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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.

Xiankun Xu
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    What is M? Please type your question in the question body, where we can see it all at once. – saulspatz Mar 31 '18 at 23:14
  • Sorry, that should be $\mathbf{A}$, just corrected it. – Xiankun Xu Mar 31 '18 at 23:38
  • It is good practice to put the full problem statement in the body of your Question, not relying on the title to carry the burden of problem statement. If you wish, I will edit the Question for you to include the information from the title in the body. – hardmath Apr 01 '18 at 14:54

3 Answers3

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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 $(0,n)^T$.

Mike
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  • Hi Mike, thank you for your kind reply. The question is that $Aa=b$, with some entries of $a$ approach infinity, but $b$ is bounded. – Xiankun Xu Apr 01 '18 at 00:39
  • It still works though, let A = A_n as above; A is a 2-by-2 matrix both off-diagonal entries 0, upper left entry of A is n (n positive w arbitrarily large absolute value), lower -right entry of A is 1/n. Having specified A specify b = (1,0)^T. Then on the one hand det(A) is 1. However on the other hand, the solution to the equation Aa = b is a = (n,0)^T. – Mike Apr 02 '18 at 16:10
  • Hi Mike, thanks for pointing out that. I suppose that you mean b=(0,1)', and a = (0, n)'? This question comes from a physical problem, where $A$ is a mass matrix. From the physical perspective, all entries in $A$ must be bounded. I will update the description of the problem in the post. – Xiankun Xu Apr 02 '18 at 18:49
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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 $\|a^*\| =1$.

copper.hat
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  • Hi @copper.hat, thank you for your inspiring reply. Now $Aa^=0$ and $||a^||=1$, can we conclude that det($A$)=0? – Xiankun Xu Mar 31 '18 at 23:50
  • If $\det A \neq 0$ then $A$ would be invertible and $a^* = 0$. – copper.hat Mar 31 '18 at 23:53
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    Thank you copper.hat, I have understood the det $\mathbf{A}$ = 0 question. – Xiankun Xu Apr 01 '18 at 00:25
  • But is the statement "Let $a^$ be an accumulation point of $\frac{a_n}{||a_n||}$, then $Aa^=0$" be always true? I have an example that A =[1, 0.999999; 0.999999, 1] and $\frac{a_n}{||a_n||}$ =[1, -1]', now $A\frac{a_n}{||a_n||}\to0$, but can we find an $a^$ that satisfies both $Aa^=0$ and $||a^*||=1$? – Xiankun Xu Apr 01 '18 at 00:31
  • Since $| {b_n | \over |a_n|} \le {L \over |a_n|}$ we must have $A {a_n \over | a_n |} \to 0$. Since the set ${ x | |x | =1 }$ is compact, there must be an accumulation point. – copper.hat Apr 01 '18 at 01:20
  • Thank you for your patience, @copper.hat. I believe the statement "Since the set {x|∥x∥=1} is compact, there must be an accumulation point" is true, but it must have a precondition. Since it can't explain the following example: $A$ =[1, 0.999999; 0.999999, 1] and $\frac{a_n}{||a_n||}=[1,-1]'$, now $A\frac{a_n}{||a_n||} \to 0$. While it is impossible to find an $a^$ which is the accumulation point of $\frac{a_n}{||a_n||}$ that makes $Aa^=0$, because det($A$) $\to 0$ but $\neq 0$. – Xiankun Xu Apr 01 '18 at 03:29
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    I have no idea what you are saying. I am assuming that $A$ is constant, so $\det A$ is constant. Also, the vector $(1,-1)^T$ does not have norm $1$, so I really don't know what you are saying. Since it is a constant and $\det A = 2 \times 10^{-6}$, $A{a_n \over |a_n|}$ is a constant, non zero vector as long as $a_n \neq 0$. – copper.hat Apr 01 '18 at 03:56
  • Thanks again. Here I used the $\infty$-norm, but if you mean 2-norm we can set $\frac{a_n}{||a_n||}=(1/\sqrt{2},-1/\sqrt{2})^{\top}$. I think my question is a little bit confusing, it should more appropriately be: $A$ is constant positive semi-definite matrix, $b=Aa$, where $||a||$ is very large (say $10^{10}$), and $||b||$ is of $\mathcal{O}(1)$, how to prove that det$A$ must have a very small magnitude(close to 0)? – Xiankun Xu Apr 01 '18 at 04:16
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    That is a completely different question than the one you asked. – copper.hat Apr 01 '18 at 05:22
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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.

Xiankun Xu
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    This is not very rigorous... – Surb Apr 01 '18 at 21:16
  • I agree that this isn't rigorous. Here you would have to show here that Aa = b; ||b|| =1, ||a|| >= K, implies det(A) <= \eps(K) for some \eps() that goes to 0 as K gets large. [And in fact that is not true, see my comment that I left a couple minutes before this one. A diagonal matrix where one diagonal entry is much larger than the other but the product of the diagonal entries is 1 will lead to a counterexample.] – Mike Apr 02 '18 at 16:22
  • I have made a new restriction on $A$: all entries of $A$ are bounded. This restriction comes from the physical problem: $A$ is a mass matrix, and all its entries must be bounded. The corresponding modifications in the proof have been made. – Xiankun Xu Apr 02 '18 at 20:55