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(possibly?) fixed a fact about eigenvectors #5

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(possibly?) fixed a fact about eigenvectors #5

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2 changes: 1 addition & 1 deletion eigenvaluesAndEigenvectors/definition.tex
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Expand Up @@ -21,7 +21,7 @@

Before proceeding with examples, we note that

\begin{proposition} If $\bf v$ is an eigenvalue of a matrix $A$, the eigenvector associated with it is unique.
\begin{proposition} If $\bf v$ is an eigenvector of a matrix $A$, the eigenvalue associated with it is unique.
\end{proposition}

\begin{proof} Suppose $\lambda_1{\bf v} = A*{\bf v} = \lambda_2{\bf v}$. Then $\lambda_1{\bf v} - \lambda_2{\bf v} = (\lambda_1 - \lambda_2){\bf v} = {\bf 0}$. But since ${\bf v}\ne {\bf 0}$, the only way this could happen is if the coefficient $(\lambda_1 - \lambda_2)$ is equal to zero, or equivalently, if $\lambda_1 = \lambda_2$.
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