Question: Given the linear transformation $T(at^2+bt+c)=(2a − c)t^2 + (2a + b − 2c)t + (−a + 2c)$, how can I find the eigenvectors and eigenvalues? I know that general idea is to solve $T(v)=\lambda v$. From there, I get:
$$\lambda at^2+ \lambda bt+ \lambda c = (2a − c)t^2 + (2a + b − 2c)t + (−a + 2c)$$

This implies,

$$2a - c = \lambda a$$
$$2a + b - 2c = \lambda b$$
$$-a + 2c = \lambda c$$

This is where I get stuck. How can I solved this system? Any help is greatly appreciated!

Answer: 
Note that $\log n + \log \log n \le 2 \log n \le (\log n)^2$ provided that $n \ge 4$ in which case $$n \log n = 2^{\log n + \log \log n} \le 2^{(\log n)^2}.$$
Actually, $n\log n=o(2^{\log^2 n})$ (little oh), which is a stronger statement.
To see the, consider the functions $f(x)=x\log x$ and $g(x)=2^{\log^2 x}$.
Clearly both functions tend to infinity as $x\to\infty$. Using L'hopital's rule,

$$\lim\_{x\to\infty}\frac{f(x)}{g(x)}=\lim\_{x\to\infty}\frac{f'(x)}{g'(x)}=\frac{
\frac{1+\ln x}{\ln 2}
}
{\frac{2\exp(\frac{\ln^2 x}{\ln 2})\ln x}{x\ln 2}}=\frac{x\ln 2(1+\ln x)}{2\ln 2\exp(\frac{\ln^2 x}{\ln 2})\ln x}$$
Since for sufficiently large $x$ we have $\ln x>2\ln 2$, we deduce that
for sufficiently large $x$ $\exp(\frac{\ln^2 x}{\ln 2})>\exp(2\ln x)=x^2$, and so the r.h.s of the displayed equation above is bounded above by
$$\frac{x\ln 2(1+\ln x)}{2\ln 2\cdot x^2\ln x}=O(\frac{1}{x})$$

Therefore $f(x)=o(g(x))$ as $x\to\infty$, and in particular, $n\log n=o(2^{\log^2 n})$ as $n\to\infty$.

**Remark**. The argument above shows that actually a much stronger statement is true, namely for every positive integer $k$,
$$\frac{f(x)}{g(x)}=o(x^{-k})\quad (x\to\infty)$$
Therefore, the ratio
$$\frac{n\log n}{2^{\log^2 n}}$$
tends to zero faster than any power of $\frac{1}{n}$ as $n\to\infty$.
In other words, for every positive integer $k$,

$$\lim\_{n\to\infty}\frac{n^k\log n}{2^{\log^2 n}}$$

This of course answers the OP's question, although its primary purpose is to show that the OP's question, is in a sense, "not the right question".