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?


\[m_T(x) = x^m\]

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 $i$.


\[J_i(\lambda) := \begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 \\\\ 0 & \lambda & 1 & \cdots & 0 \\\\ 0 & 0 & \lambda & \ddots & \vdots \\\\ \vdots & \vdots & \ddots & \ddots & 1 \\\\ 0 & 0 & \cdots & 0 & \lambda \end{pmatrix}\]

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?


\[T = D + N\]

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 $\lambda$?


\[(A - \lambda I)^k v = 0\]

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.




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