Find eigenvalues of unspecified matrix

Find all possible eigenvalues of a $2\times 2$ matrix $A$ satisfying $$\det(A^2)I-2\det(A)A+A^2=0.$$

Well, if $Av=\lambda v$ then $$\det(A^2)v-2\det(A)\lambda v+\lambda^2 v=(\det(A)^2-2\det(A)\lambda+\lambda^2) v=(\det(A)-\lambda)^2v=0$$ so $\lambda=\det(A)$. Can we go further and find $\det(A)$? Or perhaps a set of possible values? We have $$0=\det(\det(A^2)I-2\det(A)A+A^2)=(\det(\det(A)I-A))^2$$ so $$\det(A-\det(A)I)=0.$$ Can we solve for $\det(A)$?

Edit: We even have $$(A-\det(A)I)^2=0$$ which is a set of 4 equations in 4 unknowns, so perhaps this fully determines $A$. But how?

Solution 1:

You've shown that if $A$ satisfies $\det(A^2)I-2\det(A)A+A^2=0$, then any eigenvalue $\lambda$ satisfies $\lambda = \det(A)$, i.e. all the eigenvalues are equal to $\det(A)$.

Now, remember that the determinant of a matrix is the product of its eigenvalues (counting multiplicity). Since $A$ is a $2 \times 2$ matrix, it should have two eigenvalues. Thus, $\lambda \cdot \lambda = \det(A) = \lambda$.

Solving this shows that $\lambda = 0$ or $\lambda = 1$. You can easily check that $A = O_{2 \times 2}$ (the $2 \times 2$ zero matrix) and $A = I_{2 \times 2}$ (the $2 \times 2$ identity matrix) both satisfy the given equation. Hence, both $\lambda = 0$ and $\lambda = 1$ are possible eigenvalues.

Note: If you know stuff about Jordan canonical form, then you can show that the $2 \times 2$ matrices $A$ which satisfy $\det(A^2)I-2\det(A)A+A^2=0$ are precisely those which can be expressed in the form $A = VJV^{-1}$, where $V$ is an invertible $2 \times 2$ matrix and $J$ is one of the four matrices $\left\{\begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}, \begin{bmatrix}1&1\\0&1\end{bmatrix}\right\}$.