Let $A, B$ be two $3 \times 3$ (complex) symmetric matrices and suppose the equation $\det(xA - B) = 0$ has three distinct solutions. Prove that $A$ is invertible.

Any help appreciated!

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    $\begingroup$ What if $x=1, B=0$, so that one of the solutions states $\det(A)=0$. Then $A$ is not invertible. $\endgroup$ – Antonios-Alexandros Robotis Jan 22 at 19:03
  • $\begingroup$ @Antonios-AlexandrosRobotis The matrices are symmetric, but not necessarily definite positive. Besides, you don't have the identity for all $x$... $\endgroup$ – Jose Brox Jan 22 at 19:04
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    $\begingroup$ @ViktorGlombik I imagine it is a scalar. $\endgroup$ – Antonios-Alexandros Robotis Jan 22 at 19:07
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    $\begingroup$ Presumably the condition means that $\det(xA-B)=0$ has exactly three distinct roots, otherwise the problem statement is false, as illustrated by the counterexample $A=B=0$ (where $\det(xA-B)=0$ has infinitely many distinct roots). The symmetry assumption seems redundant here. By unitary triangulation, you may assume that $A$ is triangular. If $A$ is singular, then by the rule of Sarrus, $\det(xA-B)=0$ is at most a quadratic equation and hence it cannot have exactly three distinct roots. $\endgroup$ – user1551 Jan 22 at 21:07
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    $\begingroup$ @Levent It's a proof by contrapositive. $\endgroup$ – user1551 Jan 23 at 12:27

Since @user1551 didn't submit an answer, I'm giving one that is inspired on his comment. Nice one!

For starters the matrices do not have to be symmetric. They do not have to be of size $3\times 3$ either, but can be of arbitrary size $n \times n$. We do require that there are exactly $n$ distinct solutions, because otherwise $A=B=0$ is a counter example.

Let $A=U\Sigma V^*$ be a singular value decomposition (SVD) of matrix $A$, which is guaranteed to exist for any complex matrix $A$. Then $U$ and $V$ are unitary matrices and $\Sigma$ is a diagonal matrix of the singular values.

It follows that: $$\det(xA-B)=\det(xU\Sigma V^*-B)=\det(U(x\Sigma-U^*BV)V^*)=\det(x\Sigma-U^*BV)$$

Suppose $A$ is singular. Then the diagonal matrix $\Sigma$ has a $0$ on its diagonal. Consequently $(x\Sigma-U^*BV)$ has a row that does not contain $x$. When we apply the Laplace expansion to calculate it, we find that the resulting determinant cannot be of the maximum degree. Therefore there cannot be exactly $n$ distinct solutions.

Thus $A$ cannot be singular and must therefore be invertible.

  • $\begingroup$ isn't the rule of sarrus limited to $3\times 3$ matrices $\endgroup$ – Nathanael Skrepek Jan 25 at 20:21
  • $\begingroup$ Instead of Sarrus, which only holds for $3\times3$ matrices, you can invoke Laplace expansion. $\endgroup$ – egreg Jan 25 at 21:23
  • $\begingroup$ Thanks. Fixed now. $\endgroup$ – Klaas van Aarsen Jan 26 at 8:40

Firstly, if $A$ and $B$ are linearly dependent, i.e. $B=\lambda A$ for some $\lambda$ then we have $\det (xA- B)=(x-\lambda)^3\det A$ which does not have three distinct solutions. Hence $A$ and $B$ are linearly independent.

Let $V=\langle A,B\rangle$ be the plane spanned by $A$ and $B$. Passing to the projectivization of $\mathbb{C}^{3\times 3}$, $V$ becomes a line, say $L$. Since the equation has three distinct solution, $V$ contains at least $3$ lines on which every matrix is singular (namely, the lines $\mathbb{C}\cdot (\lambda_i A-B)$, $1\leq i\leq 3$ for each solution of the equation). Thus, in projective space, $L$ and the degree $3$ hyper surface of singular matrices intersect in at least three points. Since the hyper surface is of degree $3$, they intersect in exactly $3$ points. Since $A$ is not of the form $xA-B$ for any $x$, $A$ is necessarily invertible.

It is possible that I am using some facts that you are not familiar with, in this case you can just comment so I can edit the answer.

Edit : I consider the hyper surface of singular matrices, i.e. the hyper surface defined by $\det=0$ in the $8$-dimensional projective space of $3\times 3$-matrices. The degree of a hyper surface is given by the degree of the defining polynomial, in this case it is $3$. Now, for a line and a hyper surface in projective space, there are $2$ possibilities, either the line is contained in the hyper surface or they have at most the degree of the hyper surface many intersection points. The line is not contained in the hyper surface in this case since otherwise for all $x\in\mathbb{C}$ we would have $\det(xA-B)=0$. Then we pass to the case where there are at most $3$ intersection points. We already know that there are $3$ intersection points given by the solutions to $\det(xA-B)=0$, so three points, all of the form $xA-B$. Now the points on the line spanned by $A,B$ are either of the form $xA-B$ for some $x\in\mathbb{C}$ or $A$ (any point on the line is of the form $[xA+yB]\in\mathbb{P}^8$ for some $x,y\in\mathbb{C}$. All points where $y\neq 0$ can be parametrized by $xA-B, x\in\mathbb{C}$ and there is only one point where $y=0$, which is $A$).

Edit 2 : I would like to point out that this solution immediately generalises to $n\times n$ case by simply putting $n$ instead of $3$.

  • $\begingroup$ Firstly, shouldn't it be $\det (xA- B)=(x-\lambda)^3\det A$? The conclusion drawn for $A$ and $B$ stays the same. (2) What does "degree of the hypersurface" mean, and why is this degree equal to 3? (3) I do not understand how "Since $A$ is not of the form $xA-B$ for any $x$, $A$ is necessarily invertible." follows from the preceding argumentation. Could you explain? $\endgroup$ – Hanno Jan 27 at 10:59
  • $\begingroup$ @Hanno yes, it should be $(x-\lambda)^3$, thanks for pointing it out. For the rest, I edited the answer. $\endgroup$ – Levent Jan 29 at 9:38

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