Question

If A is idempotent matrix, then show that
`(A+I)^(n) = I+(2^(n)-1) A, AAn in N,` where I is the identity
matrix having the same order of A.

Answer 1

`because ` A is idempotent matrix
`therefore A^(2) = A`,
similarly `A = A^(2) = A^(3) =A^(4) =... = A^(n)` …(i)
Now, `(A+I)^(n) = (I+A)^(n)`
`I+""^(n)C_(1)A+""^(n) C_(2) A^(2)+""^(n)C_(3) A^(3) +...+ ""^(n) C_(n) A^(n) `
`I+(""^(n)C_(1)+""^(n) C_(2) +""^(n)C_(3) +...+ ""^(n) C_(n)) A ` [ from Eq.(i)]
`I+(2^(n)-1)A`
Hence, `(A+I)^(n) = I + (2^(n)-1) A, AA n in N.`