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.

  • $\begingroup$ 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

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 $A+tB$.


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$$A=\begin{bmatrix}a&a\\-a&-a\end{bmatrix}\\B=\begin{bmatrix}{1\over 4a}&{1\over 4a}\\-{1\over 4a}&-{1\over 4a}\end{bmatrix}$$


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