Stack Exchange
Necessary and sufficient conditions for an existence of an orthogonal matrix$P~$ s.t. $~P^{-1}AP~$is diagonal, using$~a~$which is one of entries of$A$
This problem is quoted from the $3$rd year transfer exam of math major in the university.
$$\begin{align}
a:=\text{real number}\\
A:=
\begin{pmatrix}
0&a&2\\
1&0&2\\
2&2&3
\end{pmatrix}
\end{align}$$

Represent the necessary and sufficient conditions for an existence of an orthogonal matrix$~P~$such that$~B:=P^{-1}AP~$is a diagonal matrix, using$~a~$

My works begin from here
$$\begin{align}
\det\left(A-\lambda I\right)&=\det\begin{pmatrix}0-\lambda&a&2\\1&-\lambda&2\\2&2&3-\lambda\end{pmatrix}\\&=\det\begin{pmatrix}-\lambda&a&2\\1&-\lambda&2\\2&2&3-\lambda\end{pmatrix}{(-1)^3\over(-1)^3}\\&=-\det\begin{pmatrix}\lambda&-a&-2\\-1&\lambda&-2\\-2&-2&\lambda-3\end{pmatrix}\\
&=-\left\{\lambda^2(\lambda-3)-4a-4-4\lambda-4\lambda-a(\lambda-3)\right\}\\&=-\left\{\lambda^2(\lambda-3)-4a-4-8\lambda-a\lambda+3a\right\}\\&=-\left\{\lambda^3-3\lambda^2-(a+8)\lambda-(a+4)\right\}
\end{align}$$
$$
\color{red}{\left(\lambda^3-3\lambda^2-(a+8)\lambda-(a+4)\right)=0}
$$
Firstly I tried setting $~ a=-4 ~$ and obtained follwing.
$$\begin{align}
\lambda^3-3\lambda^2-4\lambda&=0\\
 \lambda(\lambda^2-3\lambda-4) &=0\\
 \lambda(\lambda+1)(\lambda-4) &=0\\
 \therefore~\lambda&=\underbrace{-1,0,4}_{\text{eigenvalues} }  
\end{align}$$
$$
\begin{pmatrix}x\\y\\z\end{pmatrix}=\underbrace{\alpha\begin{pmatrix}-2\\0\\1\end{pmatrix},~\beta\begin{pmatrix}-2\\1/2\\1\end{pmatrix},~\gamma\begin{pmatrix}0\\1/2\\1\end{pmatrix}}_{\text{eigenvectors where}~(\alpha,\beta,\gamma)~\neq(0,0,0) } 
$$
BTW I don't know which of these $~  \alpha,\beta,\gamma~\not\in\mathbb{R}\setminus\{0\}~$ or $~  \alpha,\beta,\gamma~\not\in\mathbb{C}\setminus\{0\}~$ should I write.
And I checked that the following matrix  $~P~$makes $~P^{-1}AP~$ a diagonal matrix.
$$
\begin{align}
P&=\underbrace{\begin{pmatrix}
-2 & -2 & 0 \\
0 & 1/2 & 1/2 \\
1 & 1 & 1
\end{pmatrix}}_{\text{Arranged the eigenvectors} } \\
P^{-1}&=\begin{pmatrix}
0 & -2 & 1 \\
-1/2 & 2 & -1 \\
1/2 & 0 & 1
\end{pmatrix}
\end{align}
$$
So it can be concluded that putting $~a=-4~$is one of the sufficient condition(s) for the problem statement.
I am anxious about that there might be more possible value(s) of $~a~$ which makes a diagonal matrix, using the red equation. How can I verify it?
Moreover, how can I assess the necessary condition(s)? I've thought the following but cannot proceed from it.
$$
\begin{align}
\begin{pmatrix}
s & 0 & 0 \\ 
0 & t & 0 \\
0 & 0 & u 
\end{pmatrix}:&=P^{-1}AP\\Q:&=P^{-1}\\
\begin{pmatrix}
s & 0 & 0 \\ 
0 & t & 0 \\
0 & 0 & u 
\end{pmatrix}&=QAP\\
&=\begin{pmatrix}
q_{11} & q_{12} & q_{13}\\ 
q_{21} & q_{22} & q_{23}\\ 
q_{31} & q_{32} & q_{33}\\ 
\end{pmatrix}
\begin{pmatrix}
0&a&2\\
1&0&2\\
2&2&3
\end{pmatrix}
\begin{pmatrix}
p_{11} & p_{12} & p_{13}\\ 
p_{21} & p_{22} & p_{23}\\ 
p_{31} & p_{32} & p_{33}\\ 
\end{pmatrix}\\&=
\underbrace{\begin{pmatrix}
q_{12}+2q_{13} &a\cdot q_{11}+2 q_{13} & 2 (q_{11}+q_{12})+3q_{13}\\
q_{22}+2q_{23}&a\cdot q_{21}+2 q_{23} & 2 (q_{21}+q_{22})+3q_{23}\\
q_{32}+2q_{33}&a\cdot q_{31}+2 q_{33} & 2 (q_{31}+q_{32})+3q_{33}\\
\end{pmatrix}
\begin{pmatrix}
p_{11} & p_{12} & p_{13}\\ 
p_{21} & p_{22} & p_{23}\\ 
p_{31} & p_{32} & p_{33}\\ 
\end{pmatrix}}_{\text{Seems too complicated for an exam's problem.} } 
\end{align}
$$

In the first place ,I think I misunderstood the meaning of necessary and sufficient conditions. I had thought that necessary condition(s) and sufficient condition(s) might exist separately(namely these may differ).

A real square matrix is orthogonally diagonalisable if and only if it is symmetric.  So the answer is: $a=1$.
