Give an example of a symmetric $3x3$ matrix that is not diagonal.
Give an example of a symmetric $3x3$ matrix that is not diagonal.
$$\begin{pmatrix} 2 &4 & 4 \\ 4 &8 & 8 \\ 4 & 8&8 \end{pmatrix}=A $$ and $$ A^{T}=\begin{pmatrix} 2 &4 & 4 \\ 4 &8 & 8 \\ 4 & 8&8 \end{pmatrix}$$ so $$ A = A^{T} $$ i.e. $A$ is symmetric. Upon reducing A to it's canonical form we have $$ A = \begin{pmatrix} 1 &2 & 2 \\ 0 &0 & 0 \\ 0 &0 &0 \end{pmatrix}$$
Therefore $A$ is symmetric and not diagonal. So it appears that in general any symmetric matrix that has rows that are duplicates or possibly scalar multiples are not diagonal?
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$\begingroup$Clearly the matrix $$A=\begin{pmatrix} 2 &4 & 4 \\ 4 &8 & 8 \\ 4 & 8&8 \end{pmatrix} $$ Is not diagonal.
But it can be diagonalized, as any symmetric matrix:
$$A=PDP^{-1}$$ with: $$ D = \begin{pmatrix} 0 &0 & 0 \\ 0 &0 & 0 \\ 0 &0 &18 \end{pmatrix} \qquad P=\begin{pmatrix} -2 &-2 & 1 \\ 0 &1 & 2 \\ 1 &0 &12 \end{pmatrix} \qquad P^{-1}=\frac{1}{9}\begin{pmatrix} -2 &-4 & 5 \\ -2 &5 & -4 \\ 1 &2 &2 \end{pmatrix} $$ and note that the columns of $P$ are orthogonal vectors.
This is the ''canonical'' or ''normal Jordan form'' of the matrix $A$
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