diff --git a/Chapters/chapter6.tex b/Chapters/chapter6.tex
index 4c6b34a..46802bf 100644
--- a/Chapters/chapter6.tex
+++ b/Chapters/chapter6.tex
@@ -1632,7 +1632,7 @@ \subsection{Diagonalization}
 \end{eqnarray*}
 Things are not quite so simple for an arbitrary $n\times n$ matrix
 $A$. However, {\em if $A$ has a basis of eigenvectors} then it turns
-out that there exists an invertible matrix $B$ such that $AB=DB$,
+out that there exists an invertible matrix $B$ such that $AB=BD$,
 where $D$ is the diagonal matrix whose diagonal elements are the
 eigenvalues of $A$. Multiplying by $B^{-1}$ from either the left or
 right gives