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- Mar 10, 2012

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During my Mechanics of Solids course in my Mechanical Engineering curriculum I came across a certain fact about $3\times 3$ matrices.

It said that any symmetric $3\times 3$ matrix $A$ (with real entries) whose trace is zero is orthogonally similar to a matrix $B$ which has only zeroes on the diagonal.

In other words, given a symmetric matrix $A$ with $\text{trace}(A)=0$, there exists an orthogonal matrix $Q$ such that $QAQ^{-1}$ has only zeroes on the diagonal.

I think the above fact should be true not only for $3\times 3$ matrices but for matrices with any dimension.

So what I am trying to prove is that:

**Given a symmetric $n\times n$ matrix with real entries with $\text{trace}(A)=0$, there exists an orthogonal matrix $Q$ such that $QAQ^{-1}$ has only zeroes on the diagonal.**

__Problem:__I have tried to attack the problem using the spectral theorem.

Since $A$ is symmetric, we know that there exists an orthogonal matrix $S$ such that $D=SAS^{-1}$ is a diagonal matrix.

We need to show that $D$ is orthogonally similar to a matrix with only zeroes on the diagonal.

Thus we have to find an orthogonal matrix $Q$ such that $QDQ^{-1}$ has only zeroes on the diagonal.

This is equivalent to show that $\sum_{k=1}^n q^2_{ik}d_k=0$ for all $i\in \{1,\ldots,n \}$, where $q_{ij}$ is the $i,j$-th entry of $Q$ and $d_k$ is the $k$-th diagonal entry of $D$.

We also know that $d_1+\ldots+d_n=0$.

Here I am stuck.

From the above it can be seen that proposition is true for $n=2$. From $n=3$ I have taken the fact from the book but cannot easily prove it. Can anybody help.