Characterize all real-valued $2\times 2$ matrices with eigenvalues $\pm c$, for $c > 0$. 
Characterize all real-valued $2\times 2$ matrices that have as eigenvalues $\lambda_1 = c$ and $\lambda_2 = −c$, for $c > 0$. Use your result to generate a matrix that has its eigenvalues $-1$ and $1$ and does not contain any zero elements.

Where do I even start with this? I know how to compute eigenvalues/vectors and everything, but am I finding the matrix that these eigenvalues came from like matrix $A$ from $(A-\lambda I)x=0$? Or am I finding $\lambda_i$?
 A: Let's rename the desired eigenvalues to $\pm\lambda$, and consider a generic $2\times 2$ matrix
$$M =\pmatrix{a&b\\c&d}\,. $$
You simply have to translate the condition of eigenvalues to certain equations of the matrix entries.
We know the characteristic polynomial of $M$:
$$x^2 - \lambda^2 =\det(M-xI)=(a-x)(d-x) \, - \, bc$$
Can you take it from here? 
A: I think the problem is asking us to find a general expression for such matrices in terms of parameters yet to be identified; finding these parameters is part of the challenge.
We write the matrix
$A \in M_{2 \times 2}(\Bbb R) \tag 1$
in terms of its entries
$A = \begin{bmatrix} a_1 & b_1 \\ b_2 & a_2 \end{bmatrix}, \tag 2$
and recall the eigenvalues satisfy the characteristic polynomial
$\chi_A(x) = \det(A - xI) = \det \left ( \begin{bmatrix} a_1 - x &  b_1 \\ b_2 & a_2 - x \end{bmatrix} \right ) = (a_1 - x)(a_2 - x) - b_1b_2$
$= x^2 - (a_1 + a_2)x + (a_1a_2 - b_1b_2) = x^2 - \text{Tr}(A)x + \det A; \tag 3$
now if the eigenvalues of $A$ are
$\pm c, \; c > 0, \tag 4$
then 
$\chi_A(x) = (x - c)(x + c) = x^2 - c^2; \tag 5$
comparing (3) and (5) we find that
$\text{Tr}(A) = c + (-c) = 0, \tag 6$
whilst
$\det A = -c^2; \tag 7$
it follows then that
$a_1 + a_2 = \text{Tr}(A) = 0, \tag 8$
i.e., we may write
$a_1 = a = -a_2 \tag 9$
for some
$a \in \Bbb R, \tag{10}$
and also
$a_1a_2 - b_1b_2 = \det A = -c^2, \tag{11}$
which in the light of (9) yields
$-a^2 - b_1b_2 = -c^2, \tag{12}$
or
$b_1b_2 = c^2 - a^2. \tag{13}$
Based upon this equation, we may now derive the specific forms $A$ may take.  The simplest case is
$b_1 = 0 = b_2, \tag{14}$
whence via (12)
$a^2 = c^2 \Longrightarrow a = \pm c, \tag{15}$
and
$A = \begin{bmatrix} a & 0 \\ 0 & -a \end{bmatrix}; \tag{16}$
if
$b_1 \ne 0, \tag{17}$
$b_2 = \dfrac{c^2 -a^2}{b_1}, \tag{18}$
so that
$A = \begin{bmatrix} a & b_1 \\ \dfrac{c^2 -a^2}{b_1} & -a \end{bmatrix}; \tag{19}$
likewise, when
$b_2 \ne 0, \tag{20}$
the corresponding results are had:
$b_1 = \dfrac{c^2 -a^2}{b_2}, \tag{21}$
$A = \begin{bmatrix} a &\dfrac{c^2 -a^2}{b_2}   \\ b_2  & -a \end{bmatrix}; \tag{22}$
we have shown that the forms (16), (19), and (22) are necessary if the eigenvalues of $A$ are $\pm c$; they are also sufficient; this is self-evident in the case (16); in the case (19), we see that the characteristic polynomial is
$\chi_A(x) = \det \left ( \begin{bmatrix} a - x & b_1 \\ \dfrac{c^2 -a^2}{b_1} & -a -
x \end{bmatrix} \right )$
$= -(a - x)(a + x) - (c^2 - a^2) = x^2 - c^2, \tag{23}$
the zeroes of which are $\pm c$; a similar calculation applies to (22).
As a final observation, the diagonal form (16) is a one-parameter family of matrices depending solely on $a$, whereas (19), (22) are two-parameter families hinging on $a$ and $b_1$ or $b_2$.
