Question

Consider the vector space M22 of all 2*2 matrices.

2.1 Show that B ={A1,A2,A3,A4} is a basis for M22 where:
A1[3 6]; A2=[0 -1]; A3= [0 -8]; and A4= [1 0]
[3 -6] [-1 0] [-12 -4] [-1 2]

2.2 Write A=[6 2]
[5 3]

2.3 Prove that the subset

D={A element of M22: A^T+A=0}
forms a subspace of M22

Answer 1

If B is a basis, then its linearly independant

Answer 2

how would we row reduce a matrix of matrixes? :)

Answer 3

the standard, or identity of this would amount to:

10 01 00 00
00 00 10 01

or something of that sort

Answer 4

the rest of it I got no clue about :)

Answer 5

[3 6] [0 -1] [0 -8]; [1 0] 00
c1 [3 -6]+c2 [-1 0] +c3 [-12 -4] +c4 [-1 2] = 00 has only the trival solution maybe? :)

Answer 6

thank you , sorry (2.2) says write A [6 2] as a linear combination of vectors from B
[5 3]

Answer 7

\[\begin{pmatrix}3c_1+0c_2+0c_3+1c_4 = 0&6c_1-1c_2-8c_3+0c_4=0 \\
3c_1-1c_2-12c_3-1c_4 = 0&-6c_1+0c_2-4c_3+2c_4=0 \\
\end{pmatrix}\]

this is what I get when I add up all the parts :) its a system of 4 equations in for unknowns that we can form into a super matrix ... my own term nothing technical :)

Answer 8

and also with this super matrix we can evaluate the part 2

Answer 9

and keep in mind that im going at this blind, I got no idea if this is right; but it just makes sense :)

Answer 10

\[\begin{pmatrix}
3c_1+0c_2+0c_3+1c_4 = 0\\
6c_1-1c_2-8c_3+0c_4=0 \\
3c_1-1c_2-12c_3-1c_4 = 0\\
-6c_1+0c_2-4c_3+2c_4=0 \\
\end{pmatrix}\]

\[rref \begin{pmatrix}
3&0&0&1&|& 0\\
6&-1&-8&0&|&0 \\
3&-1&-12&-1&|& 0\\
-6&0&-4&2&|&0 \\
\end{pmatrix}\]

rref{{3,0,0,1,0},{6,-1,-8,0,0},{3,-1,-12,-1, 0},{-6,0,-4,2,0}} into the wolf and:

http://www.wolframalpha.com/input/?i=rref%7B%7B3%2C0%2C0%2C1%2C0%7D%2C%7B6%2C-1%2C-8%2C0%2C0%7D%2C%7B3%2C-1%2C-12%2C-1%2C+0%7D%2C%7B-6%2C0%2C-4%2C2%2C0%7D%7D

its good lol

Answer 11

so, if we simply create a new matrix with columns of [r1,r2] of the given matrixes and row reduce we can determine if it forms a basis :)

Answer 12

with this new matrix; lets determine if we can create a [b1,b2] in the same fashion

Answer 13

[6,2,5,3]
\[rref \begin{pmatrix}
3&0&0&1&|& 6\\
6&-1&-8&0&|&2 \\
3&-1&-12&-1&|& 5\\
-6&0&-4&2&|&3 \\
\end{pmatrix}\]

enter the wolf :) and since we established that the given matrix vector are a basis for M2x2, we should be able to find a solution to B

Answer 14

http://www.wolframalpha.com/input/?i=rref%7B%7B3%2C0%2C0%2C1%2C6%7D%2C%7B6%2C-1%2C-8%2C0%2C2%7D%2C%7B3%2C-1%2C-12%2C-1%2C+5%7D%2C%7B-6%2C0%2C-4%2C2%2C3%7D%7D

B even has a unique solution; just read off the rightside column vector

Answer 15

3/2 A1+25A2 -9/4 A3 +3/2 A4 = A

Answer 16

this sound right to you, or at least makes as much sense to youas it does me?

Answer 17

D={ A element of M2x2: A^T+A=0} forms a subspace of M22

Does any A matrix in M2x2 such that A transpose + A = 0 form a subspace? is how i read this

Answer 18

00^t + 00 = 00
00 00 00

so the zero component of a vector space is there

Answer 19

there is a scalar definition to test and a closure under addition to test as well but I got no idea how to implement them

Answer 20

Does A^t + A = 0 imply that C(A^t+A)=0?

Answer 21

\[
\begin{pmatrix}a&c\\b&d\end{pmatrix}+\begin{pmatrix}a&b\\c&d\end{pmatrix}=
\begin{pmatrix}0&0\\0&0\end{pmatrix}\]

a+a = 0 when a=0
d+d=0 when d=0

\[
\begin{pmatrix}0&c\\b&0\end{pmatrix}+\begin{pmatrix}0&b\\c&0\end{pmatrix}=
\begin{pmatrix}0&0\\0&0\end{pmatrix}\]

c+b = 0 and b+c = 0 when b and c are opposities

\[
\begin{pmatrix}0&-b\\b&0\end{pmatrix}+\begin{pmatrix}0&b\\-b&0\end{pmatrix}=
\begin{pmatrix}0&0\\0&0\end{pmatrix} \]

now we have a generic of the form that we can test :)

Answer 22

multiplying by a constant will not change the outcome;

\[k\left[\begin{pmatrix}0&-b\\b&0\end{pmatrix}+\begin{pmatrix}0&b\\-b&0\end{pmatrix}\right]=
\begin{pmatrix}0&0\\0&0\end{pmatrix}\]
\[k\begin{pmatrix}0&-b\\b&0\end{pmatrix}+k\begin{pmatrix}0&b\\-b&0\end{pmatrix}=
\begin{pmatrix}0&0\\0&0\end{pmatrix}\]
\[\begin{pmatrix}0&-kb\\kb&0\end{pmatrix}+\begin{pmatrix}0&kb\\-kb&0\end{pmatrix}=
\begin{pmatrix}0&0\\0&0\end{pmatrix}\]

Answer 23

when we add 2 off these together of this form; do we get the same form back?

Answer 24

\[\begin{pmatrix}0&-b\\b&0\end{pmatrix}+\begin{pmatrix}0&b\\-b&0\end{pmatrix}=
\begin{pmatrix}0&0\\0&0\end{pmatrix}\]\[+\begin{pmatrix}0&-n\\n&0\end{pmatrix}+\begin{pmatrix}0&n\\-n&0\end{pmatrix}=
\begin{pmatrix}0&0\\0&0\end{pmatrix}\]--------------------------
\[\begin{pmatrix}0&-(b+n)\\(b+n)&0\end{pmatrix}+\begin{pmatrix}0&(b+n)\\-(b+n)&0\end{pmatrix}=
\begin{pmatrix}0&0\\0&0\end{pmatrix}\]

yep, so its colsed under addition too

Answer 25

by default the rest of the requirements of a vector space are satisified by these 3 or so I have been led to believe.

Answer 26

sooo, good luck with it ;)

Answer 27

thank you

Answer 28

youre welcome, and thank you for giving me something to work on other than the distance between 2 points on a graph ;)

Answer 29

lol