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$A^2−A=0$, where A is a $9×9$ matrix. Then

(a) A must be a zero matrix (b) A is an identity matrix

(c) rank of A is $1$ or $0$ (d) A is diagonalizable.

here the eigen values of the matrix are 0 and 1. so (a) and (c) are not true. i am confused between the other two options.

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up vote 3 down vote accepted

For (d), the matrix $A$ is diagonalizable since its minimal polynomial is a divisor of $x^2 - x = x(x-1)$, which is a product of distinct linear factors.

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a) is false because $A$ can be the identity. b) is false because $A$ can be zero. c) is false because $A$ can be the identity, which has rank $9$ in this case.

for d), notice that since $A^2=A$, $\text{ker}(A) \cap \text{im}(A) = 0$. Thus for dimension reasons we have $\mathbb{R}^9 = \text{ker}(A)\oplus \text{im}(A)$. All vectors in $\text{ker}(A)$ have eigenvalue $0$ and since $A^2 = A$ it follows that all vectors in $\text{im}(A)$ have eigenvalue $1$. Hence there is a basis of $\mathbb{R}^9$ consisting of eigenvectors of $A$ so $A$ is diagonalizable.

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This is just nitpicking, but shouldn't it be "hence $\Bbb R^9$ has a basis consisting of eigenvectors of $A$ so $A$ is diagonalizable"? – Arthur Nov 16 '12 at 12:45
Yes, you are right. Thanks. – Seth Nov 16 '12 at 12:49

Hint: If you already use elimination, consider a block matrix where the first block is the identity and the second is zero. Use this to rule out $(b)$.

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