Another way of arguing existence of eigen value for matrix over Complex Number

Question

I Know that by use of fundamental theorem of algebra and by considering Characteristics polynomial we guarantee the existence of eigenvalues in case of complex field.
But Is there is any other simple way to show above by just definition or simple way?Is it possible ?
Any Help will be appreciated

Answer 1

As pointed out in one of the other answers, some form of the Fundamental Theorem of Algebra must be used in the proof that every square matrix of complex numbers has an eigenvalue. However, there are clean, insightful proofs that avoid the characteristic polynomial and determinants. For example, see the proof of Theorem 5.21 in the sample chapter titled Eigenvalues, Eigenvectors, and Invariant Subspaces at http://linear.axler.net/.

Answer 2

I don't know whether this is answers your question, but here it goes. The assertion “every endomorphism of $\mathbb{C}^n$ has at least an eigenvalue” is equivalent to the fundamental theorem of Algebra. To be more precise, we have

Theorem: If $K$ is a field, then every endomorphism of $K^n$ ($n\in\mathbb N$) has at least an eigenvalue if and only if $K$ is algebraically closed.

Proof: In fact, if $K$ is algebraically closed, you know that we can use characteristic polynomials to prove that every endomorphism of $K^n$ ($n\in\mathbb N$) has at least an eigenvalue. On the other hand, if every endomorphism of $K^n$ has at least an eigenvalue and if $P(x)\in K[x]$ is a monic polynomial, then the characteristic polynomial of the Frobenius companion matrix $M_{P(x)}$ of $P(x)$ is again $P(x)$ and if we see $M_{P(x)}$ as the matrix of an endomorphism of $K^n$ then that endomorphism will have an eigenvalue. Therefore, $P(x)$ has a root in $K$. And if every monic polynomial has a root in $K$, then every non-constant polynomial has a root in $K$.

Therefore, in order to prove that every endomorphism of $\mathbb{C}^n$ has some eigenvalue, we eill need either the fundamental theorem of Algebra or else something equivalent to it.

Answer 3

Here is a proof which uses some complex analysis instead of the fundamental theorem of algebra.

Equip $M_n(\mathbb{C})$ with some submultiplicative matrix norm. Let $T \in M_n(\mathbb{C})$.

Assume that $\sigma(T) = \emptyset$. Then the map $\mathbb{C} \to M_n(\mathbb{C}):\lambda \mapsto (T - \lambda I)^{-1}$ is well-defined. Let $\phi$ be an arbitrary (bounded) linear functional on $M_n(\mathbb{C})$. Define $f : \mathbb{C} \to \mathbb{C}$ as $f(\lambda) = \phi[(T - \lambda I)^{-1}]$. We will show that $f$ is holomorphic on $\mathbb{C}$.

For any $\lambda_0 \in \mathbb{C}$ we have

\begin{align} \lim_{\lambda \to \lambda_0} \frac{f(\lambda) - f(\lambda_0)}{\lambda - \lambda_0} &= \phi\left(\lim_{\lambda \to \lambda_0}\frac{(T - \lambda I)^{-1} - (T - \lambda_0 I)^{-1}}{\lambda - \lambda_0} \right)\\ &= \phi\left(\lim_{\lambda \to \lambda_0}\frac{(\lambda_0 - \lambda)(T - \lambda I)^{-1}(T - \lambda_0 I)^{-1}}{\lambda - \lambda_0} \right)\\ &= -\phi\left((T - \lambda_0 I)^{-2}\right) \in \mathbb{C}\\ \end{align}

Furthermore, if $|\lambda| > \|T\|$ then $$|f(\lambda)| \le \|\phi\|\|(T - \lambda I)^{-1}\| \le \frac{\|\phi\|}{\|T - \lambda I\|} \le \frac{\|\phi\|}{|\lambda | - \|T\|}$$

Therefore $f$ is a bounded entire function with $\lim_{|\lambda|\to\infty} f(\lambda) = 0$ so Liouville's theorem implies $f = 0$.

Since $\phi$ was arbitrary, we conclude $(T - \lambda I)^{-1} = 0, \forall \lambda \in \mathbb{C}$ which is a contradiction.

Therefore $\sigma(T) \ne \emptyset$.