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For scalars, given the equation $xy=a$, then $y→∞$ as $x→0$, i.e. as $x$ tends to being non-invertible.

I wanted to find an equivalent theorem for matrices. I came up with:

For matrices and vectors, given the equation $Xy=a$, then $y→∞$ as $X$ tends to being non-invertible (singular), i.e. as $X→A$, where $A$ is such that $Az=0$ has a non trivial solution.

Is this correct? Where can I find this in a book?

More interestingly, I have an idea that there are as many infinite entries in $y→∞$ as there are non-zero entries in $z$. Is this correct too? If so, how could I prove it?

Please help.

a.giannel
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    I appreciate the curiosity behind this question, but I'm really struggling to make sense of it. When you say, "$x$ tends to being invertible", at one point you mean it tending to $0$ (the only non-invertible scalar), and at another, where $X$ tends to a singular matrix $A$. Do you mean "singular" (non-invertible) when you say "invertible"? – Theo Bendit Jan 10 '19 at 01:30
  • Thank you for the corrections. – a.giannel Jan 10 '19 at 01:36
  • Suppose $X$ is $n\times m$. Then consider the $n$ rows of the matrix equation $Xy=a$. The $i$th row has the form $$\sum_{j=1}^{m}x_{i,j}\cdot y_j=a_i$$ If even one entry of $y$ explodes, say $y_k\to \infty$, then we can rewrite $$x_{i,k}\cdot y_k=a_i-\sum_{\substack{j=1\j\neq k}}^{m}x_{i,j}\cdot y_j$$ so that $x_{i,j}\to 0$ for every row $i$. But a matrix with a column of zeros is singular, and so not invertible. – nathan.j.mcdougall Jan 10 '19 at 01:45
  • Very interesting and cleaver. Thank you. Do you believe that with that I can work on to find a solution the second part of the question, regarding the number of infinite entries for $y$ and zero entries for $z$? – a.giannel Jan 10 '19 at 01:50
  • I wouldn't really consider what I gave to be a proper answer, because it considers some entries of $X$ fixed when taking the limit on others. It was point out that the property you propose is inherited from the scalar point of view. Since it's not rigorous, I doubt you could use it to prove the second part.

    I'm not exactly sure if you'd be able to formalize that second part though, because the way you count the "number of infinite entries" isn't clear. I see what you're getting at though.

     – nathan.j.mcdougall Jan 10 '19 at 01:59
  • Maybe you can tell I am not a mathematician. I am an economist. Thank you for the clarification. – a.giannel Jan 10 '19 at 02:23
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    The analogy with scalars breaks down pretty quickly. For any vector $y$ you can construct a nonzero matrix $A$ such that $Ay=0$. On the flip side, you can construct $A$ such that the $i$th element of $y$ has no effect on the value of $Ay$, so you’ll need to sharpen up what you mean by $y\to\infty$. – amd Jan 10 '19 at 03:19

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