Finding basis of spaces and dimension


by maximade
Tags: basis, matrices, space
maximade
maximade is offline
#1
Feb28-11, 08:50 PM
P: 27
1. The problem statement, all variables and given/known data
Find a basis for each of the spaces and determine its dimension:
The space of all matrices A=[a b, c d] (2x2 matrix) in R^(2x2) such that a+d=0


2. Relevant equations



3. The attempt at a solution
So I jumped at this question without knowing too much about spaces and dimensions, but:
I think a possible combination of basis can be: [1 0, 0 0]. [0 1, 0 0]. [0 0, 1 0] (not sure if [0 0, 0 -1] would be considered since d would be negative in this case) Also from that I assume the dimension is 3?
Truthfully even if I got it right, I do not even know what happened. Can someone conceptually tell me what I am doing exactly?
Thanks in advance.
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lanedance
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#2
Feb28-11, 08:54 PM
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you need to come up with a basis, that has elements, such that any 2x2 matrix with a=d can be written as a linear sum of the basis elements.

clearly the basis elements will need to satisfy being a 2x2 matrix with d=a, you first element does not
lanedance
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#3
Feb28-11, 08:55 PM
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note this is entirely equivalent to considering 4 component vectors in R^4, with x_1 = x_4

often the vector form is easier to conceptualise

maximade
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#4
Feb28-11, 09:04 PM
P: 27

Finding basis of spaces and dimension


Quote Quote by lanedance View Post
you need to come up with a basis, that has elements, such that any 2x2 matrix with a=d can be written as a linear sum of the basis elements.

clearly the basis elements will need to satisfy being a 2x2 matrix with d=a, you first element does not
Can you explain to me how d=a, since a+d=0?
Also since I am actually having a hard time learning this on my own, can you tell me what you mean by "elements" and "linear sum"?
lanedance
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#5
Mar1-11, 01:52 AM
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Quote Quote by maximade View Post
Can you explain to me how d=a, since a+d=0?
Also since I am actually having a hard time learning this on my own, can you tell me what you mean by "elements" and "linear sum"?
good pickup, should be a = -d

if you're not familair with those terms, you may need to do a bit of reading.. though i have been a little loose with terminiology

in post #4 i actually meant component and have changed accordingly

a basis, is a set of vectors that spans a vector space

an element is a member of a set, for example a vector in the basis set

a linear combination (or sum) is a vector addition with scalar multiplication
eg. if u,v are vectors, and a,b are scalars, then
w = au + bv is a linear combination

a set, S, of vectors spans a space if any vector in the space can be written as a linear combination of vectors in S
HallsofIvy
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#6
Mar1-11, 07:03 AM
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Thanks
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P: 38,877
From a+ d= 0 you get d=-a as you say. You can then write
[tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}= \begin{bmatrix}a & b \\ c & -a\end{bmatrix}[/tex]
[tex]= \begin{bmatrix}a & 0 \\ 0 & -a\end{bmatrix}+ \begin{bmatrix}0 & b \\ 0 & 0\end{bmatrix}+ \begin{bmatrix}0 & 0 \\ c & 0\end{bmatrix}[/tex]
and the dimension and a basis should be clear.


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