Q: If $\displaystyle A$ is an idempotent matrix, prove that the determinant of the matrix $\displaystyle A$ is either $\displaystyle 0$ or $\displaystyle 1$.

A: Suppose $\displaystyle A$ is idempotent. Then, by definition $\displaystyle A^{2}=A$; thus, $\displaystyle A$ has either a column of zeros, a row of zeros, or is the identity matrix (we will omit the trivial case when $\displaystyle A$ is the zero matrix).

Case 1: If $\displaystyle A$ has a row of zeros

$\displaystyle \Rightarrow det(A)= \sum_{j}a_{ij}C_{ij}=C_{i1}0+C_{i2}0+...+C_{im}=0$

Case 2: If $\displaystyle A$ has a column of zeros

$\displaystyle \Rightarrow det(A)=\sum_{i}a_{ij}C_{ij}=C_{1j}0+C_{2j}0+...+C_ {im}=0$

Case 3: If $\displaystyle A$ is the identity matrix, then each $\displaystyle a_{ij}C_{ij}$ of the determinant is equal to $\displaystyle 1$ or $\displaystyle 0$. Clearly, $\displaystyle a_{ij}C_{ij}=0$ when $\displaystyle a_{ij}=0$ or $\displaystyle M_{ij}=0$. Conversely, $\displaystyle a_{ij}C_{ij}=1$ only if $\displaystyle a_{ij}=1$ and $\displaystyle M_{ij}=1$. Observe that, $\displaystyle M_{ij}=1$ only if $\displaystyle M_{ij}$ is the minor of some $\displaystyle a_{ij}$ along the diagonal. Moreover, $\displaystyle a_{ij}=1$ only appears when $\displaystyle i=j$ (along the diagonal); thus, $\displaystyle C_{ij}=(-1)^{i+j}M_{ij}$ will always be positive. Therefore, regardless of whether you expand by a row or column, $\displaystyle a_{ij}=1$ and $\displaystyle M_{ij}=1$ only when $\displaystyle i=j$ which occurs exactly once in any row or column; as a result, $\displaystyle det(A)=1$ for any $\displaystyle A=I_{n}$.

Does that work? I know you can multiply all the terms of the diagonal to find the determinant if you have an upper or lower traingular, so it makes sense that if $\displaystyle A=I_{n} \Rightarrow det(A)=1$. Even so, I am not sure if I described everything correctly. Also, I am not sure I covered every case, given $\displaystyle A^2=A$.

Thanks