## A fundamental matrix of a linear system of first-order differential equations

All computations take place over $\mathbb{R}$.

Let $y_i$, $1 \leq i \leq n$, be differentiable functions of the real variable $t$ which are related as follows: $\frac{d}{dt} y_j = \sum_i a_{i,j} y_i$. Letting $Y = [y_1\ \ldots\ y_n]^\mathsf{T}$ and letting $A = [a_{i,j}]$, we can write this system as $\frac{d}{dt} Y = AY$. Any matrix $M$ whose columns are linearly independent solutions of this equation is called a fundamental matrix of the equation. (From the theory of differential equations, we know that the solutions of this equation are an $n$-dimensional $\mathbb{R}$-vector space, of which the columns of a fundamental matrix form a basis.)

Prove that $\mathsf{exp}(At)$ is a fundamental matrix of $\frac{d}{dt} Y = AY$. Show also that if $C = [c_1\ \ldots\ c_n]^\mathsf{T}$ is an $n \times 1$ constant matrix (i.e., entries not dependent on $t$) then $Y = \mathsf{exp}(At)C$ is the particular solution of this equation satisfying $Y(0) = C$.

Recall from this previous exercise that $\mathsf{exp}(At)$ is nonsingular, and so its columns are linearly independent. Now write $\mathsf{exp}(At) = [Y_1\ \ldots\ Y_n]$ as a column matrix. From this previous exercise, we have $\frac{d}{dt} \mathsf{exp}(At) = A \mathsf{exp}(At)$, and so $[\frac{d}{dt} Y_1\ \ldots\ \frac{d}{dt} Y_n] = [AY_1\ \ldots\ AY_n]$. In particular, the columns of $\mathsf{exp}(At)$ are solutions of $\frac{d}{dt} Y = AY$. So by definition $\mathsf{exp}(At)$ is a fundamental matrix of $\frac{d}{dt}Y = AY$.

Now let $Y_0(t) = \mathsf{exp}(At)C$. Certainly $Y_0(0) = C$, and using part 2 of this previous exercise, we have $\frac{d}{dt} Y_0 = \frac{d}{dt} \mathsf{exp}(At)C$ $= (\frac{d}{dt} \mathsf{exp}(At)) C$ $= A \mathsf{exp}(At) C$ $= AY_0$.