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Let $A(t)$ be a symmetrical matrix such that each element $A_{ij}(t)$ is a continuous function of a vector $t\in T\subset \mathbb{R}^m$. Are the eigenvalues of $A(t)$ continuous functions of $t$?

I found a few answers that show that this is the case when $T \subset \mathbb{R}$ (for instance this one and this one). Can these results be generalized to $T\subset \mathbb{R}^m$ for $m>1$?

user_lambda
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    The characteristic polynomial is a continuous function of $t$. – Hagen von Eitzen Jul 11 '18 at 20:06
  • Yes but are the zeros of a polynomial continuous in $t$ if $t$ is a vector? I suspect that they are but all the results I can find assume that $t\in T\subset \mathbb{R}$. – user_lambda Jul 11 '18 at 20:08
  • What does it change? The important ingredient is that the characteristic polynomial is a continuous function of matrix entries. If the matrix entries are continuous functions of whatever (scalar or vector variable), eigenvalues are continuous functions of the same whatever. – Algebraic Pavel Jul 13 '18 at 11:49

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Here is my thought on your problem. Basically, your matrix which is a square and symmetric has elements which depend on the multi-variable vector $\mathbf{t}\in\mathbb{R}^m$. The characteristic polynomial this will generate in $\lambda$ will have coefficients which depend on your vector $\mathbf{t}$. Since finding the eigenvalues comes down to finding the roots of your polynomial there that, to me, would imply that the eigenvalues will be continuous function of $\mathbf{t}$ too. On a simple case thinking of a $2\times 2$ matrix with elements $A_{ij}(\mathbf{t})$, the polynomial would have the form

$$ \lambda^2 -Tr(\mathbf{M}(\mathbf{t}))\lambda + \det\mathbf{M}(\mathbf{t})$$

I think shows clearly that the Eigenvalues would, of course depend on $\mathbf{t}$ since the coefficient clearly depend on $\mathbf{t}$.

jean-francois Niglio
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  • Yes, the eigenvalues would of course depend on $t$. But are they continuous in $t$? There are many results that show that if $t\in\mathbb{R}$ then the answer is yes. I am not aware however of a similar result when $t\in\mathbb{R}^m$ but I suspect that it is true. – user_lambda Jul 12 '18 at 13:53
  • I see. Yes I would suspect that you are right, I don't see any reason why they shouldn't be continuous in it as well. – jean-francois Niglio Jul 12 '18 at 14:43