Symmetric matrix operator

Let $$A\in\mathbb{R}^{n\times n}$$ be a symmetric matrix with eigen values $$\lambda_1\geq \lambda_2\geq\cdots\geq\lambda_n$$ and spectral decomposition given by $$A = PDP^{\top},$$ where $$D = \text{diag}(\lambda_1,...,\lambda_n)$$ and $$P$$ is orthogonal.

Let $$\mathcal{S}\subset\mathbb{R}$$ be a set such that $$\lambda_i\in\mathcal{S}$$ for all $$i$$, and define some function $$f:\mathcal{S}\to\mathbb{R}$$.

We define the matrix $$f(A)$$ as $$f(A) = Pf(D)P^{\top},$$ where $$f(D) = \text{diag}(f(\lambda_1),...,f(\lambda_n)).$$

Is $$f(A)$$ well defined?

I mean, the spectral decomposition $$A = PDP^{\top}$$ is not unique since there could be $$i,j$$ s.t. $$\lambda_i=\lambda_j$$, which provokes that $$P$$ is not unique.

Any insight will be apreciated!

It is well defined. The reason is that $$f(D)$$ can be expressed as a polynomial in $$D$$ (whose degree and coefficients depend on $$f$$ and $$D$$), which makes $$f(A)$$ expressible as a polynomial in $$A$$. In fact, if $$\mu_1,\mu_2,\ldots,\mu_m$$ are those distinct eigenvalues of $$A$$ and $$g$$ is the Lagrange interpolation polynomial $$g(x)=\sum_{i=1}^m\prod_{j\ne i}\frac{x-\mu_j}{\mu_i-\mu_j}f(\mu_i),$$ then $$g(\mu_i)=f(\mu_i)$$ for each $$i$$. (In the corner case where $$A=\mu I$$, we define $$g$$ as the constant polynomial $$f(\mu)$$.) Hence $$g(\lambda_i)=f(\lambda_i)$$ for each $$i$$ (because each $$\lambda_i$$ is some $$\mu_k$$), $$f(D)=g(D)$$ and $$Pf(D)P^T=Pg(D)P^T=g(PDP^T)=g(A)$$. Since $$g$$ depends only on the unordered multi-set of eigenvalues of $$A$$ but not on $$P$$ or on the order of the $$\lambda_i$$s on the diagonal of $$D$$, the product $$Pf(D)P^T$$ always evaluates to the same value, namely $$g(A)$$, regardless of spectral decomposition performed.

• What if $\lambda_i=\lambda_j$ for some $i\neq j$? Is $g$ well defined?
– RLC
Commented Oct 17, 2020 at 15:06
• @RLC Yes, but I have forgotten to adapt my answer to that case. Will edit in a moment. Commented Oct 17, 2020 at 15:12
• Thanks fo the edit! I don't understand the equality $Pg(D)P^T=g(PDP^T)$. Why is that true? (I suppose that it is not by definition)
– RLC
Commented Oct 17, 2020 at 15:23
• @RLC Since $P^TP=I$, we have $PD^kP^T=(PDP^T)(PDP^T)\cdots(PDP^T)=(PDP^T)^k$ for each $k$. So, if $g(x)=\sum_kc_kx^k$, we have $$Pg(D)P^T=P(\sum_kc_kD^k)P^T=\sum_kc_kPD^kP^T=\sum_kc_k(PDP^T)^k=g(PDP^T).$$ Commented Oct 17, 2020 at 15:43
• Thanks for clarifaying it!
– RLC
Commented Oct 17, 2020 at 16:01

Because $$P^\perp P=I$$, $$A^2 = (PDP^\perp)^2=PDP^{\perp}PDP^{\perp}=PD^2P^\perp \\ A^n = PD^nP,\;\;\; n=1,2,3,\cdots.$$ Define $$P^0=I$$. Then, if $$p$$ is a polynomial, $$p(A)=Pp(D)P^\perp.$$ For the diagonal matrix $$D$$ with diagonal $$d_1,d_2,\cdots,d_N$$, this definition of $$p(D)$$ gives the diagonal matrix with diagonal $$p(d_1),p(d_2),\cdots,p(d_N)$$. This is consistent with standard evaluation of a polynomial with the matrix $$A$$. And you can see that $$(pq)(A)=p(A)q(A)$$ for example.