Problem: find a fundamental set of solutions for the system $y'=Ay$, where $A=\begin{bmatrix} -3 & 14 \\ 0 & 4 \end{bmatrix}$
I find the eigenvalues $\lambda_1=-3$, $\lambda_2=4$
For $\lambda=-3$, I have $\begin{bmatrix} 0 & 14 \\ 0 & 7 \end{bmatrix}$, row reduce it to get $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Then $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
Question: I know $b=0$, but how to explain a=1 conceptually? Is that because the first column is 0, then it is a free variable, so we pick 1?
An eigenvector to the eigenvalue $\lambda = - 3$ is a non-trivial solution of the homogeneous linear equation system
$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.$$
The second equation tells us $b = 0$ and the first equation $0 \cdot a = 0$ is fulfilled for every $a$. Since we are looking for a non-trivial solution $(a, b) \neq (0, 0)$, we can choose every $a \neq 0$ in order to get an eigenvector. In particular, the solution is not unique. For simplicity, we can take for e.g. $a = 1$.