# Find the eigenvalues for a matrix which is a product of matrices

Suppose I have a matrix $A \in \mathbb{R}^{2, 2}$ which is the product of $3$ other matrices, lets call them $A_1 = \left(\begin{matrix} cosx & -sinx \\ sinx & cos x\end{matrix}\right)$, $A_2= \left(\begin{matrix} 3 & 0 \\ 0 & 7\end{matrix}\right)$ and $A_3 = \left(\begin{matrix} cosx & sinx \\ -sinx & cos x\end{matrix}\right)$. So the situation is: $$A = A_1 \cdot A_2 \cdot A_3$$

Now, I am asked to find the eigenvalues for this matrix $A$ without multiplying $A_1$, $A_2$ and $A_3$.

In general, if we assume that we don't want zero eigenvectors, then we can find the eigenvalues solving the characteristic polynomial, which can be found by assuming (actually it is a consequence) that: $$\det{(\lambda \cdot I - A)} = 0$$

Then we solve for $\lambda$.

Now, my situation is the following:

$$\det{(\lambda \cdot I - A_1 \cdot A_2 \cdot A_3)} = 0$$

I have found the determinant of $A$, using a property of the determinants, but I don't see any property that can help me to find the determinant just above...

• I can't see how you can make any progress without there being a relationship between $A_1, A_2$ and $A_3$ of some sort. Is $A_3$ the inverse of $A_1$ or anything of that sort? – Paul May 18 '15 at 18:49

Hint $$\det (\lambda I -A_1A_2A_1^{-1})=\det (\lambda A_1A_1^{-1} -A_1A_2A_1^{-1}) = \det ( A_1 (\lambda I -A_2) A_1^{-1}) \\= \det (A_1) \det (\lambda I -A_2) \det (A_1^{-1})$$

• I have found the eigenvalues $\lambda = 3$ and $\lambda = 7$. Now, again without expanding $A$, how would you find the eigenvectors? I know we have to find the eigenspaces by of the null space of the general $\lambda I - A$, but in this case I am not seeing how, without computing $A$ first... – nbro May 18 '15 at 21:35
• @Xenomorph Since $A_2$ is diagonal, its eigenvectors are $e_1, e_2$. Now, if $v$ is an eigenvector of $A_1A_2A_1^{-1}$ we have $$A_1A_2A_1^{-1}v= \lambda v \Rightarrow A_2 (A_1^{-1}v)= \lambda A_1^{-1}v$$ This shows that $A_1^{-1}v$ is an eigenvector for $A_2$ or equivalently that $$v= A_1u$$ where $u$ is an eigenvector for $A_2$. – N. S. May 18 '15 at 23:21

Similar to comment above, I don't think that the question is properly well defined.

For instance, based on the example in the post below, the eigenvalue of AB cannot be determined by the eigenvalues of A and B. This is related to the diagonalization of a matrix, which basically tells us that a matrix is only uniquely determined when BOTH its eigenvectors and eigenvalues are specified.

Eigenvalues of Matrices and Eigenvalue of product of Matrices

On a related note, the question could be solved without multiplying the matrices if we diagonalize every matrix separately. In 2 dimensions, we have the time property that all rotations commute ($SO_2$ is abelian). From there, I think we can express the eigenvalues of ($A_1 A_2 A_3$) as a function of the individual eigenvalues and the angle of the eigenvectors of the composite matrices.

Is this the form of the solution that you are looking for? If so, I can work out the details.

• I have updated my question with the concrete matrices... – nbro May 18 '15 at 20:13