- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.21 Miscellaneous review exercises on matrices
4. $\quad$ The matrix $A = \begin{bmatrix} a & i \\ i & b \end{bmatrix},$ where $i^2 = -1,$ $a = \frac{1}{2}(1 + \sqrt{5}),$ and $b = \frac{1}{2}(1 - \sqrt{5}),$ has the property that $A^2 = A.$ Describe completely all $2 \times 2$ matrices $A$ with complex entries such that $A^2 = A.$
Solution. $\quad$ First, we note that the $2 \times 2$ zero matrix $O$ and $2 \times 2$ identity matrix $I$ satisfy $A^2 = A$ trivially. For the general case, we wish to find the set of all $2 \times 2$ matrices $A = \begin{bmatrix} a & b \\ c & d\end{bmatrix}$ with complex entries $a, b, c, d$ satisfying \begin{align*} \begin{bmatrix} a & b \\ c & d\end{bmatrix} \begin{bmatrix} a & b \\ c & d\end{bmatrix} &= \begin{bmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{bmatrix} \\ &= \begin{bmatrix} a & b \\ c & d\end{bmatrix} \end{align*} This gives us the following relations between the entries of $A$ \begin{align*} \\ a^2 + bc &= a \\ b(a + d) &= b \\ c(a + d) &= c \\ d^2 + bc &= d \end{align*} From this, we find that $a + d = 1,$ or $d = 1 - a.$ Then, the set of $2 \times 2$ matrices satisfying $A^2 = A$ are $\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix},$ and those of the form \begin{align*} A &= \begin{bmatrix} a & b \\ c & 1 - a \end{bmatrix} \end{align*} where $b, c$ are arbitrary complex numbers and $a$ is the solution to $a^2 + bc - a = 0. \quad \blacksquare$