Suppose the 2x2 constant matrix $A = \begin{bmatrix} a&b\\ c&d\\ \end{bmatrix}$ has real, positive, and distinct eigenvalues $\lambda_{1}$, $\lambda_{2}$. We know that A is diagonalizable since the eigenvalues are distinct and nonzero, so we may write $C^{-1}AC= [v_1|v_2]^{-1}A[v_1|v_2] =\begin{bmatrix} \lambda_{1}&0\\ 0&\lambda_{2}\\ \end{bmatrix}$ where $v_1$ and $v_2$ are the eigenvectors of $\lambda_{1}$ and $\lambda_{2}$, respectively.
My question is, how do we obtain the following suitable C?
If $b \neq 0$, then $C = \begin{bmatrix} -b&-b\\ a-\lambda_{1}&a-\lambda_{2}\\ \end{bmatrix}$.
If $b = 0$ and $c \neq 0$, then $C = \begin{bmatrix} -c&-c\\ d-\lambda_{1}&d-\lambda_{2}\\ \end{bmatrix}$.
If $b=c=0$, then the equations are already in suitable form.
I am certain this is just a linear algebra problem, but here is a little background in case I missed anything.
I am currently studying and working through the Fourth Edition Nonlinear Ordinary Differential Equations textbook by D.W. Jordan and P. Smith. I am in section 10.8 Stability and the linear approximation in two dimensions. In a specific part of this section, the authors are trying to establish conditions in which the instability of the zero solution of the first order two-dimensional system ${\mathbf{\dot{x}}} = \textbf{Ax}$ implies the same property for the zero solutions of the two-dimensional autonomous system ${\mathbf{\dot{x}}} = \textbf{Ax+h(x)}$. The authors only state these suitable $C$ matrices under the aforementioned conditions on $b$ and $c$, but I fail to see how vectors $v_1$ and $v_2$ solve the equations $(\textbf{A}-\lambda_1*\textbf{I})\textbf{v}=0$ and $(\textbf{A}-\lambda_2*\textbf{I})\textbf{v}=0$, respectfully. This is only a step in the construction part of a proof to be used later.
You could always verify these equations directly. The key to doing so is to use the identities $\det A=\lambda_1\lambda_2$ and $\operatorname{tr}A=\lambda_1+\lambda_2$. However, I think you’re more interested in how one arrives at these particular eigenvectors.
Suppose that $A\mathbf v=\lambda_1\mathbf v$, $\mathbf v\ne0$. Then $(A-\lambda_2I)\mathbf v = (\lambda_1-\lambda_2)\mathbf v$. On the other hand, we know that $A-\lambda_2I$ has rank one, therefore all of its nonzero columns are eigenvectors of $A$ with eigenvalue $\lambda_1$. Similarly, all of the nonzero columns of $A-\lambda_1I$ are eigenvectors of $A$ with eigenvalue $\lambda_2$.
We have $$A-\lambda I = \begin{bmatrix}a-\lambda&b\\c&d-\lambda\end{bmatrix},$$ so if $b\ne0$, we can take the last column to get the linearly-independent eigenvectors $(b,d-\lambda_2)^T$ and $(b,d-\lambda_1)^T$. Since $\lambda_1+\lambda_2=\operatorname{tr}A=a+d$, we can write these as $(b,\lambda_1-a)^T$ and $(b,\lambda_2-a)^T$, respectively. Of course, any nonzero scalar multiple of an eigenvector is also an eigenvector with the same eigenvalue, so these two vectors can be scaled arbitrarily. I expect that negating them makes later manipulations in the text more convenient or eventually leads to a “nice-looking” result.
However, it looks to me like the eigenvectors for the case $b=0$, $c\ne0$ are incorrect. Using the first column of $A-\lambda I$ instead of the second as above, we end up with $$C=\begin{bmatrix}d-\lambda_1&d-\lambda_2\\-c&-c\end{bmatrix}.$$ The rows of this matrix are transposed from what you’ve got in your question. You can directly verify for yourself that $A(d-\lambda_1,-c)^T=\lambda_1(d-\lambda_1,-c)^T$, but that $A(-c,d-\lambda_1)^T$ doesn’t produce $\lambda_1(-c,d-\lambda_1)^T$.
Incidentally, these considerations lead to a convenient decomposition of $A$ into a linear combination of projectors onto its eigenspaces. Let $$P_1={A-\lambda_2I\over \lambda_1-\lambda_2} \\ P_2 = {A-\lambda_1I\over \lambda_2-\lambda_1}.$$ Then $A=\lambda_1P_1+\lambda_2P_2$. These two matrices are projectors—$P_1^2=P_1$ and $P_2^2=P_2$—but they also have the handy property that $P_1P_2=P_2P_1=0$, which allows you to compute powers of $A$ as $A^n=\lambda_1^nP_1+\lambda_2^nP_2$. In fact, for any analytic function $f$, we have $f(A)=f(\lambda_1)P_1+f(\lambda_2)P_2$, so in particular $$e^{tA} = e^{\lambda_1t}P_1+e^{\lambda_2t}P_2.$$ Thus, you can write down a solution to $\dot{\mathbf x}=A\mathbf x$ once you know the two distinct eigenvalues of $A$ without having to compute any eigenvectors explicitly. You can use this when the eigenvalues are complex, too, but for that case there’s a more convenient way to compute the exponential that doesn’t require simplifying an expression with complex exponentials.