Let $A\in \mathbb{R}^{n \times n}$ be a symmetric matrix such that $A^3=A^2$. Prove that $A^2=A$.
Solution
Since the matrix $A$ is symmetric then it is known that $A$ is diagonizable. The polynomial $P(x)=x^3-x^2$ has $A$ as a root (from the assumptions). This means that $P$ divides the minimal polynomial of $A$.
However the minimal polynomial of $A$ (since $A$ is symmetric) is known to be $m_A(x)=x(x-1)$. On the other hand $A$ is a root of the minimal polynomial, that is $m_A(A)=0$. Hence:
$$m_A(A)=0 \Leftrightarrow A(A-1)=0 \Leftrightarrow A^2=A$$
and this is the conclusion we wanted.
Solution
Since the matrix $A$ is symmetric then it is known that $A$ is diagonizable. The polynomial $P(x)=x^3-x^2$ has $A$ as a root (from the assumptions). This means that $P$ divides the minimal polynomial of $A$.
However the minimal polynomial of $A$ (since $A$ is symmetric) is known to be $m_A(x)=x(x-1)$. On the other hand $A$ is a root of the minimal polynomial, that is $m_A(A)=0$. Hence:
$$m_A(A)=0 \Leftrightarrow A(A-1)=0 \Leftrightarrow A^2=A$$
and this is the conclusion we wanted.
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