#linearAlgebra

2026-02-05

TIL "Linear Algebra Done Right" is now open access [1], and seems updated since the 2015 edition. I love that it gives proofs for the things that, in the words of Victor Klee, are obvious but also true.

library.oapen.org/bitstream/ha

#Textbooks #linearalgebra #openaccess

2026-01-18

@typeswitch
tl;dr solution at bottom

I guess we need to clarify what the allowed operation is.
e.g. if I am allowed to whether any (natural?) linear combination of outcomes is probability 1/2, then I can just ask is 3*P(rolling i) = 1/2.

If the question I'm allowed to ask is
P(rolling i, j, or k) = 1/2, then we are essentially asking for an invertible 6x6 matrix with one row of 1s, and five rows containing a permutation of [1, 1, 1, 0, 0, 0]. In which case, again you are using 5 tests

That is certainly brute forceable by computer (at most (6 choose 3) choose 5 = 15,504 meaningfully different possibilities), but there may be a clever rank argument that it's possible or not.

Okay, I did some more thinking while doing my laundry, and here's a strategy:

If a + b + c = 0.5 and d + b + c = 0.5, then a = d. Repeat with modifications four more times to show that that all are equal to one another.

The problem with that is that doing it naively requires six tests, so seven equations, if we include that they all sum to one, my bad. So you need to take advantage of all the information you've gathered in the first four iterations and choose smartly in the last one.

Here is a solution written as a matrix equation

\(\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 0 & 0\\
1 & 1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 1 & 0\\
1 & 1 & 0 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 & 0 & 0\\
\end{pmatrix}
\cdot \begin{pmatrix}
1/6 \\ 1/6 \\ 1/6 \\ 1/6 \\ 1/6 \\ 1/6
\end{pmatrix}
= \begin{pmatrix}
1 \\ 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \\ 1/2
\end{pmatrix}
\)

One final comment, because I can't help myself. It shouldn't be too hard to generalise this pattern to other *even* sizes of dice.

#linearalgebra #probability

Jason Bowen 🇺🇦jbowen@mast.hpc.social
2025-12-30

When I switched my major in college from physics to mathematics, I met with the undergraduate advisor for the department to sketch out courses, she (Kathy Davis at the University of Texas) said "you can never learn enough linear algebra."

As time has gone on, I keep going back to that as probably the deepest truth I've ever been told.

#mathematics #mathematicseducation #linearalgebra #universityoftexas

Ebokifyebokify
2025-12-18
⏚ Antoine Chambert-Loirantoinechambertloir@mathstodon.xyz
2025-11-30

The determinant of transvections (an update). New blog post on Freedom math dance.

In the previous post, I had explained how I could prove a general version of the classic fact that transvections have determinant 1. Here, I show than one can do more with less effort!

freedommathdance.blogspot.com/

#math #LinearAlgebra

N-gated Hacker Newsngate
2025-11-14

Ah, the age-old enigma: 🧙‍♂️ words that defy . Clearly, the only solution is to whip out some 🧮 linear algebra! Because why struggle with when you can just slam it on a and call it a day? 🙄
aethermug.com/posts/linear-alg

2025-11-08

just in case anyone is interested in how to do complex linear algebra, including complex tensor products, without using complex numbers - using real linear operators \(J\) with \(J\circ J=-I\) instead of complex scalars - here are some links.  

The TLDR version on a poster:
mwaa.math.indianapolis.iu.edu/

Using a basis, matrices, and summing over indices - sections 4&5 of this paper:
users.pfw.edu/CoffmanA/pdf/bas

Without a basis (but still a lot of notation) - Chapter 5 starting on page 195, with tensor products starting with Example 5.74 on page 208:
users.pfw.edu/CoffmanA/pdf/boo
#LinearAlgebra #ComplexNumbers #NotASubToot

⏚ Antoine Chambert-Loirantoinechambertloir@mathstodon.xyz
2025-11-06

The determinant of transvections. — New blog post on Freedom Math Dance

A transvection in a K-vector space V is a linear map T(f,v) of the form x↦x+f(x)v, where f is a linear form and v is a vector such that f(v)=0. It is known that such a linear map is invertible, with inverse given by f and −v. More precisely, one has T(f,0)=id and T(f,v+w)=T(f,v)∘T(f,w). In finite dimension, these maps have determinant 1 and it is known that they generate the special linear group SL(V), the group of linear automorphisms of determinant 1.

When I started formalizing in Lean the theory of the special linear group, the question raised itself of the appropriate generality for such results. In particular, what happens when one replaces the field K with a ring R and the K-vector space V with an R-module?

freedommathdance.blogspot.com/

#math #LinearAlgebra #AbstractAlgebra

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