# 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 $$\lambda_i$$?

• Hint: $A^2-c^2I=0$. – Anurag A Aug 14 at 20:11
• What can you say about its trace and its determinant? And what about a matrix with te same trace and determinant? – Paul Aug 14 at 20:14
• Hint: The eigenvalues are distinct, so the matrices are all diagonalizable. What does the common diagonal matrix look like? – amd Aug 14 at 20:39

## 2 Answers

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 $$b_2$$.

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?