## Nonsingular, constant multiples of a fundamental matrix are fundamental

Let $\frac{d}{dt} Y = AY$ be a linear system of first order differential equations over $\mathbb{R}$ as in this previous exercise. Suppose $M$ is a fundamental matrix of this system (i.e. a matrix whose columns are linearly independent solutions) and let $Q$ be a nonsingular matrix over $\mathbb{R}$. Prove that $MQ$ is also a fundamental matrix for this differential equation.

Write $Q = [Q_1\ \ldots\ Q_n]$ as a column matrix. Now $\frac{d}{dt} MQ = \frac{d}{dt} [MQ_1\ ldots\ MQ_n]$ $= [(\frac{d}{dt}M)Q_1\ \ldots\ (\frac{d}{dt} M)Q_n]$ by the definition of our matrix derivative and part 2 of this previous exercise. This is then equal to $(\frac{d}{dt} M)Q = AMQ$. In particular, the columns of $MQ$ are solutions of $\frac{d}{dt} Y = AY$. Moreover, since $Q$ is nonsingular, $MQ$ is nonsingular. So $MQ$ is also a fundamental matrix.