# Example of $2\times 2$ matrices with common eigenvalues

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.

• Have you tried just writing up the general characteristic polynomial, setting the roots and seen what happens? – Arthur Aug 16 at 19:46
• What does it mean for negative $t$? – mathcounterexamples.net Aug 16 at 19:47
• @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. – Inter Veridium Aug 16 at 19:52
• 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 $B$? – TZakrevskiy Aug 16 at 20:09
• @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. – 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$$A=\begin{bmatrix}a&a\\-a&-a\end{bmatrix}\\B=\begin{bmatrix}{1\over 4a}&{1\over 4a}\\-{1\over 4a}&-{1\over 4a}\end{bmatrix}$$