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Setup: Let $S$ be a vector space endomorphism of $\mathbb{Q}^3$. Suppose that given some basis $\beta$ for the vector space and that we have matrix representation $[T]_\beta$ be given in the edit.

Question: Show that $T$ is invertible.

Edit: Suppose that $$ [T]_\beta = \begin{pmatrix} a_{11}409 & a_{12}346 & a_{13}18 \\ a_{21}996 & a_{22}847 & a_{23}142 \\ a_{31}892 & a_{32}304 & a_{33}717 \end{pmatrix} $$ where $a_{ij} \in \{0,1,...,9\}$. So these coefficients are the thousands and hundreds place of the entries.

Arturo Magidin
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Irving Rabin
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    A vector space endomorphism $T$ is an isomorphism if and only if you can find a basis for the space such that the matrix representation of $T$ in that basis is invertible. – azif00 Apr 25 '22 at 05:40
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    Strictly positive elements are not enough. Consider the matrix with every entry equal to $1$. – Dietrich Burde Apr 25 '22 at 07:47
  • I've edited the question for more context. Maybe there's something now? – Irving Rabin Apr 25 '22 at 13:58
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    $S$ is an isomorphism if and only if the matrix is invertible. – Arturo Magidin Apr 25 '22 at 14:25
  • Is there any way to argue that $[T]_\beta$ is invertible in this case? Like any slick approach I could use? Because I only need the determinant to be non-zero right, and the determinant is an alternating multilinear function, is there any way to row reduce the matrix into something that certainly has non-zero determinant? – Irving Rabin Apr 25 '22 at 14:49
  • I know that I could brute force it with a computer and check each possible value for $a_{ij}$ and see whether or not the determinant is zero, but is there a slick argument which would work to argue it on paper? Or a hint? – Irving Rabin Apr 25 '22 at 14:53
  • @ArturoMagidin is there a way to show that the determinant is non-zero in this case? – Irving Rabin Apr 27 '22 at 18:53
  • @ArturoMagidin Apparently it's possible, I just don't see it. – Irving Rabin Apr 27 '22 at 19:21
  • Can you expand on the context of how you got this transformation? Precisely, in what context does $T$ arise, and what is your motivation for working in the basis $\beta$? I'm wondering if there is a key detail that might allow us to argue that $T$ is (not) invertible without resorting to using coordinates. – ml0105 Apr 28 '22 at 04:04
  • @ml0105 I can't give you anything more than this unfortunately. All I know is that given some basis $\beta$ we have the matrix representation that I have in the question. – Irving Rabin Apr 28 '22 at 04:15

2 Answers2

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Since $\mathbb Q$ is a field, what you want to do is confirm that the $\det\big(T\big)\neq 0$. We do this by showing the determinant is odd.

The easiest way to look at this is view this matrix over the integers (not $\mathbb Q$) then do a ring homomorphism $\phi:\mathbb Z\longrightarrow\mathbb F_2$ giving you the identity matrix $I_3$ since all diagonal entries are odd and all off diagonal entries are even. Thus when working over $\mathbb F_2$ the determinant is $=\det\big(I_3\big)=1$

Conclusion: $\det\big(T\big)\%2 = 1 \neq 0$ i.e. the determinant of $T$ is odd hence non-zero.

user8675309
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  • Oh because zero is even!? That's a neat trick! But do we even need to pass to this ring homomorphism? Because we know how even and odd numbers interact via addition and multiplication, couldn't we just do a cofactor expansion and like you said the entries along the diagonal are odd and others are even, so we could check if the result is even that way? I guess I don't see why switching the field we're working over is allowed? – Irving Rabin Apr 28 '22 at 06:17
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    It depends on background and preference. I find the method here quite useful in other settings when dealing with integer matrices -- see e.g. https://math.stackexchange.com/questions/3867140/how-can-you-calculate-the-rank-of-an-nxn-matrix-with-the-given-conditions/3867230#3867230 . In the case of your problem specifically you one could write $\det\big(A\big)=\sum_{\sigma \in \text{Perm(n)}} a_{\sigma(1),1}a_{\sigma(2),2}a_{\sigma(3),3}\cdot \text{sign}(\sigma)$ and observe that the diagonal product (identity permutation) is odd but all other terms are even hence the sum is odd $\neq 0$. – user8675309 Apr 28 '22 at 16:07
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    "I guess I don't see why switching the field we're working over is allowed?" reducing the entries mod $p$ gives a lower bound on the rank and since the lower bound here is 3, and the upper bound is of course 3, your matrix has rank 3- see here: https://math.stackexchange.com/questions/4144758/rank-of-integer-matrix-reduced-bmod-p-is-at-most-that-of-the-original-matrix/ – user8675309 May 01 '22 at 19:41
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This matrix $T$ is in fact invertible for all $a_{ij}\in\mathbb Z$, not just for digits. Indeed, if we reduce the coefficients of $T$ modulo $100$, we have $\det(T)\equiv d \ ( \mod{100}) $ where

$$ d = \left|\begin{matrix} 9 & 46 & 18 \\ 96 & 47 & 42 \\ 92 & 4 & 17 \end{matrix}\right| $$

We then compute $d$ using the usual row and column operations, combined with reduction modulo $100$ :

$$ \begin{array}{lcl} d &=& \left|\begin{matrix} 9 & 46 & 18 \\ 96 & 47 & 42 \\ 92 & 4 & 17 \end{matrix}\right| \\ &=& \left|\begin{matrix} 9 & 46 & 18 \\ -4 & 47 & 42 \\ -8 & 4 & 17 \end{matrix}\right| \ (\textrm{reduce modulo}\ 100)\\ &=& \left|\begin{matrix} 1 & 50 & 35 \\ -4 & 47 & 42 \\ -8 & 4 & 17 \end{matrix}\right| \ (\textrm{using}\ (R_1) \gets (R_1)+(R_3) )\\ &=& \left|\begin{matrix} 1 & 50 & 35 \\ 0 & 247 & 182 \\ -8 & 4 & 17 \end{matrix}\right| \ (\textrm{using}\ (R_2) \gets (R_2)+4(R_1) ) \\ &=& \left|\begin{matrix} 1 & 50 & 35 \\ 0 & 247 & 182 \\ 0 & 404 & 297 \end{matrix}\right| \ (\textrm{using}\ (R_3) \gets (R_3)+8(R_1) ) \\ &=& \left|\begin{matrix} 247 & 182 \\ 404 & 297 \end{matrix}\right| \ (\textrm{expand first column} ) \\ &=& \left|\begin{matrix} 47 & -18 \\ 4 & -3 \end{matrix}\right| \ (\textrm{reduce modulo}\ 100 ) \\ &=& 47\times(-3)-4\times(-18)=-141+72=-69. \end{array} $$

This finishes the proof.

Ewan Delanoy
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  • Why is it fair game to reduce everything mod 100? I've only seen that determinants are invariant to elementary operations? – Irving Rabin Apr 28 '22 at 06:14
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    @AndreyYanyuk A determinant is polynomial in its coefficients, i.e. made up of additions, substractions and multiplications. All those operations are compatible with "modulo $m$" for any $m$. – Ewan Delanoy Apr 28 '22 at 06:19