If $A$ is a square matrix

Question

If $A$ is a square matrix, and $A=A^2$, then what would the possible values of $|A|$?

I've tried to calculate it through basic mathematics, however I feel it's not appropriate... $$A=A^2$$ $$A-A^2=0$$ $$A=0 \text { or } A=1$$ So $A$ will be a zero matrix or identity matrix. Hence $|A|$ will be $0$ & $1$ respectively.

But I doubt if we can perform such operations in case of matrices... Also, I feel there may be more possible solutions, rather than 0 or unity. Please Guide Accordingly.

Answer 1

I assume that $|A|$ is the determinant of the matrix $A$

If $A=A^2$ then what can you say about $|A|$ and $|A^2|$ ?

Also remember that for any two square matrices of the same size A and B we have : $|AB| = |A||B|$

Hope it helps

Answer 2

In general, $\det(AB)=\det(A)\det(B)$, so $\det(A^2)=\det(A)^2$. We are given that $A^2=A$, so $\det(A)=\det(A)^2$, which implies that $\det(A)=0$ or $\det(A)=1$. As Arthur mentions in the comments, you need to be careful about the logic: what we have proven is

For all square matrices $A$, if $A^2=A$ then $\det(A)=0$ or $\det(A)=1$.

This is not the same as saying that

For all square matrices $A$, if $\det(A)=0$ or $\det(A)=1$, then $A^2=A$.

Your argument is not correct because from $A-A^2=0$ you cannot infer that $A$ is the zero matrix or the identity matrix. You can infer from $A-A^2=0$ that $A(I-A)=0$, by distributivity; however, there are many matrices that satisfy this equation which are not the zero matrix or the identity matrix, e.g. $\begin{bmatrix}3 & -6 \\ 1 & -2\end{bmatrix}$. This shows that, for matrix multiplication, $AB=0$ does not imply that $A=0$ or $B=0$. (In the language of abstract algebra, we can say that the ring of $n\times n$ matrices with coefficients in $\mathbb R$ contains nontrivial zero divisors, for $n>1$.)