- 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
2.16 Exercises
10. $\quad$ Find all 2 x 2 matrices $A$ such that $A^2 = O.$
Solution. $\quad$ We wish to find a matrix $A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ such that:
\begin{align*}
A^2 &= \begin{bmatrix}a & b \\ c & d\end{bmatrix} \begin{bmatrix}a & b \\ c & d\end{bmatrix}
\\
&=
\begin{bmatrix}
a^2 + bc & ab + bd
\\
ca + dc & cb + d^2
\end{bmatrix}
\\
&= \begin{bmatrix}
0 & 0
\\
0 & 0
\end{bmatrix}
\end{align*}
This gives us the following system of equations:
\begin{align*}
\\
a^2 + bc &= 0
\\
(a + d)b &= 0
\\
(a + d)c &= 0
\\
bc + d^2 &= 0
\end{align*}
In particular, we have $(a + d)b = (a + d)c = 0.$ From this, we can derive two cases.
(1) $\quad$ $(a + d) \neq 0. \quad$ In this case, we can divide the equation by $(a + d)$ to give $b = c = 0.$ But as we can see, plugging this into the equations $a^2 + bc = 0$ and $bc + d^2 = 0,$ we find that $a = d = 0,$ leading to a contradiction. Thus, $(a + d)$ must be zero.
(2) $\quad$ $(a + d) = 0. \quad$ As a result of the contradiction reached in case (1), we find that $(a + d)$ must be zero. As such, we have $d = -a$ and $a^2 = d^2 = -bc.$ This gives us the following set of $2 \times 2$ matrices such that $A^2 = O:$
\begin{align*}
A &= \begin{bmatrix}
a & b
\\
c & -a
\end{bmatrix}
\end{align*}
where $b$ and $c$ are arbitrary scalars, with $a$ satisfying $a^2 = -bc. \quad \blacksquare$