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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 $\lambda_i$?

Blue
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Morgan Campbell
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2 Answers2

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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?

Berci
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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 $b_2$.

Robert Lewis
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