Question-37
Let \(A\in \mathbb{R}^{n\times n}\) be such that \(A^{3} =A\). Which one of the following statements is always correct?
Let us work with the equation given:
\[ \begin{array}{ r l l } & A^{3} & =A\\ \Longrightarrow & A\left( A^{2} -I\right) & =0\\ \Longrightarrow & A( A+I)( A-I) & =0 \end{array} \]
A few observations follow:
- \(A=0\) satisfies the above equation. This suggests that \(A\) is not always invertible.
- \(A=I\) satisfies the above equation. This suggests that the determinant of \(A\) is not always zero. This also suggests that the trace of \(A\) is not always \(1\).
By elimination, we are left with the last option. However, let us go ahead and prove this. Note that matrix multiplication can never increase the rank. That is, if \(A\) and \(B\) are compatible for multiplication, then \(\text{rank}( AB) \leqslant \underset{}{\min}\left(\text{rank}( A) ,\text{rank}( B)\right)\). From this, we have:
\[ \text{rank}\left( A^{3}\right) \leqslant \text{rank}\left( A^{2}\right) \leqslant \text{rank}( A) \]
Since \(A=A^{3}\), we see that \(A\) and \(A^{2}\) have the same rank.