0
$\begingroup$

Give an example of $2\times 2$ matrices $\mathbf{A, B}$ such that $\forall t \in \mathbb{R}$ the matrix $\mathbf{A} + t\mathbf{B}$ has the eigenvalues $\pm\sqrt{t}$.

I presume that there are no solutions, although I would be glad to hear the tips.

$\endgroup$
  • $\begingroup$ Have you tried just writing up the general characteristic polynomial, setting the roots and seen what happens? $\endgroup$ – Arthur Aug 16 at 19:46
  • $\begingroup$ What does it mean for negative $t$? $\endgroup$ – mathcounterexamples.net Aug 16 at 19:47
  • $\begingroup$ @Arthur I need to find a solution with respect to matrix entries. Brute-force approach with characteristic polynomial produces quadratic forms, which I found difficult to solve. $\endgroup$ – Inter Veridium Aug 16 at 19:52
  • $\begingroup$ Hint: take $t=0$, what can you deduce from here? What is the simpliest matrix $A$ that satisfies your conclusion? Can we now find a matrix $B$? $\endgroup$ – TZakrevskiy Aug 16 at 20:09
  • $\begingroup$ @Arthur In the problem's description the domain of $t$ is in $\mathbb{R}$, therefore, if we set $t = 0$, that would mean $\mathbf{A} + t\mathbf{B}$ has the eigenvalues of zero, but $\mathbf{A} + 0\mathbf{B} = \mathbf{A}$. It assumes that in some basis $\mathbf{A} = \begin{bmatrix} 0 & a \\ a & 0 \end{bmatrix}$. Let $a = 1$, and so on. $\endgroup$ – Inter Veridium Aug 16 at 20:16
3
$\begingroup$

Hint: The polynomial $\lambda^2-t$ happens to have $\lambda=\pm \sqrt t$ as roots. Find $A$ and $B$ such that this is the characteristic polynomial of $A+tB$.

$\endgroup$
1
$\begingroup$

First of all, the eigenvalues of $A$ are zero. Suppose $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$therefore$$|\lambda I-A|=\lambda^2-(a+d)\lambda+ad-bc=0$$since the latter equation has only zero roots, we must have $$a=-d\\ad=bc$$Note that if the eigenvalues of $A+tB$ are $\pm\sqrt t$, then those of ${1\over t}(A+tB)={A\over t}+B$ are $\pm {\sqrt{t}\over t}=\pm{1\over \sqrt t}$by tending $t\to \infty$, we conclude that all the eigenvalues of $B$ are also $0$. Therefore both $A$ and $B$ possess a similar form as$$A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&-a_{11}\end{bmatrix}$$and$$B=\begin{bmatrix}b_{11}&b_{12}\\b_{21}&-b_{11}\end{bmatrix}$$where$$a_{11}^2=-a_{12}a_{21}\\b_{11}^2=-b_{12}b_{21}$$from which we conclude that$$A+tB=\begin{bmatrix}a_{11}+tb_{11}&a_{12}+tb_{12}\\a_{21}+tb_{21}&-a_{11}-tb_{11}\end{bmatrix}$$so the characteristic equation becomes$$|\lambda I-A-tB|=\lambda^2-(a_{11}+tb_{11})^2-(a_{12}+tb_{12})(a_{21}+tb_{21})\\=\lambda^2-t[2a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}]$$therefore we obtain the following set of equations $$2a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}=1\\a_{11}^2=-a_{12}a_{21}\\b_{11}^2=-b_{12}b_{21}$$which give the most general conditions on $A$ and $B$. As an interesting special case take$$A=\begin{bmatrix}a&a\\-a&-a\end{bmatrix}\\B=\begin{bmatrix}{1\over 4a}&{1\over 4a}\\-{1\over 4a}&-{1\over 4a}\end{bmatrix}$$

$\endgroup$

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.