Find a diagnalizable matrix Find a non-diagonal $4 \times 4$ matrix with eigenvalues 2, 3 and 6 which can be orthogonally diagonalized.
I know since there are 3 eigenvalues for the $4 \times 4$ matrix, one of the values must have a multiplicity of 2, but I do not know how to find the matrices.
Can someone help me, please?
 A: Let's just pick 2 to be the one with multiplicity 2.
$$
A = \begin{pmatrix}
2 & 0 & 0 & 0\\
0 & 2 & 0 & 0\\
0 & 0 & 3 & 0\\
0 & 0 & 0 & 6\\
\end{pmatrix}
$$
This is a matrix with the specified eigenvalues but it's not diagonal.
If we want a matrix $B$ that orthogonally diagonalizes to $A$, then we just need to pick some orthogonal matrix $O$ because $B=OAO^T$ will work as long as our answer as long as $B$ is not diagonal.
O can be quite simple. For example, it might only affect the last two row's and columns.
$$
O = \begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & \cos \theta & \sin \theta\\
0 & 0 & -\sin \theta & \cos \theta \\
\end{pmatrix}
$$
This is orthogonal. In fact it reduces to the orthogonal rotation matrix in just those last two dimensions.
$$
B = O A O^T = \begin{pmatrix}
2 & 0 & 0 & 0\\
0 & 2 & 0 & 0\\
0 & 0 & 3 \cos^2(θ) + 6 \sin^2(θ) & 3 \cos(θ) \sin(θ) \\
0 & 0 & 3 \cos(θ) \sin(θ) & 6 \cos^2(θ) + 3 \sin^2(θ))\\
\end{pmatrix}
$$
As long as $3 \cos \theta \sin \theta \neq 0$, this is non-diagonal and orthogonally diagonalizes to give the specified eigenvalues.
If you want you can plug in a value for $\theta$ but just make sure the result is still not diagonal.
A: The matrix $$A=\begin{bmatrix}{3}&{0}&{0}&0\\{0}&{3}&{0}&0\\{0}&{0}&{4}&2\\{0}&{0}&{2}& 4\end{bmatrix}$$
has eigevalues 3, 2, 6, is non-diagonal and symmetric, so can be orthogonally diagonalized.
