Notes - Linear Algebra MT23, Primary decomposition theorem

\[V = \ker a(T) \oplus \ker b(T)\]

Furthermore, each $\ker a(T)$ and $\ker b(T)$ is a $T$-invariant subspace.

• Since coprime, $\exists s, t \in \mathbb F[x]$ such that $a(x)s(x) + b(x)t(x) = 1$. But then $a(T)s(T) + b(T)s(T) = I$, which implies $v = a(T)s(T)v + b(T)s(T)v$. To see that each of these vectors are in the respective kernels, note that since $f$ annihilates, $b(T)a(T)s(T)v = f(T)v = 0$, so $a(T)s(T)v \in \ker b(T)$ and likewise $b(T)s(T)v \in \ker a(T)$.
• To see that the intersection is trivial, pick $v \in \ker a(T) \cap \ker b(T)$. Then $v = a(T)s(T)v + b(T)s(T)v$ but because these are kernels both summands are $0$.

Suppose $v \in \ker a(T)$. Then $a(T)(T(v)) = T(a(T)v) = T(0) = 0$, with the first step justified since $a(T)$ is a polynomial is $T$. The other case is similar.

\[T = \left(\begin{array}{c|c} \ast & 0 \\\\ \hline 0 & \ast \end{array}\right)\]

\[a(x) = m_{(T \vert_{\ker(a(T))})}\]

and

\[b(x) = m_{(T \vert_{\ker(b(T))})}\]

Shorthand, let $m _ 1 = m _ {(T \vert _ {\ker(a(T))})}$ and $m _ 2 = m _ {(T \vert _ {\ker(b(T))})}$. Then $m _ 1 \mid a$ since

\[a(T \vert_{\ker(a(T))}) = a(T)|_{\ker(a(T))} = 0\]

and likewise $m _ 2 \mid b$, so $m _ 1 m _ 2 \mid ab$.

But also $ab \mid m _ 1 m _ 2$ since $m _ T = ab$, and $m _ 1(T)m _ 2(T) = 0$ for any $v$.

Hence $m _ 1 = a$, $m _ 2 = b$ since if they were not equal, the degree of $ab$ would not equal the degree of $m _ 1 m _ 2$, a contradiction (also note that $a, b$ are monic, so they can’t be a constant multiple of $m _ 1$ or $m _ 2$).

Suppose:

• $T : V \to V$
• $m _ T(x) = f _ 1^{q _ 1}(x) \cdots f _ r^{q _ r}(x)$
• Each $f _ i$ distinct, monic (leading coeff is $1$), and irreducible

Then define $W _ i = \ker(f^{q _ i} _ i (T))$. Then

• $V = W _ 1 \oplus \cdots \oplus W _ r$.
• $W _ i$ is $T$-invariant.
• $m _ {T \vert _ {W _ i}\,} = f _ i^{q _ i}$

Suppose:

• $T : V \to V$
• $f \in \mathbb F[x]$, $f(T) = 0$
• $f(x) = a(x)b(x)$, where $a, b$ coprime

Then:

\[V = \ker a(T) \oplus \ker b(T)\]

$m _ T(x)$ splits into distinct linear factors.

\[m_T(x) \text{ splits into distinct linear factors} \iff T \text{ diagonalisable}\]

Suppose $m _ T(x) = \prod^{s} _ {i=1} (x - \lambda _ i)$. Then by the primary decomposition theorem,

\[V = \ker (T - \lambda_1I) \oplus \cdots \oplus \ker (T - \lambda_sI)\]

So it follows we can find a basis of eigenvectors.

Alternative, more explicit proof: Suppose

\[m_T(x) = \prod^k_{i = 1}(x - \lambda_i)\]

We induct on $k$. Write

\[m_T(x) = (x - \lambda) g(x)\]

By definition, $g(T) \ne 0$ and $g(\lambda) \ne 0$. Define

\[\begin{aligned} U_1 &= \ker(T - \lambda I) \\\\ U_2 &= \ker g(T) \end{aligned}\]

We want to show $V = U _ 1 \oplus U _ 2$. First suppose $v \in U _ 1 \cap U _ 2$. Then

\[0 = g(T)v = g(\lambda)v\]

which implies $v = 0$, since $g(\lambda) \ne 0$. Why is $g(T)v = g(\lambda) v$? Because $v$ is in $U _ 1$, so an eigenvector of $T$. For arbitrary $v \in V$ write

\[\begin{aligned} v_1 &= g(\lambda)^{-1} g(T)v \\\\ v_2 &= v - v_1 \end{aligned}\]

Then $v _ 1 \in U _ 1$, and $v _ 2 \in U _ 2$ since

\[\begin{aligned} g(T) (v - v_1) &= g(T)v - g(T)v_1 \\\\ &= g(T)v - g(\lambda) v_1 \\\\ &= g(T)v - g(T)v \\\\ &= 0 \end{aligned}\]

Furthermore, $T$ acts as $Tv = \lambda v$ in $U _ 1$, and since the degree of $g(x)$ is less than $k$, we can find a basis diagonalising $T$ on $U _ 2$.

Let $\lambda _ 1, \cdots, \lambda _ s$ be the distinct eigenvalues of $T$. Then consider

\[M(x) = \prod^s_{i=1}(x - \lambda_i)\]

Note that since $T$ is diagonalisable, there exists invertible $P$ such that $P^{-1}TP =D$ where $D$ is diagonal with diagonal entries from $\lambda _ 1, \cdots, \lambda _ s$. Then

\[M(T) = M(PDP^{-1}) = P \prod^s_{i=1} (D - \lambda_i) P^{-1} = P \cdot 0 P^{-1}\]

Since $m _ T \mid M$, and $M$ is the product of distinct linear factors, $m _ T(x) = M(x)$.

Proof sketch:

• Induct on the baby version of the primary decomposition theorem about $m _ T(x) = a(x) b(x)$ where $a, b$ coprime.
• $T$-invariance follows from the fact that $f _ i^{q _ i}(T)(Tv) = T(f _ i^{q _ i}(v))$, i.e. $T$ and polynomials in $T$ communte
• The fact $m _ {T \vert _ {W _ i}\,} = f _ i^{q _ i}$ also follows from induction (in the original proof, this is to do with noting that $m _ T = ab \mid m _ 1 m _ 2$ and aklso $m _ 1 \mid a$ and $m _ 2 \mid b$ where $m _ 1 = m _ {(T \vert _ {\ker(a(T))})}$ and $m _ 2 = m _ {(T \vert _ {\ker(b(T))})}$).