📂Linear Algebra

The Characteristic Polynomial of a Diagonalizable Linear Transformation Is Factorizable

Definition¹

The polynomial $P(F)$ being split over $F$ means that there exists a constant $c, a_{1}, \dots, a_{n} \in F$ that satisfies the following.

$$f(t) = c(t - a_{1})(t - a_{2})\cdots(t - a_{n})$$

If the polynomial $f(t)$ that is decomposed is the characteristic polynomial of some linear transformation $T$ or matrix $A$, we say that $T$(or $A$) is decomposed.

Explanation

By definition, if the characteristic polynomial of $T: V \to V$ is decomposed, then $T$ has $n = \dim(V)$ eigenvalues.(It is not said that they are distinct)

An easy example is that $f(t) = t^{2} + 1 = (t-i)(t+i)$ cannot be decomposed over $\mathbb{R}$.

Theorem

The characteristic polynomial of any diagonalizable linear transformation is decomposed.

Proof

Let $V$ be a $n$-dimensional vector space, and $T : V \to V$ be a diagonalizable linear transformation. Let $\beta$ be an ordered basis that makes $\begin{bmatrix} T \end{bmatrix}_{\beta} = D$ a diagonal matrix.

$$D = \begin{bmatrix} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_{1} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n} \end{bmatrix}$$

And let $f(t)$ be the characteristic polynomial of $T$. Then, since the characteristic polynomial does not depend on the choice of the ordered basis,

$$\begin{align*} f(t) &= \det(D - tI) = \det \begin{bmatrix} \lambda_{1} - t & 0 & \cdots & 0 \\ 0 & \lambda_{1} - t & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n} - t \end{bmatrix} \\ &= (\lambda_{1} - t)(\lambda_{2} -t)\cdots(\lambda_{n}-t) \\ &= (-1)^{n}(t-\lambda_{1})(t-\lambda_{2})\cdots(t-\lambda_{n}) \end{align*}$$

■