If \(A\) is an \(n\times n\) symmetric matrix with eigenvectors \(\{v_1,\ldots,v_n\}\) which form an orthonormal basis of \(R^n\) corresponding to eigenvalues \(\lambda_1,\ldots,\lambda_n\), then

\[\begin{equation} e^{tA}=e^{\lambda_1t}v_1v_1^T+\ldots+e^{\lambda_nt}v_nv_n^T \end{equation}\]

Derivation:

The spectral decomposition of \(A\) is

\[\begin{equation} A=\lambda_1v_1v_1^T+\ldots+\lambda_nv_nv_n^T \end{equation}\]

Define the matrix

\[\begin{equation} V_i=v_iv_i^T \end{equation}\]

There are a few properties of these matrices which will be important later. Proofs are left as an exercise for the reader (they’re not that bad). Using the fact that the vectors \(v_1,\ldots,v_n\) form an orthonormal basis:

\[\begin{align} &V_iV_j=0,\quad i\neq j\\ &V_iV_i=V_i\\ \implies &V_iV_j=V_jV_i \end{align}\]

Now we can rewrite \(A\) as a sum of commuting matrices.

\[\begin{equation} A=\lambda_1V_1+\ldots+\lambda_nV_n \end{equation}\]

We establish a few important properties of matrix exponentials

\[\begin{align} AB=BA&\implies e^{A+B}=e^Ae^B\\ P^2=P&\implies e^{tP}=I+(e^t-1)P\\ \end{align}\]

The latter, I believe, deserves proof.

Suppose \(P^2=P\) which trivially implies that for any natural number \(n\), \(P^n=P\).

\[\begin{equation} e^{tP}=\sum_{n=0}^\infty \frac{(tP)^n}{n!}=I+\left(\sum_{n=1}^\infty \frac{t^n}{n!}\right)P=I+(e^t-1)P \end{equation}\]

Using the established properties of matrix exponentials

\[\begin{gather} e^{tA}=e^{t(\lambda_1V_1+\ldots+\lambda_nV_n)}\\ e^{tA}=e^{\lambda_1tV_1}\ldots e^{\lambda_ntV_n}\\ e^{tA}=(I+(e^{\lambda_1t}-1)V_1)\ldots(I+(e^{\lambda_nt}-1)V_n)\\ e^{tA}=I+(e^{\lambda_1t}-1)V_1+\ldots+(e^{\lambda_nt}-1)V_n \end{gather}\]

The last line definitely requires proof. We will use induction.

Base case \(n=2\):

\[\begin{gather} (I+(e^{\lambda_1t}-1)V_1)(I+(e^{\lambda_2t}-1)V_2)\\ =I^2+(e^{\lambda_1t}-1)V_1+(e^{\lambda_2t}-1)V_2+(e^{\lambda_1t}-1)(e^{\lambda_2t}-1)V_1V_2\\ =I+(e^{\lambda_1t}-1)V_1+(e^{\lambda_2t}-1)V_2 \end{gather}\]

Now suppose it is true for \(n=k\):

\[\begin{equation} (I+(e^{\lambda_1t}-1)V_1)\ldots(I+(e^{\lambda_kt}-1)V_k)=I+(e^{\lambda_1t}-1)V_1+\ldots+(e^{\lambda_kt}-1)V_k \end{equation}\]

It follows

\[\begin{gather} (I+(e^{\lambda_1t}-1)V_1)\ldots(I+(e^{\lambda_kt}-1)V_k)(I+(e^{\lambda_{k+1}t}-1)V_{k+1})\\ =\bigg[(I+(e^{\lambda_1t}-1)V_1)\ldots(I+(e^{\lambda_kt}-1)V_k)\bigg] (I+(e^{\lambda_{k+1}t}-1)V_{k+1})\\ =\bigg[I+(e^{\lambda_1t}-1)V_1+\ldots+(e^{\lambda_kt}-1)V_k\bigg] (I+(e^{\lambda_{k+1}t}-1)V_{k+1})\\ =\bigg[I+(e^{\lambda_1t}-1)V_1+\ldots+(e^{\lambda_kt}-1)V_k\bigg] +\bigg[I+(e^{\lambda_1t}-1)V_1+\ldots+(e^{\lambda_kt}-1)V_k\bigg] (e^{\lambda_{k+1}t}-1)V_{k+1}\\ =I+(e^{\lambda_1t}-1)V_1+\ldots+(e^{\lambda_kt}-1)V_k +(e^{\lambda_{k+1}t}-1)V_{k+1}+ (e^{\lambda_{k+1}t}-1)((e^{\lambda_1t}-1)V_1V_{k+1}+\ldots+(I+(e^{\lambda_kt}-1)V_kV_{k+1})\\ =I+(e^{\lambda_1t}-1)V_1+\ldots+(e^{\lambda_kt}-1)V_k +(e^{\lambda_{k+1}t}-1)V_{k+1}\\ \end{gather}\]

Nearing the end we rewrite the exponential,

\[\begin{gather} e^{tA}=I+(e^{\lambda_1t}-1)V_1+\ldots+(e^{\lambda_nt}-1)V_n\\ e^{tA}=e^{\lambda_1t}V_1+\ldots+e^{\lambda_nt}V_n+I-(V_1+\ldots+V_n) \end{gather}\]

Proposition:

If \(v_1,\ldots,v_n\) form an orthonormal basis of \(R^n\), then

\[\begin{equation} v_1v_1^T+\ldots+v_nv_n^T=I \end{equation}\]

Proof:

Any vector \(w\in R^n\) can be written as a linear combination of \(v_1,\ldots,v_n\) because it is a basis.

\[\begin{equation} w=c_1v_1+\ldots+c_nv_n \end{equation}\]

Multiply \(w\) by \(v_1v_1^T+\ldots+v_nv_n^T\):

\[\begin{gather} (v_1v_1^T+\ldots+v_nv_n^T)w=(v_1v_1^T+\ldots+v_nv_n^T)(c_1v_1+\ldots+c_nv_n)\\ =v_1v_1^T(c_1v_1+\ldots+c_nv_n)+\ldots+v_nv_n^T(c_1v_1+\ldots+c_nv_n)\\ =v_1(c_1v_1^Tv_1+\ldots+c_nv_1^Tv_n)+\ldots+v_n(c_1v_n^Tv_1+\ldots+c_nv_n^Tv_n)\\ =v_1(c_1(v_1\cdot v_1)+\ldots+c_n(v_1\cdot v_n))+\ldots+v_n(c_1(v_n\cdot v_1)+\ldots+c_n(v_n\cdot v_n))\\ =v_1(c_1(1)+0)+\ldots+(0+c_n(1))\\ =c_1v_1+\ldots+c_nv_n=w\\ (v_1v_1^T+\ldots+v_nv_n^T)w=w \end{gather}\]

Since this transformation leaves all vectors in \(R^n\) unchanged it’s clear that \(v_1v_1^T+\ldots+v_nv_n^T\) is the identity transformation.

Therefore,

\[\begin{equation} I-(V_1+\ldots+V_n)=0 \end{equation}\]

and so it follows

\[\begin{equation} e^{tA}=e^{\lambda_1t}v_1v_1^T+\ldots+e^{\lambda_nt}v_nv_n^T \end{equation}\]

or more generally for an orthogonal set of vectors,

\[\begin{equation} e^{tA}=\frac{e^{\lambda_1t}}{|v_1|^2}v_1v_1^T+\ldots+\frac{e^{\lambda_nt}}{|v_n|^2}v_nv_n^T \end{equation}\]

An Application

This would make solving nonhomogeneous systems of first-order linear differential equations with a symmetric coefficient matrix much easier.

If \(A\) is an \(n\times n\) symmetric matrix with eigenvectors \(\{\vec{v}_1,\ldots,\vec{v}_n\}\) which form an orthogonal basis of \(R^n\) corresponding to eigenvalues \(\lambda_1,\ldots,\lambda_n\).

Then the nonhomogeneous system of first order linear differential equations

\[\begin{equation} \vec{x}'=A\vec{x}+\vec{g}(t),\quad \vec{x}(t_0)=\vec{x}_0 \end{equation}\]

has the solution

\[\begin{equation} \vec{x}(t) =\sum _{k=1}^n \left[\frac{e^{\lambda_kt}}{|\vec{v}_k|^2}\left\langle \vec{v} _k\cdot\left(\int _{t_0}^te^{-\lambda_ku}\,\vec{g}(u)\,du+\vec{x}_0\right)\right\rangle\vec{v}_k\right] \end{equation}\]

Proof:

Suppose the solution to the differential equation is of the form

\[\begin{equation} \vec{x}(t)=e^{tA}\vec{z}(t)+e^{tA}\vec{x} _0 \end{equation}\]

Subsitituting into the differential equation,

\[\begin{gather} \vec{x}'(t)=Ae^{tA}\vec{z}(t)+e^{tA}\vec{z}'(t)+Ae^{tA}\vec{x} _0=A(e^{tA}\vec{z}(t)+e^{tA}\vec{x} _0)+\vec{g}(t)\\ e^{tA}\vec{z}'(t)=\vec{g}(t)\\ \vec{z}'(t)=e^{-tA}\vec{g}(t)\\ \vec{z}(t)=\int _{t_0}^te^{-uA}\vec{g}(u)du\\ \vec{x}(t)=e^{tA}\int_{t_0}^te^{-uA}\vec{g}(u)du+e^{tA}\vec{x} _0\\ \vec{x}(t)=\int _{t_0}^te^{(t-u)A}\vec{g}(u)du+e^{tA}\vec{x} _0\\ \vec{x}(t)=\int _{t _0}^t\left(\sum _{k=1}^n\left[e^{\lambda _k(t-u)}\:\vec{v} _k\:\vec{v} _k^T\right]\right)\vec{g}(u)\,du +\sum _{k=1}^n\left[e^{\lambda _kt}\:\vec{v} _k\:\vec{v} _k^T\right]\vec{x} _0\\ \vec{x}(t)=\sum _{k=1}^n\left[\int _{t _0}^te^{\lambda _k(t-u)}\:\vec{v} _k\:\vec{v} _k^T\vec{g}(u)\,du\right] +\sum _{k=1}^n\left[e^{\lambda _kt}\vec{v} _k\:\vec{v} _k^T\vec{x} _0\right]\\ \vec{x}(t)=\sum _{k=1}^n\left[e^{\lambda _kt}\int _{t _0}^te^{-\lambda _ku}\:\vec{v} _k\:\vec{v} _k^T\vec{g}(u)\,du\right] +\sum _{k=1}^n\left[e^{\lambda _kt}\:\vec{v} _k\:\left\langle\vec{v} _k^T\vec{x} _0\right\rangle\right]\\ \vec{x}(t)=\sum _{k=1}^n\left[e^{\lambda _kt}\int _{t _0}^t\vec{v} _k\left\langle\vec{v} _k^T\left(e^{-\lambda _ku}\:\vec{g}(u)\right)\right\rangle\,du\right] +\sum _{k=1}^n\left[e^{\lambda _kt}\:\vec{v} _k\:\left\langle\vec{v} _k^T\vec{x} _0\right\rangle\right]\\ \vec{x}(t)=\sum _{k=1}^n\left[e^{\lambda _kt}\int _{t _0}^t\vec{v} _k\left\langle\vec{v} _k\cdot\left(e^{-\lambda _ku}\:\vec{g}(u)\right)\right\rangle\,du\right] +\sum _{k=1}^n\left[e^{\lambda _kt}\:\vec{v} _k\:\left\langle\vec{v} _k\cdot\vec{x} _0\right\rangle\right]\\ \vec{x}(t)=\sum _{k=1}^n\left[e^{\lambda _kt}\int _{t _0}^t\left\langle\vec{v} _k\cdot\left(e^{-\lambda _ku}\:\vec{g}(u)\right)\right\rangle\vec{v} _k\,du\right] +\sum _{k=1}^n\left[e^{\lambda _kt}\left\langle\vec{v} _k\cdot\vec{x} _0\right\rangle\:\vec{v} _k\right]\\ \vec{x}(t)=\sum _{k=1}^n\left[e^{\lambda _kt}\left(\int _{t _0}^t\left\langle\vec{v} _k\cdot\left(e^{-\lambda _ku}\:\vec{g}(u)\right)\right\rangle\,du\right)\vec{v} _k\right] +\sum _{k=1}^n\left[e^{\lambda _kt}\left\langle\vec{v} _k\cdot\vec{x} _0\right\rangle\vec{v} _k\right]\\ \vec{x}(t)=\sum _{k=1}^n\left[e^{\lambda _kt}\left\langle\vec{v} _k\cdot\left(\int _{t _0}^te^{-\lambda _ku}\:\vec{g}(u)\,du\right)\right\rangle\vec{v} _k\right]+\sum _{k=1}^n\left[e^{\lambda _kt}\left\langle\vec{v} _k\cdot\vec{x} _0\right\rangle\vec{v} _k\right]\\ \vec{x}(t)=\sum _{k=1}^n\left[e^{\lambda _kt}\left(\left\langle\vec{v} _k\cdot\left(\int _{t _0}^te^{-\lambda _ku}\:\vec{g}(u)\,du\right)\right\rangle+\left\langle\vec{v} _k\cdot\vec{x} _0\right\rangle \right)\vec{v} _k\right]\\ \vec{x}(t)=\sum _{k=1}^n\left[e^{\lambda _kt}\left\langle\vec{v} _k\cdot\left(\int _{t _0}^te^{-\lambda _ku}\:\vec{g}(u)\,du+\vec{x} _0\right)\right\rangle\vec{v} _k\right] \end{gather}\]

Alternatively, if the vectors \(\vec{v}_1,\ldots,\vec{v}_n\) are orthogonal but not necessarily normalized,

\[\begin{equation} \vec{x}(t)=\sum _{k=1}^n\left[\frac{e^{\lambda _kt}}{|\vec{v}_k|^2}\left\langle\vec{v} _k\cdot\left(\int _{t _0}^te^{-\lambda _ku}\:\vec{g}(u)\,du+\vec{x} _0\right)\right\rangle\vec{v} _k\right] \end{equation}\]