Exercise 4 If A is diagonalizable we have a fast way to comp

Exercise 4. If A is diagonalizable we have a fast way to compute A^n. Prove that if B^-1AB = D is diagonal A^n = BD^nB^-1 for all n >= 1.

Solution

Let A be diagonalisable

Then For A thre exists a B matrix and B inverse such that

B^-1AB = D is a diagonal matrix

Or A = BDB^-1

A*A = BDB^-1 *BDB^-1 = BD²B^-1

As the statement is true for n =2

Let us prove this by induction.

Let the statement be true for n =k

i.e. A^k = BD^kB^-1

Consider

A^k+1 = A^k*A = BD^kB^-1(A)

But A = BDB^-1

Substitute to get

A^k+1 = BD^kB^-1 BDB^{-1 =} BD^kDB-1 = BD^k+1 B-1

Thus if true for k, it is true for k+1

Already true for k =2

Hence true for n =2,3,4......