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March 27

Linear algebra

I have two questions. First, can the rank — $dim (range (T))$ where $T$ is a linear transformation — of a linear transformation be zero? My question originates from this problem:

Use the given information to find the rank of the linear transformation

T

where

T : V \to W

. The null space of

T : P 5 \to P 5

is

P 5

.

I used the rank–nullity theorem and produced the following:

rank(T) + nullity (T) = dim(V) nullity(T) = 6, dim(V) = 6 rank(T) + 6 = 6 rank(T) = 0

Is this result correct? I feel like I erred somewhere.

My second question is the following problem:

Let

T : R 3 \to ℝ 3

be the transformation defined as

T(v) = w

where

w

is the orthogonal projection of

v

onto the plane

x + y + z = 0

. You can take as a fact that

T

is a linear transformation. Find the matrix

A

such that

T = T A

.

Thank you in advance for any and all help. 147.126.10.21 (talk) 04:40, 27 March 2017 (UTC)[reply]

@147.126.10.21: Yes, zero rank is precisely the rank of the zero map from V to W. Exercise: Prove that the only linear transformation with zero rank is the zero map.

As for the second question, I assume you want the expression of the projection map in terms of a basis. So first you need a basis. Without one specified I assume you wish to work with the standard basis. Now what are the images of the standard basis vectors under this map? Additional exercise: show that projection is indeed a linear mapping.--Jasper Deng (talk) 08:06, 27 March 2017 (UTC)[reply]

Heptomino tiling

The heptomino article's section on tiling with heptominos needs to be made more interesting. Using the 107 non-hole heptominos, can anyone make a 750-square grid with just one square removed?? Georgia guy (talk) 13:38, 27 March 2017 (UTC)[reply]

Well, I see that you can make the 7x107 from the 107 non-hole heptominoes (https://books.google.com/books?id=D8KbnTGXDWEC&pg=PA123) , not sure if that helps get you there. There are also choices in the dimensions of the 750 which would be different problems. 750 = 2x3x5^3, and eliminating those factors below 7, that gives 10x75, 15x50, and 25x30.Naraht (talk) 14:05, 27 March 2017 (UTC)[reply]

Inscribe regular n+1 gon in n-gon percentage?

Is there a formula for the maximum percentage of a regular n-gon's area that can be covered by an inscribed regular n+1 gon? (note, I'm not sure whether this maximum always forces all of the n+1 gon's vertices to be on the edges of the n-gon.Naraht (talk) 13:53, 27 March 2017 (UTC)[reply]

By Triangle#Squares, for the n=3 case the maximum covered fraction of the area is 1/2, occurring for the isosceles right triangle with a side of the square coinciding with part of the hypotenuse and all four of its vertices on the triangle's edges. Loraof (talk) 21:03, 27 March 2017 (UTC)[reply]

Whoops, you said regular polygons. The answer for the n=3 case can be inferred from the equation there. Loraof (talk) 21:24, 27 March 2017 (UTC)[reply]

Agreed, no clue on a Pentagon inscribed in a square or beyond.Naraht (talk) 17:52, 30 March 2017 (UTC)[reply]

Consider rotating the inner polygon, all the time keeping it as large as possible within the outer one. For some angles of rotation the figure as whole (both polygons taken together), will have bilateral symmetry. The size of the inner polygon must be a local extremum at these angles. This is because rotating the inner polygon in either direction away from the symmetrical state will have the same effect on its size (it will grow or shrink by the same amount, because the rotations will produce mirror image configurations).

There are two classes of symmetrical configuration; either (A) one side of the inner polygon is parallel (or even collinear) with the neighbouring side of the outer polygon, or (B) one corner of the inner polygon points directly into a corner of the outer polygon (the vertices are at the same angle relative to centres of the polygons).

In a full rotation of the inner polygon, there are n(n+1) of the (A) configurations, interleaved with n(n+1) of the (B) configurations. It's conceivable that these are not the only extrema, but it's far more plausible that either the (A)s are the global maximum and the (B)s are the global minimum or vice-versa.

This means that for each n, there are only really two configurations to consider in order to maximize the size of the inner polygon. For a square in a triangle, the (A) configuration is the answer, and for a pentagon in a square it's (B). --catslash (talk) 23:37, 30 March 2017 (UTC)[reply]

pentagram in a pentagon question.

An inscribed regular pentagram in a pentagon would have both figures with the same vertices. Is there a name for the pentagram in a pentagon whose vertices are the midpoints of the pentagon's sides?Naraht (talk) 13:55, 27 March 2017 (UTC)[reply]

Intuitively I'd want to call it "alternating inscribed pentagons". StuRat (talk) 20:42, 29 March 2017 (UTC)[reply]

Alternating Pentagons and Pentagrams?Naraht (talk) 17:53, 30 March 2017 (UTC)[reply]

Yes, "alternating inscribed pentagons and pentagrams", but perhaps the "alternating" might just be taken to mean swapping one shape for the other, leaving us without a short term for it. StuRat (talk) 17:58, 30 March 2017 (UTC)[reply]

Is this the one where the Pentagram's points are in the corners of the Pentagon on in the middle of the sides?Naraht (talk) 18:59, 30 March 2017 (UTC)[reply]

As per Inscribed figure, a polygon is said to be inscribed in another polygon if all the vertices of the former polygon are on the edges of the latter one. So both the case of sharing vertices and the case of vertices on the midpoints of the sides are "an inscribed regular pentagram in a pentagon". I don't know of any specific names that distinguish the cases. Loraof (talk) 20:31, 30 March 2017 (UTC)[reply]