# The eigenvalues of a linear operator are the roots of its characteristic polynomial

Let $$T$$ be a linear operator on an $$n-$$dimensional vector space $$V$$ with characteristic polynomial $$f(t)$$. Then

$$a)$$ A scalar $$\lambda$$ is an eigenvalue of $$T$$ if and only if $$\lambda$$ is a zero of the polynomial $$f(t)$$ (i.e., if and only if $$f(\lambda)=0$$).

$$b)$$ $$T$$ has at most $$n$$ distinct eigenvalues.

Attempt:

$$a)$$ $$\lambda$$ is an eigenvalue $$\Leftrightarrow$$ there is $$v\neq \vec{0}$$ and $$T(v) = \lambda v$$ $$\Leftrightarrow$$ there is $$v\neq \vec{0}$$, $$v \in ker(T − \lambda I)$$ $$\Leftrightarrow$$ $$ker(T − \lambda I)\neq \vec{0}$$ $$\Leftrightarrow$$ $$ker([T]_\beta −\lambda I) \neq \vec{0}$$ for some basis $$\beta$$ $$\Leftrightarrow$$ $$f(\lambda) = det([T]_\beta − \lambda I) = 0$$

• b) follows from a) since $f$ has degree $n$. Aug 13, 2021 at 5:07
• Using polynomial long division it can be proved that a polynomial of degree $n$ (with coefficients in a field say) has at most $n$ zeroes. Aug 13, 2021 at 5:42

Suppose that $$m$$ is a minimal polynomial for $$T$$, and suppose that $$\lambda$$ is an eigenvalue of $$T$$. Then there is a non-zero vector $$x$$ such that $$(T-\lambda I)x=0$$. Consequently, $$(m(T)-m(\lambda)I)x=g(T)(T-\lambda I)x=0,$$ where $$g$$ is a polynomial. Because $$m(T)=0$$ and $$x\ne 0$$, it follows that $$m(\lambda)=0$$ for any eigenvalue $$\lambda$$ of $$T$$.
Conversely, suppose that $$m(\lambda)=0$$, where $$m$$ is the minimal polynomial for $$T$$. By polynomial division, there is a polynomial $$g$$ of lower order than $$m$$ such that $$0=m(T)=m(T)-m(\lambda)I=(T-\lambda I)g(T).$$ Because $$m$$ is minimal, it follows that $$g(T)\ne 0$$. And, from the above, every non-zero vector in the range of $$g(T)$$ must be an eigenvector of $$T$$ with eigenvalue $$\lambda$$. So $$\lambda$$ is an eigenvalue of $$T$$.