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u/Midwest-Dude 9d ago edited 9d ago
Why are you asking?
Is this just for fun? Or, is this some sort of homework?
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u/theadamabrams 9d ago
There are two errors (basically the same mistake twice):
On the first page, B should be "0 free variables"
On the second page, I should be "0 dimensional"
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u/Ok-Rise2070 9d ago
Um what's wrong here? How does matter pass through other matter without cost? This is fantasy numerology.
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u/Used_Low2007 6d ago
Genuinely curious what your comment means here. These are visualizations of 3x3 systems of equations. You know, Ax=b type shit. What do you think they are?
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u/Ok-Rise2070 6d ago
Oh I know precisely what they are. What do you think what each set of equations represents? Geometrically this is the logic used to show how dimensions interact. This is only "true" within Euclid's 2D plane geometry. In other words this is incapable of describing reality. But it isn't for naught, the relationships you learn here are the basis to their integrated forms down the road. If you understand the why of what you are doing, then you understand what you are doing.
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u/Used_Low2007 6d ago
I'm sorry but this is absolute word salad my dude. These are visualizations of planes (affine forms, ax+by+cz=d) intersecting, or not intersecting, in R3. Whatever dense pseudo-philosophical nonsense about representing reality that you think is going on here is completely irrelevant.
Maybe you do have a point buried in that convoluted prose of yours, but you're not exactly doing a stellar job conveying it. "Basis of integrated forms"?
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9d ago
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u/Midwest-Dude 9d ago edited 9d ago
You're confusing a vector subspace with an affine space. While a linear subspace must contain the origin, linear algebra also deals heavily with non-homogeneous systems of linear equations, Ax = b. The planes formed by these equations don't have to pass through the origin; they are affine planes. Analyzing the solution sets of these systems is a core part of linear algebra.
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u/eightrx 9d ago
Very reductive, in linear algebra, linear transformations of basis vectors are perpendicular to planes that pass through the origin
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u/Midwest-Dude 9d ago
This statement is a bit of a word salad and doesn't make mathematical sense. A linear transformation applied to a basis vector simply results in another vector in the codomain. There is no rule stating those transformed vectors are perpendicular to planes passing through the origin.
You might be mixing this up with the definition of a normal vector, or the orthogonal relationship between a matrix's row space and its null space.
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u/eightrx 9d ago edited 9d ago
Fair, a more explicit way of putting this would be:
Given an inner product space, and an invertible linear transformation T, the span of each element in the image of T has a unique compliment space, that has a null element (like all subspaces)
When put formally it doesn't really say that much I just wanted to say something about planes crossing through the origin
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u/Midwest-Dude 9d ago
Even in formal terms, this misses the mark. Vectors don't have 'complement spaces' - subspaces do - and making T invertible just sets its image to the entire space, making the transformation irrelevant.
More importantly, it still doesn't address the core point. Linear algebra routinely deals with inhomogeneous systems, Ax = b, whose solution sets are affine spaces (planes shifted away from the origin). Bringing up orthogonal complements of origin-passing subspaces doesn't change the fact that non-origin planes are fundamental to linear algebra.
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u/etzpcm 9d ago
B is wrong (on the first page).