# Bounding determinants of the following form from above

To bound a determinant of a matrix from above it's quite common to apply Hadamard's inequality. Unfortunately, in the following problem Hadamard's inequality isn't good enough. Are there other methods to bound determinants from above?

For $t\in [0,1]$, let $A_t$ be an $(n\times n)$-matrix over the complex numbers. Assume that $A_t$ is an invertible matrix for $t>0$ and that $A_0 = A_{t=0}$ is a singular matrix. Moreover, we are given that the limit $$L=\lim_{t\to 0^+} \frac{\vert\det A_t\vert}{t}$$ exists. Also, the limit has the following form. Write $A_t = (a_{ij,t})_{i,j=1}^n$. Thus, the $(ij)$-th entry of $A_t$ is $a_{ij,t}$. Then $a_{ij,0} = a_i$, where $a_i$ is a non-zero complex number. (So the $i$-th row of $A_0$ is $(a_i,a_i,\ldots,a_i)$.)

For example, if $n=1$, we have that $A_t$ is a function going to zero.

If $n=2$, we can take the matrix $$A_t=\left( \begin{array}{cc} 1+\sqrt{t} & 1+2\sqrt{t} \\ 2+\sqrt{t} & 2+2\sqrt{t}\end{array}\right).$$ Note that $A_t$ converges to a singular matrix and that $A_t$ is invertible for any $0<t\leq 1$.

Goal. Bound $L$ from above.

Unfortunately, Hadamard's inequality is useless in this case. In fact, Hadamard's inequality states that $\vert \det A_t\vert$ is bounded from above by a certain number. This certain number does not go to $0$ as $t\to 0$. Therefore, Hadamard's inequality gives the trivial upper bound $L \leq \infty$.

Therefore I am led to ask you.

Question. Are there other ways one can obtain upper bounds on absolute values of determinants?

Remark. Note that the example given by user3296 below $A_t = \lambda I_{n} t$ does not fulfill the conditions of my $A_t$. His matrix converges to the zero matrix.

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I'm sure there are other methods for matrices of special form, but since you're not telling us anything about your matrices it's hard to give you any specific advice. So I'll give you some general advice: type determinant bound into Google. A lot of references to Hadamard come up, of course, but so do some other things. –  Gerry Myerson Oct 6 '11 at 23:39
Tell us more about the map $A_t$. –  Olivier Bégassat Oct 8 '11 at 1:10
Ok I'll try to tell you a bit more about the map $A_t$. –  shaye Oct 8 '11 at 11:16
@shaye I think you statement about Hadamard's inequality is a little bit too harsh since it is an upper bound on det$A$ for $t=1$, it is just not sharp. Note that the rank of a matrix is semicontinous. Hence the limits might not agree. Also, it might be a good idea to give at least an example of $A_t$. I don't understand your double index-comma-index notation. –  user13838 Oct 8 '11 at 11:53
Consider the matrices $$A_t:=\left[\matrix{a\sqrt{t} & 0\cr 0 & \sqrt{t} \cr}\right]$$ for fixed $a>0$. Then $L=a$, which can be arbitrarily large. –  Christian Blatter Oct 8 '11 at 13:16

I'm not sure that I'm understanding your question correctly -- are you looking for $M$ such that $$\lim_{t \rightarrow 0^+} \frac{|\det A_t|}{t} \leq M$$ for every path $A : [0, 1] \rightarrow M_{n \times n}(\mathbb{C})$ with $\det A_t = 0 \iff t = 0$? In this case, clearly the trivial bound of infinity is the best you can do: consider $A_t = \lambda I_{n \times n} t$ with $|\lambda|$ arbitrarily large. (If you have additional information about your path $A_t$, we can't help much unless you tell us what that information is...)