How to simplify the following to find the eigen values? I have the following matrix
$$M = T^2_1 \begin{pmatrix} A^2_{x} \;\;A_xA_y \;\; A_xA_z \\
                            A_yA_x \;\; A^2_y \;\; A_yA_z \\
                            A_zA_x \;\; A_zA_y \;\; A^2_z\end{pmatrix}
            + T^2_2 \begin{pmatrix} B^2_{x} \;\;B_xB_y \;\; B_xB_z \\
                            B_yB_x \;\; B^2_y \;\; B_yB_z \\
                            B_zB_x \;\; B_zB_y \;\; B^2_z\end{pmatrix}
         +T^2_3 \begin{pmatrix} C^2_{x} \;\;C_xC_y \;\; C_xC_z \\
                            C_yC_x \;\; C^2_y \;\; C_yC_z \\
                            C_zC_x \;\; C_zC_y \;\; C^2_z\end{pmatrix}$$
$A=(A_x,A_y,A_z),B=(B_x,B_y,B_z),C=(C_x,C_y,C_z)$. The conditions are 


*

*$A,B,C$ are
unit vectors with real components ( ie. all entries are real ). 

*$A,B,C$ are orthogonal to each other. 


I want to find the eigen values of $M$ , I suspect they are $T^2_1,T^2_1,T^2_3$ but can't prove it.  ( Although I think it is general, in my case $T_3=1,T_1=T_2=sin(2\theta)$ for some angle $\theta$. )
 A: Maybe this can help
Consider eigenvector $v = \frac{1}{{{B_x}{B_y}{B_z}{C_x}{C_y}{C_z}}}A$, then apply orthogonality of vectors and unity.
$Mv = \frac{{T_1^2}}{{{B_x}{B_y}{B_z}{C_x}{C_y}{C_z}}}\left[ {\begin{array}{*{20}{c}}
{{A_x}\left( {A_x^2 + A_y^2 + A_z^2} \right)}\\
{{A_y}\left( {A_x^2 + A_y^2 + A_z^2} \right)}\\
{{A_z}\left( {A_x^2 + A_y^2 + A_z^2} \right)}
\end{array}} \right] + T_2^2\left[ {\begin{array}{*{20}{c}}
{\frac{1}{{{B_y}{B_z}{C_x}{C_y}{C_z}}}\left( {{A_x}{B_x} + {A_y}{B_y} + {A_z}{B_z}} \right)}\\
{\frac{1}{{{B_x}{B_z}{C_x}{C_y}{C_z}}}\left( {{A_x}{B_x} + {A_y}{B_y} + {A_z}{B_z}} \right)}\\
{\frac{1}{{{B_x}{B_y}{C_x}{C_y}{C_z}}}\left( {{A_x}{B_x} + {A_y}{B_y} + {A_z}{B_z}} \right)}
\end{array}} \right] + T_2^2\left[ {\begin{array}{*{20}{c}}
{\frac{1}{{{B_x}{B_y}{B_z}{C_y}{C_z}}}\left( {{A_x}{C_x} + {A_y}{C_y} + {A_z}{C_z}} \right)}\\
{\frac{1}{{{B_x}{B_y}{B_z}{C_x}{C_z}}}\left( {{A_x}{C_x} + {A_y}{C_y} + {A_z}{C_z}} \right)}\\
{\frac{1}{{{B_x}{B_y}{B_z}{C_x}{C_y}}}\left( {{A_x}{C_x} + {A_y}{C_y} + {A_z}{C_z}} \right)}
\end{array}} \right]$
$M\frac{1}{{{B_x}{B_y}{B_z}{C_x}{C_y}{C_z}}}A = \frac{{T_1^2}}{{{B_x}{B_y}{B_z}{C_x}{C_y}{C_z}}}A$
which shows $T_1^2$ is an eigenvalue. You can proceed for other eigenvalues.
A: The easiest way is to work matricially.
Let us consider $A,B,C$ as element of $M_{3,1}(\mathbb{R})$ (vectors viewed as column matrices). Then, it is easy to check that :
$$M = T_1^2 A A^t + T_2^2 B B^t + T_3^2 C C^t$$
where $P^t \in M_{q,p}(\mathbb{R})$ design the transpose of the matrix $P \in M_{p,q}(\mathbb{R})$ (so that here $A^t,B^t,C^t$ are row matrices, i.e. elements of $M_{1,3}(\mathbb{R})$).
The orthonormality condition then reads :
$$ \begin{array}{ccc}
A^t A=1 & A^t B=0 & A^t C=0 \\
B^t A=0 & B^t B=1 & B^t C=0 \\
C^t A=0 & C^t B=0 & C^t C=1 \\
\end{array}$$
Then, it is easy to check, using the associativity of matrix multiplication that :
$$MA=T_1^2A, \; \; \; MB=T_2^2B, `\; \; \; MC=T_3^2C$$
Thus, the eigenvectors are $A,B,C$ and the eigenvalues are the $T_i^2$.
For example, for A :
$$MA = T_1^2 A (A^tA) + T_2^2 B (B^tA) + T_3^2 C (C^tA)=T_1^2A$$
