Linear Algebra: finding eigenvalues and eigenvectors Problem number 15:

I tried doing this problem without using any software or graphing calculator. I got my eigenvalues as 1 and -1. and got the eigenvector for $-1$ to be a $2\times 1$ matrix $[0, 0]$. I checked the book at this point and it says that my eigenvalues are $-3$ and $1$. 
I solved for my eigenvalues by using the characteristic equation, and for the eigenvectors by using (characteristic equation) $x=0$. What am I doing wrong?
 A: In addition to using the characteristic polynomial, when you have small matrices (especially $2\times 2$), the following two facts are often useful:


*

*The product of the eigenvalues of $A$ equals $\det(A)$.

*The sum of the eigenvalues of $A$ equals $\mathrm{trace}(A)$.


With $2\times 2$ matrices, both the trace and determinant can be calculated easily in your head, so if you can find two numbers that multiply to the determinant and add to the trace, you've got your eigenvalues. This is essentially the same as factoring the characteristic polynomial "by eye" rather than via the quadratic formula.
For the matrix
$$\left(\begin{array}{rr}
2 & 5\\-1 & -4
\end{array}\right),$$
the trace is $-2$ and the determinant is $-8+5 = -3$. So you want two numbers that add up to $-2$ and multiply to $-3$; the answer is $-3$ and $1$, so the two eigenvalues are $-3$ and $1$.
For the matrix
$$\left(\begin{array}{rr}
4 & 3\\-3 & -2
\end{array}\right),$$
the trace is $2$ and the determinant is $-8+9 = 1$, so having both eigenvalue equal to $1$ will do it.
A: Hint:  The eigenvalues of $A$ are given as the roots of the polynomial $\det\left(A-\lambda I\right)$.
Lets go over number $15$, I'll leave $16$ for you.  Then $$A-\lambda I=\left[\begin{array}{cc}
2-\lambda & 5\\
-1 & -4-\lambda\end{array}\right]$$ so that $$\det\left(A-\lambda I\right)=(2-\lambda)(-4-\lambda)+5.$$Expanding then factoring we get $$\lambda^{2}+2\lambda-3=\left(\lambda+3\right)\left(\lambda-1\right),$$so the eigenvalues are $-3$ and $1$.
Hope that helps,
A: For the $15$ the characteristic polynomial $\text{det}(A -\lambda I)$ is: $(2-x)(-4-x) + 5 =0$ this says $x^{2}+2x-3=0$ which gives $(x+3)(x-1)=0$ which says $x=-3$ or $x=1$.
Similarly work out the next one. Now for finding the eigenvectors (one corresponding to $3$) you simply have to solve $$\left[\begin{array}{cc} 2 & 5 \\\ -1 & -4 \end{array}\right] \cdot \left[\begin{array}{c} x_{1} \\\ x_{2} \end{array}\right] = -3 \left[\begin{array}{c} x_{1} \\\ x_{2}\end{array}\right]$$
