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Given that a matrix $A_{3\times3},$which is not idempotent matrix.And $A^{3}=A.$
Then find them,
$1)$ Eigen Values
$2)$ Trace of the matrix$=$Sum of Leading Diagonal Elements$=\sum a_{ij},$ where $i=j$
$3)Det(A)$

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0
is it correct

A^3= A

A^3-A=0

A(A^2-I)=0

A(A+I)(A-I)=0

A= 0, A=I OR A=-I am going in right way ?
+3

$A^{3}=A$

$|A^{3}|=|A|$--->(1)

for eigen values:Characteristic equation $|A-\lambda I|=0$

$\Rightarrow|A|=|\lambda I|$

$\Rightarrow|A|=|\lambda |$

Cayley–Hamilton Theorem: A square matrix satisfies its own characteristic equation.

Put the value in the equation $(1),$

$|\lambda|^{3}=|\lambda|$

$\lambda^{3}-\lambda=0$

$\lambda(\lambda^{2}-1=0$

$\lambda(\lambda+1).(\lambda-1)=0$

so $,\lambda=0,-1,1$

0

If   A3=A then what do you mean by not idempotent  matrix???

+1
$A^{2}\neq A$

1