How to prove that the determinant of a matrix must be very small (close to 0)?

Question

In equation $\mathbf{b=Aa}$, if all the entries in $\mathbf{A}$ and $\mathbf{b}$ are bounded, and $||\mathbf{a}|| \to \infty$. Then how to prove that det$(\mathbf{A})$ $\to 0$ (be very close to 0 but not equal to 0)?

$||\mathbf{a}|| \to \infty$ means some entries in $\mathbf{a}$ have very big magnitude of values.

$\mathbf{A}$ is a positive definite $n \times n$ matrix, $\mathbf{a}$ and $\mathbf{b}$ are all $n$-dimensional vectors.

Answer 1

If $A a_n = b_n$ with $\|b_n\| \le L$ and $\|a_n \| \to \infty$, then $A {a_n \over \|a_n\|} = {b_n \over \|a_n\|}$. In particular, $A {a_n \over \|a_n\|} \to 0$. Let $a^*$ be an accumulation point of ${a_n \over \|a_n\|}$, then $A a^* = 0$, and since $\| {a_n \over \|a_n\|} \| = 1$ for all $n$ we have $\|a^*\| =1$.

Answer 2

This is not a well-formed question, as written. If $A$ is a single finite matrix and $b$ is one single vector w a finite number of coordinates, then the entries of $b$ do not "approach infinity".

Meanwhile even besides that, consider $A_n$ which is $\frac{1}{n}$ times the square identity matrix (say $2 \times 2$). Let the upper-left entry of $A_n$ be $n$ and the lower-right entry of $A_n$ be $\frac{1}{n}$. Then the matrix is positive definite and has determinant 1. Now let $a$ be the vector $(0,1)^T$. Then $b_n$ satisfying the equation $A_nb_n = a$ is $(0,n)^T$.

Answer 3

Thanks for all the replies. I myself finally got a solution, the proof is as follows:

Since $A$ positive definite, $A$ can be uniquely factorized as $A=U^{\top}U$, where $U$ is an upper triangular matrix.

Because $||b||$ is finite, and $||a|| \to \infty$, then $A\frac{a}{||a||}=\frac{b}{||a||} \to 0$ (this part was inspired by coppert.hat's post, much appreciate that), then $\left(\frac{a}{||a||}\right)^{\top}A\left(\frac{a}{||a||}\right) \to 0$, then $\left(U\frac{a}{||a||}\right)^{\top}\left(U\frac{a}{||a||}\right) \to 0$.

For simplicity, assume that $a$ has only its $i$-th entry $a_i\to \infty$, and the other entries are finite. Set $||a||=||a||_{\infty}=|a_i|$, then $\frac{a}{||a||}$ is a vector with its $i$-th entry be 1, and all other entries be $\to 0$.

Since the sizes of $U$ and $\frac{a}{||a||}$ are all finite, $U\frac{a}{||a||} \to U_i$, where $U_i$ is the $i$-th column of $U$. Then $\left(U\frac{a}{||a||}\right)^{\top}\left(U\frac{a}{||a||}\right) \to 0 \Rightarrow U_i^{\top}U_i \to 0 \Rightarrow U_i \to 0 \Rightarrow U_{ii} \to 0$.

{Because $U$ is upper triangular, det$U$=$\sum_{j=1}^{n} U_{jj} \to 0$ due to that $U_{ii}\to 0$ and $n$ finite.

Note: This part is not rigorous, we also need the other diagonal elements $U_{jj} (j\neq i)$ be bounded to support the conclusion. The modification will be made in the following. }

Because $A$ positive definite, a basic Cholesky decomposition step can be made as follows:

$$ A= \left[ \begin{matrix} A_{11} & v^{\top}\\ v & B\\ \end{matrix} \right] = \left[ \begin{matrix} \sqrt{A_{11}} & 0^{\top}\\ \frac{v}{\sqrt{A_{11}}} & I_{n-1}\\ \end{matrix} \right] \left[ \begin{matrix} 1 & 0^{\top}\\ 0 & B-\frac{vv^{\top}}{A_{11}}\\ \end{matrix} \right] \left[ \begin{matrix} \sqrt{A_{11}} & \frac{v^{\top}}{\sqrt{A_{11}}}\\ 0 & I_{n-1}\\ \end{matrix} \right] $$

1. $U_{11}=\sqrt{A_{11}}$ is bounded because all entries in $A$ is bounded .
2. Denote $B'=B-\frac{vv^{\top}}{A_{11}}$, then $B'_{11}$ is nonzero and bounded. It is nonzero because $B'$ must be positive definite; it is bounded because $B'_{11}=B_{11}-v_1^2/A_{11}$, all the values on the right hand side are finite, and $A_{11} \neq 0$. As a result, $U_{22}=\sqrt{B'_{11}}$ is nonzero and finite.
3. By induction, if the matrix size $n$ is finite, then all the diagonal entries of $U$ are nonzero and finite.

Now because $U$ is upper triangular, det$U$=$\sum_{j=1}^{n} U_{jj} \to 0$, due to that 1) $U_{ii}\to 0$; 2) $U_{jj} (j\neq i) $ finite; 3) $n$ finite.

Finally, det$A$=(det$U$)$^2 \to 0$.

If $a$ has more than one entries that have very large magnitude, e.g. $\left\{ a_{i1}, a_{i2}, \cdots, a_{il} \right\} \to \infty$. Follow the above procedure, we can still prove that $U_{il,il}\to 0$, then $\det (U) \to 0$.

ps. Here the notation $\to 0$ only means that the value has a very small magnitude.