Step 1

According to the given information, it Let A and B be \(n \times n\) similar matrices.

If A is idempotent the show that B is idempotent.

\(A \text{ is idempotent } \Rightarrow A^2=A\)

\(\text{ to show } B \text{ is idempotent } B^2=B\)

Step 2

A and B are similar matrices so, there exist an invertible matrix P such that: \(B=P^{-1}AP ...(A)\)

Step 3

Square both sides:

\(B^2=(P^{-1}AP)(P^{-1}AP)\)

\(B^2=P^{-1}A(PP^{-1})AP\left[PP^{-1}=I\right]\)

\(B^2=P^{-1}A(I)AP\)

\(B^2=P^{-1}A^2P\)

\(B^2=P^{-1}AP \left[ \text{by given condition } A^2=A \right]\)

\(B^2=B \left[ \text{from equation } (A) \right]\)

Therefore, B is also an idempotent matrix.

According to the given information, it Let A and B be \(n \times n\) similar matrices.

If A is idempotent the show that B is idempotent.

\(A \text{ is idempotent } \Rightarrow A^2=A\)

\(\text{ to show } B \text{ is idempotent } B^2=B\)

Step 2

A and B are similar matrices so, there exist an invertible matrix P such that: \(B=P^{-1}AP ...(A)\)

Step 3

Square both sides:

\(B^2=(P^{-1}AP)(P^{-1}AP)\)

\(B^2=P^{-1}A(PP^{-1})AP\left[PP^{-1}=I\right]\)

\(B^2=P^{-1}A(I)AP\)

\(B^2=P^{-1}A^2P\)

\(B^2=P^{-1}AP \left[ \text{by given condition } A^2=A \right]\)

\(B^2=B \left[ \text{from equation } (A) \right]\)

Therefore, B is also an idempotent matrix.