Proof of Theorem EMHE

Suppose that $A$ has size $n$, and choose $\vect{x}$ as any nonzero vector from $\complex{n}$. (Notice how much latitude we have in our choice of $\vect{x}$. Only the zero vector is off-limits.) Consider the set \begin{equation*} S=\set{\vect{x},\,A\vect{x},\,A^2\vect{x},\,A^3\vect{x},\,\ldots,\,A^n\vect{x}} \end{equation*}

This is a set of $n+1$ vectors from $\complex{n}$, so by Theorem MVSLD, $S$ is linearly dependent. Let $a_0,\,a_1,\,a_2,\,\ldots,\,a_n$ be a collection of $n+1$ scalars from $\complex{\null}$, not all zero, that provide a relation of linear dependence on $S$. In other words, \begin{equation*} a_0\vect{x}+a_1A\vect{x}+a_2A^2\vect{x}+a_3A^3\vect{x}+\cdots+a_nA^n\vect{x}=\zerovector \end{equation*}

Some of the $a_i$ are nonzero. Suppose that just $a_0\neq 0$, and $a_1=a_2=a_3=\cdots=a_n=0$. Then $a_0\vect{x}=\zerovector$ and by Theorem SMEZV, either $a_0=0$ or $\vect{x}=\zerovector$, which are both contradictions. So $a_i\neq 0$ for some $i\geq 1$. Let $m$ be the largest integer such that $a_m\neq 0$. From this discussion we know that $m\geq 1$. We can also assume that $a_m=1$, for if not, replace each $a_i$ by $a_i/a_m$ to obtain scalars that serve equally well in providing a relation of linear dependence on $S$.

Define the polynomial \begin{equation*} p(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_mx^m \end{equation*}

Because we have consistently used $\complex{\null}$ as our set of scalars (rather than ${\mathbb R}$), we know that we can factor $p(x)$ into linear factors of the form $(x-b_i)$, where $b_i\in\complex{\null}$. So there are scalars, $\scalarlist{b}{m}$, from $\complex{\null}$ so that, \begin{equation*} p(x)=(x-b_m)(x-b_{m-1})\cdots(x-b_3)(x-b_2)(x-b_1) \end{equation*}

Put it all together and \begin{align*} \zerovector&=a_0\vect{x}+a_1A\vect{x}+a_2A^2\vect{x}+\cdots+a_nA^n\vect{x}\\ &=a_0\vect{x}+a_1A\vect{x}+a_2A^2\vect{x}+\cdots+a_mA^m\vect{x}&&\text{$a_i=0$ for $i>m$}\\ &=\left(a_0I_n+a_1A+a_2A^2+\cdots+a_mA^m\right)\vect{x}&&\knowl{./knowls/theorem.MMDAA.knowl}{\text{Theorem MMDAA}}\\ &=p(A)\vect{x}&&\text{Definition of $p(x)$}\\ &=(A-b_mI_n)(A-b_{m-1}I_n)\cdots(A-b_2I_n)(A-b_1I_n)\vect{x} \end{align*}

Let $k$ be the smallest integer such that \begin{equation*} (A-b_kI_n)(A-b_{k-1}I_n)\cdots(A-b_2I_n)(A-b_1I_n)\vect{x}=\zerovector. \end{equation*}

From the preceding equation, we know that $k\leq m$. Define the vector $\vect{z}$ by \begin{equation*} \vect{z}=(A-b_{k-1}I_n)\cdots(A-b_2I_n)(A-b_1I_n)\vect{x} \end{equation*}

Notice that by the definition of $k$, the vector $\vect{z}$ must be nonzero. In the case where $k=1$, we understand that $\vect{z}$ is defined by $\vect{z}=\vect{x}$, and $\vect{z}$ is still nonzero. Now \begin{equation*} (A-b_kI_n)\vect{z}=(A-b_kI_n)(A-b_{k-1}I_n)\cdots(A-b_3I_n)(A-b_2I_n)(A-b_1I_n)\vect{x}=\zerovector \end{equation*} which allows us to write \begin{align*} A\vect{z} &=(A+\zeromatrix)\vect{z}&&\knowl{./knowls/property.ZM.knowl}{\text{Property ZM}}\\ &=(A-b_kI_n+b_kI_n)\vect{z}&&\knowl{./knowls/property.AIM.knowl}{\text{Property AIM}}\\ &=(A-b_kI_n)\vect{z}+b_kI_n\vect{z}&&\knowl{./knowls/theorem.MMDAA.knowl}{\text{Theorem MMDAA}}\\ &=\zerovector+b_kI_n\vect{z}&&\text{Defining property of $\vect{z}$}\\ &=b_kI_n\vect{z}&&\knowl{./knowls/property.ZM.knowl}{\text{Property ZM}}\\ &=b_k\vect{z}&&\knowl{./knowls/theorem.MMIM.knowl}{\text{Theorem MMIM}} \end{align*}

Since $\vect{z}\neq\zerovector$, this equation says that $\vect{z}$ is an eigenvector of $A$ for the eigenvalue $\lambda=b_k$ (Definition EEM), so we have shown that any square matrix $A$ does have at least one eigenvalue.