Exercise 939Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

Powers via diagonalization

#diagonalization#powers#matrix

Statement

Let $A=PDP^{-1}$ with $D=\textrm{diag}(\lambda_1,\ldots,\lambda_n)$ . Show that for all $k\ge1$ ,
$A^k=PD^kP^{-1}=\textrm{diag}(\lambda_1^k,\ldots,\lambda_n^k)$
in the appropriate basis.

Hint

Prove by induction using $A=PDP^{-1}$ .

Show solutionHide solution

Solution

Induction. For $k=1$ , $A^1=PDP^{-1}$ . Suppose $A^k=PD^kP^{-1}$ . Then
$A^{k+1}=A^k\cdot A=PD^kP^{-1}\cdot PDP^{-1}=PD^kDP^{-1}=PD^{k+1}P^{-1}.$

Diagonal power. $D^k=\textrm{diag}(\lambda_1^k,\ldots,\lambda_n^k)$ .

Conclusion. $A^k=PD^kP^{-1}$ . This formula is used to compute powers efficiently once $A$ is diagonalized.

Application. If $D=\textrm{diag}(2,3)$ and $P=I$ , then $A^k=\textrm{diag}(2^k,3^k)$ .

Powers via diagonalization

Statement

Solution

More problems in Eigenvalues and diagonalisation