5
$\begingroup$

$a\in \mathbb{R}^n$ $a=(a_1,a_2,\dots,a_n)$ lets define this to be equivalent to $(a_1,a_2,\dots,a_n,0,0,0 \dots,0)$ (finite many zeros) by this I think we can make an $n \times m$ matrix $A$ equivalent to any $ p \times p$ $B$ matrix with $p\ge \max\{n,m\}$ by $B_{ij}= A_{ij}$ if $1\le i\le n$ and $1\le j \le m$ otherwise all the other entries of the matrix $B$ will be $0$ now we can multiply any two matrices $A, B$ by making them in the form of $P_1$, $P_2$ such that $P_1, P_2$ are $p \times p$ matrices where $p is \max\{ n,m,r,s\} $ where $A$ is an $n \times m$ matrix and $B$ is a $r \times s$ matrix .

my question is What problems will arise if we define matrix multiplication this way ? There should be a lot of problems because mathematicians because if there aren't any problems someone would probably generalise matrix multiplication this way

Another question is: is this definition agrees with the usual one in the cases where both methods of multiplying are defined? l examples that I tried result to the "same" matrix but I couldn't prove that.

$\endgroup$
8
  • 4
    $\begingroup$ Mathematicians define things if the definitions turn out to be useful, not if they do not make problems. Why would this be useful? $\endgroup$
    – Martin R
    Commented Apr 23 at 12:52
  • $\begingroup$ @MartinR maybe it is not useful but is it wrong ? Is there any useful properties of matrices will be lost? $\endgroup$
    – pie
    Commented Apr 23 at 12:57
  • 3
    $\begingroup$ I agree with Martin. It's a known fact that 0.4(9) = 0.5, but no one tends to write a half as either 0.50000000000000000000000000, or 0.499999999999999.... Moreover, in which way is your way a generalisation of matrix multiplication? $\endgroup$ Commented Apr 23 at 13:07
  • 1
    $\begingroup$ @EgorLarionov you can multiply any two matrices $\endgroup$
    – pie
    Commented Apr 23 at 13:10
  • 1
    $\begingroup$ In one dimension is $(x,0)*(y,0)=(xy,0)$ useful? Is $(x,egg)*(y,egg)=(xy,egg)$ useful? $\endgroup$
    – Paul
    Commented Apr 23 at 13:40

3 Answers 3

6
$\begingroup$

Yes, this is of interest, and it is studied (but not necessarily presented in this way). https://en.wikipedia.org/wiki/Direct_limit

We obey the convention that for an ordinal $\xi$, $$\xi=[0,\xi).$$ So for a natural number $n$, $$n=\{0,1,\ldots, n-1\}.$$ We let $\omega$ be the first infinite ordinal, so $$\omega=\{0,1,2,\ldots \}.$$ For $a,b\leqslant \omega$, let $M^a_b$ denote the set of all $a\times b$ matrices. For each $A\in M^a_b$ and each $i\in a$, $j\in b$, we let $[A]^i_j$ be the row $i$, column $j$ entry of $A$ (row/column numbering starts at $0$).

For each $a,b<\omega$ and $p,q\leqslant \omega$, we can define an inclusion map $$j^{a,a+p}_{b,b+q}:M^a_b\to M^{a+p}_{b+q}$$ by letting $$[j^{a,a+p}_{b,b+q}A]^i_j = \left\{\begin{array}{ll} [A]^i_j & : i\in a, j\in b \\ 0 & : \text{otherwise.} \end{array}\right.$$ Pictorially, we have $$A\mapsto \begin{pmatrix} A & 0 \\ 0 & 0\end{pmatrix}.$$

Then for any $a,a',b,b'<\omega$ and $A\in M^a_b$, $B\in M^{a'}_{b'}$, we can choose $\max\{a,a',b,b'\}\leqslant n<\omega$, include $A,B$ in $M^n_n$ as $j^{a,n}_{b,n}A$ and $j^{a',n}_{b',n}B$, and then multiply these last two matrices.

We can also map all of $M^a_b$, $a,b\in \omega$, into $M^\omega_\omega$. So we don't have to pick a specific $n$: Just extend everything by $0$ to have $\omega$ rows and $\omega$ columns, and multiply the resulting $\omega\times\omega$ matrices.

It should be noted that for any $a,b,c<\omega$, if $A\in M^a_b$ and $B\in M^b_c$, then $$j^{a,\omega}_{b,\omega}(AB)=(j^{a,\omega}_{b,\omega}(A))(j^{b,\omega}_{c,\omega}(B)),$$ so this does really generalize the usual matrix multiplication. However, the construction above runs into certain ambiguities that make it less common than the usual matrix multiplication (for example, the matrices $\begin{pmatrix} 1 & 0 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ become the same $\omega\times \omega$ matrix when mapped into $M^\omega_\omega$. So while there certainly are instances where this kind of construction is of significant interest, those instances usually don't come up in the standard examples of linear algebra problems that we encounter, and in those situations the added generality is not worth the added effort used to define things and the added ambiguity associated with working with these objects. So you'd likely only find this kind of construction when studying more specialized topics.

$\endgroup$
5
  • 2
    $\begingroup$ Nice answer! It's perhaps also worth mentioning that a shortcoming of this particular limit is that the inclusion maps preserve very little of the algebraic structure - in particular, they do not preserve identities, so the direct limit is only a semigroup. The construction of the related object $\mathrm{GL}_\infty(\Bbb R)$ gives a group, essentially by extending the main diagonal to consist of infinitely many $1$s. $\endgroup$ Commented Apr 23 at 16:21
  • $\begingroup$ Relatedly to @IzaakvanDongen's comment, as the other answer shows, in this approach one loses the determinant (in your words, it would become "ambiguous" i.e. not well-defined), one of the most powerful tools in linear algebra. $\endgroup$ Commented Apr 23 at 18:38
  • $\begingroup$ That being said, @IzaakvanDongen, what algebraic structure gets lost besides the existence of a unit? If I'm not mistaken, the direct limit here is still an associative algebra over the ground field (nothing depends on it being $\mathbb R$), just non-unital. $\endgroup$ Commented Apr 24 at 5:47
  • $\begingroup$ Thank you both for your additional insight. $\endgroup$
    – user469053
    Commented Apr 24 at 9:47
  • 1
    $\begingroup$ @TorstenSchoeneberg, yes that's true. I perhaps over-exaggerated. I just meant that if you're looking at the operation of multiplication by itself, you lose many of its nice properties. I see the loss of an identity as the tip of the iceberg, leading to the loss of other useful things, like "a matrix is invertible iff its determinant is nonzero iff multiplication by that matrix is an injective map", which doesn't even make sense in the absence of an identity. (So concretely, the inclusion maps do not preserve the property "multiplication on the left by $x$ is an injective map", for example) $\endgroup$ Commented Apr 24 at 14:31
5
$\begingroup$

Let $A=[\begin{array}{c}1 & 2\end{array}]$ and $B=A^T=\left[\begin{array}{c}1 \\ 2\end{array}\right]$. The usual multiplication yields AB as defined, resulting in the 1 by 1 matrix $[4]$, which is invertible with inverse $[\frac{1}{4}]$.

Now suppose we generalize the matrix product as you said, since we can pick any $p\geq max\{1,2\}$ we can pick $p=2$, making

$AB = \left[\begin{array}{c}1 & 2 \\ 0 & 0\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 2 & 0\end{array}\right] = \left[\begin{array}{cc}4 & 0 \\ 0 & 0\end{array}\right]$.

Now $AB$ has determinant $0$, being non-invertible. So the determinant (and all the properties that come with it such as invertibleness) are lost.

Also I think it's pretty weird taking $p\geq max\{1,2\}$ because picking 3 the result would be a different matrix: $\left[\begin{array}{ccc}4 & 0 & 0\\ 0 & 0& 0\\ 0 & 0 & 0\end{array}\right],$ so even the size would be ambiguous...

$\endgroup$
1
  • 1
    $\begingroup$ OP's method is of interest. It's just inductive limits. $\endgroup$
    – user469053
    Commented Apr 23 at 13:58
3
$\begingroup$

The construction you describe is very similar to the group $GL(R)$ for a ring $R$, which is one of the central objects of study in the field of K-theory.

For $n\geq 0$, we denote by $GL_n(R)$ the group of $n$ by $n$ invertible matrices with entries in $R$.

Now suppose we want to drop the $n$ and define $GL(R)$. That is we want to be able to multiply square invertible matrices of different sizes. Your construction is not quite the right one, because if we just add rows and columns of $0$'s, then the matrix will no-longer be invertible.

Instead we define a matrix $A$ to be equivalent to the matrix obtained by extending $A$ by the same number of rows and columns of $0$'s, but then setting the new diagonal elements to $1$. Pictorially: $$\left(\begin{array}{cc}A&0\\0&I \end{array}\right)$$

K-theory originated from two sources: vector bundles over manifolds and algebraic number theory. In both cases it was realised that summands of $R^n$ were of interest. For vector bundles, $R$ was the ring of smooth functions on a manifold. In number theory $R$ was the ring of integers of a number field. In the former case the summands corresponded to vector bundles. In the latter case the rank 1 summands corresponded to "hidden" primes that were needed to make unique factorisation work. Fermat's Last theorem was a big motivation for this study, as you can factorise Fermat's equation as $$y^n=(z-x)(z-\zeta x)\cdots (z-\zeta^{n-1} x)$$ where $n$ is a prime number and $\zeta$ is a primitive $n$'th root of unity. People realised quite early on that if you have unique factorisation for these numbers, then any solution to Fermat's Last theorem would lead to a smaller one, resulting in an impossible infinitely descending tower of solutions.

Thus both topologists and number theorists took an especial interest in summands of $R^n$, and it was natural to study these up to equivalence, where $P$ is regarded as equivalent to $P\oplus R^n$. It was then natural to study linear maps up to the analogous equivalence, which brings us back to $GL(R)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .