Question

If $ Z $ is an idempotent matrix, then $ {{\left( I+Z \right)}^{n}}= $ \[\]
A. $ I+{{2}^{n}}Z $ \[\]
B. $ I+\left( {{2}^{n}}-1 \right)Z $ \[\]
C. $ I-\left( {{2}^{n}}-1 \right)Z $ \[\]
D. None of these \[\]

Answer 1

Hint: We recall the definitions of idempotent matrix, identity matrix and the binomial formula. We use the dentition of idempotent matrix to conclude $ Z={{Z}^{2}}=...={{Z}^{n}} $ for all $ n\in N $ . We expand $ {{\left( I+Z \right)}^{n}} $ using binomial formula for two terms $ x=I,y=Z $ in $ {{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{0}}+...{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}} $ and simplify. \[\]

Complete step by step answer:
We know that binomial is the algebraic expression involving two terms and each term with a distinct variable. We know that we can use the binomial theorem (or binomial expansion) to describe the algebraic expansion of power of a binomial. If $ x,y $ are the two terms of binomial with some positive integral power $ n $ then the binomial expansion is given by;
\[{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{0}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{0}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}\]
The above expression is called binomial formula or binomial identity. Here $ {}^{n}{{C}_{0}},{}^{n}{{C}_{1}},...,{}^{n}{{C}_{n}} $ are called binomial coefficients and their sum given by
\[{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+...+{}^{n}{{C}_{n}}={{2}^{n}}\]

We know that an idempotent matrix is a square matrix that when multiplied with itself yields itself. If $ Z $ is an idempotent matrix then $ Z\times Z=Z $ which means
\[{{Z}^{2}}=Z\]
We multiply $ Z $ with $ {{Z}^{2}} $ and use previously obtained $ {{Z}^{2}}=Z $ to have;
\[{{Z}^{3}}={{Z}^{2}}\cdot Z=Z\cdot Z=Z\]
We can on multiplying $ Z $ with $ {{Z}^{n-1}} $ for all $ n\in N $ and have
\[{{Z}^{n}}=Z\]
 We know that $ I $ is the identity matrix with the same order as $ Z $ . . An identity matrix or unit matrix is a square matrix that has all the entries in the main diagonal as 1 and zeros in the rest of the positions. The identity matrix when multiplied with any matrix results that matrix which means for any matrix $ A $ we have;
\[I\times A=A\times I=A\]
We are asked to find the value of $ {{\left( I+Z \right)}^{n}} $ , We expand $ {{\left( I+Z \right)}^{n}} $ using binomial formula for $ x=I,y=Z $ and have;

\[\begin{align}
  & {{\left( I+Z \right)}^{n}}={}^{n}{{C}_{0}}{{I}^{n}}{{Z}^{0}}+{}^{n}{{C}_{1}}{{I}^{n-1}}{{Z}^{1}}+{}^{n}{{C}_{2}}{{I}^{n-2}}{{Z}^{2}}+...+{}^{n}{{C}_{n}}{{I}^{0}}{{Z}^{n}} \\
 & \Rightarrow {{\left( I+Z \right)}^{n}}={}^{n}{{C}_{0}}I\cdot I+{}^{n}{{C}_{1}}I\cdot Z+{}^{n}{{C}_{1}}I\cdot {{Z}^{2}}+...+{}^{n}{{C}_{2}}I\cdot {{Z}^{n}} \\
 & \Rightarrow {{\left( I+Z \right)}^{n}}={}^{n}{{C}_{0}}I+{}^{n}{{C}_{1}}Z+{}^{n}{{C}_{2}}{{Z}^{2}}+...+{}^{n}{{C}_{n}}{{Z}^{n}} \\
\end{align}\]

We put previously obtained $ Z={{Z}^{2}}=...={{Z}^{n}}=Z $ and have;
\[\begin{align}
  & \Rightarrow {{\left( I+Z \right)}^{n}}=1\cdot I+{}^{n}{{C}_{1}}Z+{}^{n}{{C}_{2}}Z+...+{}^{n}{{C}_{n}}Z \\
 & \Rightarrow {{\left( I+Z \right)}^{n}}=I+\left( {}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}...+{}^{n}{{C}_{n}} \right)Z \\
 & \Rightarrow {{\left( I+Z \right)}^{n}}=I+\left( {{2}^{n}}-1 \right)Z \\
\end{align}\]
Hence the correct option is C. \[\]

Note:
We note that the identity matrix is also an idempotent matrix since $ {{I}^{2}}=I $ and its is the only idempotent matrix that is invertible. A square matrix has the same number of rows and columns. We can add two matrices when they are of the same order. We also note that matrix multiplication of two matrices is not commutative unless one of them is an identity matrix.