Linear Algebra MT23, Minimal polynomials

The monic polynomial of least degree such that $p(A) = 0$.

Note that $\dim _ {\mathbb F} A = n^2$, so the $n^2 + 1$ elements $I, A, A^2, \ldots, A^{(n^2)}$ are linearly dependent. Then there exists a non-trivial linear combination to get $0$.

\((P^{-1}AP)^i = P^{-1}A^iP\) so \(f(P^{-1}AP) = P^{-1}f(A)P = 0\)

\[m _ A(x) \mid f(x)\]

Use the Euclidean algorithm, and note that \(f(A) = q(A)m _ A(A) + r(A)\) and as $\deg r < \deg m _ A$, it must be that $r(x) = 0$.

\(\mathbb F[x] \to M _ {n \times n} (\mathbb F)\) where \(f(x) \mapsto f(A)\)

\[\lambda \text{ root of } \chi _ A(t) \iff \lambda \text{ root of } m _ A(t)\]

Forward direction: Since $\lambda$ is a root of $\chi _ A(\lambda)$, there exists $v \ne 0$ such that $Av = \lambda v$. Applying $m _ A$ to both sides and using the fact it’s a polynomial, we see $m _ A(A)v = m _ A(\lambda)v$ but $m _ A(A) = 0$, so it must be be the case that $m _ A(\lambda) = 0$ (otherwise we somehow get $0$ by multiplying by a non-zero scalar).

Backward direction: Suppose $\lambda$ is a root of $m _ A(x)$. Then $m _ A(x) = (x- \lambda) g(x)$ where $g(A) \ne 0$ (by minimality). Then $\exists w \ne 0$. Then note

\[(A - \lambda I) g(A)w = m _ A(A)w = 0\]

so $g(A)w$ is an eigenvector for $A$, hence a root of $\chi _ A(x)$.

\[f(A)v = f(\lambda)v\]

\(f(A^\top) = (f(A))^\top\) Thus \(f(A) = 0 \iff f(A^\top) = 0\) Hence \(m _ A(x) = m _ {A^\top}(x)\)

Consider \(\begin{pmatrix}1 & 0 & \cdots & 0 \end{pmatrix} \sum^{n-1} _ {k=0} b _ k C^k = \begin{pmatrix}b _ 0 & b _ 1 & \cdots & b _ {n-1} \end{pmatrix}\) So that $g(C) \ne 0$ for all polynomials $g$ with $\deg g < n$. Since \(\begin{pmatrix}1 & 0 & \cdots & 0 \end{pmatrix} C^n = \begin{pmatrix}-a _ 0& -a _ 1 & \cdots & -a _ {n-1} \end{pmatrix}\) We need $b _ i = a _ i$, so the minimal polynomial and $f$ agree.

Let $\lambda _ 1, \cdots, \lambda _ k$ be the eigenvalues of $T _ 1$. Then we have the decomposition \(V = V _ {\lambda _ 1} \oplus \cdots \oplus V _ {\lambda _ 2}\) If $v \in V _ {\lambda _ i}$, then $T _ 1(T _ 2 v) = T _ 2(T _ 1 v) = \lambda _ i T _ 2(v)$.

In other words, $T _ 2(V _ {\lambda _ i}) \subseteq V _ {\lambda _ i}$. So we may consider the restriction \(T _ 2 \vert _ {V _ {\lambda _ i} } : V _ {\lambda _ i} \to V _ {\lambda _ i}\) But this is a diagonalisable linear transformation (consider the fact that the minimal polynomial must still be a product of linear factors), so $T _ 2$ is diagonaliable on $V _ {\lambda _ i}$.

But on $V _ {\lambda _ i}$, $T _ 1 + T _ 2$ is of the form $\lambda _ i I + T _ 2$ for all bases, so by diagonalising $T _ 2$, we can diagonalise $T _ 1 + T _ 2$ on $V _ {\lambda _ i}$.

By doing this for each subspace in the decomposition $V = V _ {\lambda _ 1} \oplus \cdots \oplus V _ {\lambda _ 2}$, we see that $T _ 1 + T _ 2$ is diagonalisable.

Note that $(x-i)^2 (x+i)^2 = (x^2 + 1)^2$ is a real polynomial that annihilates $T$. So the minimal polynomial must divide $(x^2 + 1)^2$, and in fact it is equal since otherwise $T^2 + I$ would be $0$, which it is not.

• The first part follows from the division algorithm applied to $f / m _ A$, and noticing that $r = 0$ for it to annihilate
• The second part follows from the fact that if $m _ A$ and $m _ A’$ are two minimal polynomials, $m _ A \mid m _ A’$ and $m _ A’ \mid m _ A$.