Given a general matrix $A(t), t>0$, with real entries, I would like to know if the eigenvalues of $A(t)$ are continuous functions of $t$. These eigenvalues may be real or complex.
What about the spectral radius?
A classical result from complex analysis states that the roots of a polynomial vary continuously with the coefficients. Can we use the theorem directly to prove the above? or there are other cases where the eigenvalues are actually discontinuous?
What I'm actually doing is trying to prove that there exists a $t$ for which the spectral radius of $A(t)$ is in $(0,1)$, and I'm doing that by proving that the spectral radius is 1 if $t\rightarrow 0$ and 0 if $t\rightarrow \infty$ (which I already know). Then I would invoke the continuity of the spectral radius to say that there must exist a value of $t$ for which the spectral radius is in $(0,1)$
Thank you.
Best Answer
All of these are continuous, since they are the compositions of continuous functions.
The function from the matrix to any coefficient of the polynomial is itself a polynomial on the entries of the matrix, which is continuous. Thus, the function from a matrix to the vector listing the coefficients of the polynomial is continuous. So, the function from a matrix to its characteristic polynomial is continuous.
The function from a characteristic polynomial to its roots is continuous. So, by the continuity of composition, the function from a matrix to its eigenvalues is continuous.
The function $x \mapsto |x|$ is continuous, as is $(x_1,\dots,x_n) \mapsto \max\{x_1,\dots,x_n\}$. So, the function that yields the largest absolute value of an entry of a vector is continuous. So, by composition, the spectral radius function is continuous.