambda _ i$ relate to its Jordan normal form?
- Algebraic multiplicity: the sum of the sizes of the Jordan blocks with $\lambda _ i$ on the diagonal.
- Geometric multiplitiy: the number of Jordan blocks with $\lambda _ i$ on the diagonal.
- Multiplicity as a root of minimal polynomial: size of largest Jordan block with $\lambda _ i$ on the diagonal.
How can you tell from a matrix’s Jordan normal form whether it is diagonalisable?
All Jordan blocks have size $1$.
How can you use the Jordan normal form to decompose any linear transformation $T$ into a diagonal and nilpotent part?
where $D$ is nilpotent, $N$ is diagonal.
Quickly prove using Jordan normal form that every square matrix $A$ is similar to its transpose.
Write $J _ 1, \cdots, J _ n$ for each of the Jordan blocks in the JNF $J$ of $A$. Then for each of the Jordan blocks
\[\begin{pmatrix} 0 & 0 & 0 & \cdots & 1 \\\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\\ 0 & 0 & 1 & \cdots & 0 \\\\ 0 & 1 & 0 & \cdots & 0 \\\\ 1 & 0 & 0 & \cdots & 0 \end{pmatrix} \begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 \\\\ 0 & \lambda & 1 & \cdots & 0 \\\\ 0 & 0 & \lambda & \ddots & \vdots \\\\ \vdots & \vdots & \ddots & \ddots & 1 \\\\ 0 & 0 & \cdots & 0 & \lambda \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & \cdots & 1 \\\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\\ 0 & 0 & 1 & \cdots & 0 \\\\ 0 & 1 & 0 & \cdots & 0 \\\\ 1 & 0 & 0 & \cdots & 0 \end{pmatrix} = \begin{pmatrix} \lambda & 0 & 0 & \cdots & 0 \\\\ 1 & \lambda & 0 & \cdots & 0 \\\\ 0 & 1 & \lambda & \ddots & \vdots \\\\ \vdots & \vdots & \ddots & \ddots & 0 \\\\ 0 & 0 & \cdots & 1 & \lambda \end{pmatrix}\]so by taking a block diagonal matrix $B$ to conjugate each of the blocks in this way, we see that
\[B^{-1} J B = J^\top\]and so it follows that $A$ is similar to its transpose.
What does it mean for $v$ to be a generalised eigenvector with value $\lambda$?
for some $k$.
Suppose you are given a matrix $A$. Quickly describe the procedure you would use to actually calculate the Jordan normal form of $A$.
For each eigenvalue of $A$, we need to work out what Jordan blocks we have corresponding to $A$.
Suppose $\lambda$ is an eigenvector of algebraic multiplicity $a$ and geometric multiplicity $g$. Then define $g _ k$ to be the number of solutions to
\[(A - \lambda I)^k v = 0\]There is a result that says
\[g_1 < g_2 < g_3 < \cdots < g_N = a\]for some $N$. Define
\[\begin{aligned} L_1 &= g_1 \\\\ L_2 &= g_2 - g_1 \\\\ &\vdots \\\\ L_N &= g_n - g_{n-1} \\\\ \end{aligned}\]This is the number of Jordan blocks of size at $L$east $k\times k$. Then define
\[\begin{aligned} R_1 &= L_1 - L_2 \\\\ R_2 &= L_2 - L_3 \\\\ &\vdots \\\\ R_N &= L_{N-1} - L_N \end{aligned}\]This is the number of Jordan blocks that are exactly $R$ight of size $k \times k$.
Sticking together these for each eigenvector gives the required result.
Proofs
Prove that if $T$ is a nilpotent linear map, then its minimal polynomial is given by $x^m$ for some $m$ and there exists a basis of $T$ where it’s zero, but with $0$ or $1$ entries on the superdiagonal (actually don’t prove this last part, it’s non-examinable).
Todo.