10 - Invariant Subspaces

Invariant under a linear transformation

Let $T : V \to V$ be a linear transformation
Let subspace $U \subseteq V$ then
$U is T -invariant if T (U) \subseteq U$

Properties of Invariant Linear Transformations

Let $T : V \to V, S : V \to V$ be a linear transformation
Let subspace $U \subseteq V$ and $U$ is $T$ - and $S$ - invariant then

$U$ is invariant under

Zero Map

Identity Map

$a T$ , $\forall a \in F$

$S + T$

$S \circ T$

$p (T)$ for any polynomial $p (x) \in F [x]$ (combining 2, 3, 4, 5)

Map of Quotients from Invariant Polynomials

Let $T : V \to V$ be a linear transformation and $U$ is invariant under $t$
So $U$ is also invariant under polynomial $p (x)$ evaluated at $T$ then
$\overline{p (T)} : V / U \to V / U$
is a map of quotients

Characteristic Polynomial of an Invariant Subspace

Let $T : V \to V$ be a linear transformation on a finite dimensional space
Assume $U \subseteq V$ is $T -$ invariant then
$χ_{T} (x) = χ_{T ∣_{U}} (x) \times χ_{\overline{T}} (x)$

Proof

Extend a basis $E$ for $U$ to a basis $B$ of $V$
Let $\overline{B}$ be the associated basis for $V / U$
By Block Matrix Decomposition of a Linear Map then
$_{B} [T]_{B} = (_{E} [T ∣_{U}]_{E} 0 *_{\overline{B^{'}}} [\overline{T}]_{\overline{B}})$
Then the determinant of a triangular block matrix is the product of the determinants of diagonal blocks

Block Diagonalisation via Invariant Subspaces

Let $V$ be a finite dimensional vector space
Following the setup from direct sum of vector space

Let $T : V \to V$ be a linear map

If each $W_{i}$ is $T -$ invariant then $T$ with respect to basis $B$ is block diagonal
$_{B} [T]_{B} = A_{1} ⋱ A_{r} with A_{i} =_{B_{i}} [T ∣ W_{i}]_{B_{i}}$
Hence
$χ_{T} (x) = χ_{T ∣_{W_{1}}} (x) \dots χ_{T ∣ W_{r}} (x)$

Decomposition Theorem of a Vector Space

Let $V$ be a finite dimensional vector space
Let $T : V \to V$ be a linear map

Assume $f (x) = a (x) b (x)$ with $g cd (a, b) = 1$ and $f (T) = 0$ then
$V = Ker (a (T)) \oplus Ker (b (T))$
is a $T -$ invariant direct sum decomposition

Especially if $f = m_{T}$ and $a, b$ are monic then
$m_{T ∣_{Ker (a (T))}} (x) = a (x) and m_{T ∣_{Ker (b (T))}} (x) = b (x)$

Proof

By Bezout’s Identity for Polynomials then there exist $s, t$ such that $a s + b t = 1$ but then
$a (T) s (T) + b (T) t (T) = Id_{v}$
So for all $v \in V$
$a (T) s (T) v + b (T) t (T) v = v$
As $f$ is annihilating then
$a (T) (b (T) t (T) v) = f (T) t (T) v = 0 and b (T) (a (T) s (T) v) = 0$
Hence
$V = Ker (a (T)) + Ker (b (T))$
To show that this is a direct sum decomposition, assume that $v \in Ker (a (T)) \cap Ker (b (T))$
So then we get $v = 0 + 0 = 0$ hence
$V = Ker (a (T)) \oplus Ker (b (T))$
To show that both factors are $T$ -invariant then for $v \in Ker (a (T))$ then
$a (T) (T (v)) = T (a (T) v) = T (0) = 0$
Similarly $b (T) T (v) = 0$ for $v \in Ker (b (T))$

Assume that $f = m_{T}$ and let
$m_{1} = m_{T ∣_{Ker (a (T))}} and m_{2} = m_{T_{Ker (b (T))}}$
Then $m_{1} ∣ a$ as
$m_{1} = m_{T ∣_{Ker (a (T))}}$
And similarly $m_{2} ∣ b$ and
$m_{T} = ab ∣ m_{1} m_{2} as m_{1} (T) m_{2} (T) = 0$
Any $v \in V$ can be written as $v = w_{1} + w_{2}$ wh ere $w_{1} \in Ker (a (T))$ and $w_{2} \in Ker (b (T))$ so
$m_{1} (T) m_{2} (T) v = m_{2} (T) (m_{1} (T) w_{1}) + m_{1} (T) (m_{2} (T) w_{2}) = 0 + 0 = 0$
Hence, for degree reasons and because all these polynomials then $m_{1} = a, m_{2} = b$

Invariant Factor Decomposition of the Minimal and Characteristic Polynomials

There exist unique distinct irreducible monic polynomials

$f_{1}, \dots, f_{r} \in F [x]$
and positive integers
$n_{i} \geq q_{i} > 0 for 1 \leq i \leq r$
such that
$m_{T} (x) = f_{1}^{q_{1}} \dots f_{r}^{q_{r}} and χ_{T} = \pm f_{1}^{n_{1}} \dots f_{r}^{n_{r}}$

Proof $m_{T} = f_{1}^{q_{1}} \dots f_{r}^{q_{r}}$ into distinct monic irreducibles over $F [x]$ By Cayley-Hamilton then $m_{T} ∣ χ_{T}$ so

Factor

$χ_{T} = f_{1}^{n_{1}} \dots f_{r}^{n_{r}} \cdot b (x) for some n_{i} \geq q_{i} and b (x) \in F [x]$
As $m_{T}$ and $χ_{T}$ share roots by eigenvalue-polynomial characterisation then
$b (x)$ has no roots and must be constant so by comparing coefficients then
$b (x) = (- 1)^{n} where n = dim V$

Oxford Maths Notes

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10 - Invariant Subspaces

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