Let $P$ be an orthogonal matrix, $R$ be an upper triangular matrix, and $A=PRP^{-1}$. Show that if $A$ is symmetric, then $R$ is symmetric, and hence a diagonal matrix.
My approach:
Suppose $A=PRP^{-1}$. Since $P$ is an orthogonal matrix, it has orthogonal columns, and these columns are linearly independent, so $P$ is invertible and we can express $R$ as $R=P^{-1}AP$.
Since $P$ is orthogonal, $P^{-1}=P^T$ and since $A$ is symmetric $A=A^T$, transposing $R$ we show that $R$ is symmetric:
$$R^T=(P^{-1}AP)^T=P^TA^T(P^{-1})^T=P^TA^T(P^T)^T=P^{-1}AP=R$$
From here how do I show that $R$ is a diagonal matrix?
Best Answer
Since $R$ is an upper matrix then it has the form $$ R = \left( {\begin{array}{*{20}c} {a_{11} } & {a_{12} } & \cdots & {a_{1n} } \\ 0 & {a_{22} } & \cdots & {a_{2n} } \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & {a_{nn} } \\ \end{array}} \right) $$ and so that $R^T$ has the form $$ R^T = \left( {\begin{array}{*{20}c} {a_{11} } & 0 & \cdots & 0 \\ {a_{12} } & {a_{22} } & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ {a_{1n} } & {a_{2n} } & \cdots & {a_{nn} } \\ \end{array}} \right) $$ Now, since $R$ is symmetric $R=R^T$ then by comparing the corresponding terms in $R$ and $R^T$ then we find that $$ R = \left( {\begin{array}{*{20}c} {a_{11} } & {a_{12} } & \cdots & {a_{1n} } \\ 0 & {a_{22} } & \cdots & {a_{2n} } \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & {a_{nn} } \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} {a_{11} } & 0 & \cdots & 0 \\ {a_{12} } & {a_{22} } & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ {a_{1n} } & {a_{2n} } & \cdots & {a_{nn} } \\ \end{array}} \right) = R^T $$ so that $$ \begin{array}{l} a_{12} = 0,a_{13} = 0, \ldots ,a_{1n} = 0 \\ a_{21} = 0,a_{23} = 0, \ldots ,a_{2n} = 0 \\ \vdots \\ \end{array} $$ which means that $ R = \left( {a_{ii} } \right)$, $i=1,2,\cdots,n$ is a diagonal matrix.