1
$\begingroup$

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 $\lambda_i$?

$\endgroup$
  • 2
    $\begingroup$ Hint: $A^2-c^2I=0$. $\endgroup$ – Anurag A Aug 14 at 20:11
  • 2
    $\begingroup$ What can you say about its trace and its determinant? And what about a matrix with te same trace and determinant? $\endgroup$ – Paul Aug 14 at 20:14
  • 1
    $\begingroup$ Hint: The eigenvalues are distinct, so the matrices are all diagonalizable. What does the common diagonal matrix look like? $\endgroup$ – amd Aug 14 at 20:39
2
$\begingroup$

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 $b_2$.

$\endgroup$
3
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.