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.

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