Question

Let $A$ be an $n \times n$ matrix such that $A^3 = A$. Then which one of the following is true?

• A. $A$ must be an identity matrix.
• B. $A^2$ must be an identity matrix.
• C. $A$ is invertible.
• D. $A$ is diagonalisable.

Hint:

Any matrix satisfying an equation of the form $A^k = A$ (for $k > 1$) is diagonalizable because the polynomial $x^k - x = 0$ has only distinct, single-multiplicity roots.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question is from linear algebra. It concerns the properties of an $n \times n$ matrix $A$ satisfying the algebraic equation $A^3 - A = 0$.

Step 2: Key Formula or Approach:
A matrix is diagonalizable over a field if and only if its minimal polynomial $m(x)$ factors into distinct linear factors over that field.

Step 3: Detailed Explanation:

• The given equation is $A^3 = A$, which can be written as:
\[ A^3 - A = 0 \]
• This means that the polynomial $p(x) = x^3 - x$ annihilates the matrix $A$.

• We can factor this polynomial completely over the real numbers:
\[ p(x) = x(x^2 - 1) = x(x - 1)(x + 1) \]
• The minimal polynomial $m(x)$ of $A$ must divide the annihilating polynomial $p(x)$.

• Since $p(x)$ has only distinct linear roots ($0$, $1$, and $-1$), any divisor of $p(x)$ must also have only distinct linear roots.

• According to the diagonalization theorem, because the minimal polynomial of $A$ has only distinct linear factors, the matrix $A$ is diagonalizable.

• Let's check why the other options are not always true:

• If $A = 0$ (the zero matrix), then $A^3 = 0 = A$, but $A$ is not the identity matrix. Thus, option (A) is not necessarily true.

• If $A = 0$, then $A^2 = 0 \neq I$. Thus, option (B) is not necessarily true.

• If $A = 0$, the matrix is singular (not invertible). Thus, option (C) is not necessarily true.

Step 4: Final Answer:
The matrix $A$ must be diagonalizable, which corresponds to option (D).