Characteristic Polynomial of a Diagonalisable Transformation

Topics: Eigenvalues and Eigenvectors - Algebra

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(theorem)

Let $T:V\to V$ be a diagonalisable on a vector space $V$, and let $f(t)$ be the characteristic polynomial of $T$.

Then, $f(t)$ is broken down into a product of $n$ factors, all with a degree of 1. That is, there exist scalars ${\lambda }_1,\dots ,{\lambda }_2$ (not necessarily distinct) such that:

$$f(t)=(-1{)}^n(t-{\lambda }_1)\dots (t-{\lambda }_2)$$

(note)

If $f(t)$ has no distinct eigenvalues, then $f(t)$ has multiple zeroes.

Example

Let $A=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 3\\ 0& 0& 2\end{array}\right)$

We have that:

$$f(t)=\det (A-\lambda I)$$

$$f(t)=\det \left(\begin{array}{ccc}1-\lambda & 1& 0\\ 0& 1-\lambda & 3\\ 0& 0& 2-\lambda \end{array}\right)$$

$$f(t)=(1-\lambda )(1-\lambda )(2-\lambda )$$

$$f(t)=(-1)(\lambda -1)(\lambda -1)(\lambda -2)$$

$$f(t)=(-1)(\lambda -1{)}^2(\lambda -2)$$

With that, we get that the zeroes of the characteristic polynomial are:

• ${\lambda }_1=1$ (with a multiplicity of 2)
• ${\lambda }_2=2$ (with a multiplicity of 1)