Maths WorkshopMaths problems with full solutions
Exercise 1658Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

Functions of a diagonalizable matrix

#functions of matrices#diagonalization#polynomial

Statement

Let A=PDP−1A=PDP^{-1} with D=diag(λ1,…,λn)D=\textrm{diag}(\lambda_1,\ldots,\lambda_n). Define f(A)=Pf(D)P−1f(A)=P f(D) P^{-1} where f(D)=diag(f(λ1),…,f(λn))f(D)=\textrm{diag}(f(\lambda_1),\ldots,f(\lambda_n)) for a function ff defined on the eigenvalues. Justify this definition for polynomials ff.

Hint

For a polynomial f(X)=∑kakXkf(X)=\sum_k a_k X^k, f(A)=∑kakAkf(A)=\sum_k a_k A^k.

Show solutionHide solution

Solution

Polynomial case. For f(X)=∑kakXkf(X)=\sum_k a_k X^k:
f(A)=∑kakAk=∑kakPDkP−1=P(∑kakDk)P−1=Pf(D)P−1.f(A)=\sum_k a_k A^k=\sum_k a_k PD^kP^{-1}=P\left(\sum_k a_k D^k\right)P^{-1}=Pf(D)P^{-1}.
So the definition is consistent with the polynomial case.

Extension. For other functions (e.g. exp⁡\exp, sin⁡\sin, log⁡\log), the same formula defines f(A)f(A) provided ff is defined on the spectrum of AA. This requires AA to be diagonalizable.

Example. etA=Pdiag(etλ1,…,etλn)P−1e^{tA}=P\textrm{diag}(e^{t\lambda_1},\ldots,e^{t\lambda_n})P^{-1}.

Conclusion. Functions of a diagonalizable matrix are defined via the diagonal form. For non-diagonalizable matrices, one uses Taylor or Jordan form instead.