How much larger can the largest singular value of a matrix be, relative to the largest eigenvalue?
Specifically, given some square matrix $A$ with spectral radius greater than $0$, can one derive a general bound $C(A) < \infty$ such that $$\sigma_1(A) \le C(A) \lvert \lambda_1(A)\rvert$$ where $\lambda_1(A)$ is the largest eigenvalue, and $\sigma_1(A)$ is the largest singular value? Obviously for this bound to be interesting $C(A)$ must be finite.
For symmetric matrices we can take C(A) to be $1$, but, generally, does there exist some (non-trivial) uniform bound that depends on, say the rank of $A$, or its dimensions, or its determinant, or its maximum entry, or smallest entry, etc.? If so, how would this bound scale? Polynomially, exponentially, etc.?
For simplicity, what about the case where $A$ is just a $2 \times 2$ matrix.
Edit:
So I guess, besides the fact the largest eigenvalue is smaller than the largest singular value, there isn't really any meaningful relationship between the magnitude of the largest eigenvalues and that of the largest singular value, for non-symmetric matrices. As pointed out in the comments, the singular values can always be trivially bounded by the square root of the product of the 1-norm and $\infty$-norm ($\sigma_1(A) \leq \sqrt{\lVert A \rVert_1 \lVert A \rVert_\infty}$) but there doesn't seem to exist any meaningful relationship between the largest singular value and the spectral radius.