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2 changes: 1 addition & 1 deletion Chapters/chapter6.tex
Original file line number Diff line number Diff line change
Expand Up @@ -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
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