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Math Help - Linear Algebra has me stumped

  1. #1
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    Linear Algebra has me stumped

    Hi I am new to this forum and I would like to say that normally I am very good at math, I actually too Calculus 1 and 2 in High school and would have taken more if I had the time.

    Anyways onto my question(s). I have 2.

    The first I am trying to decide of the transformation is linear or not. The transformation is

    T: F -> R, T(x) = Integral(x'(t)/(t2+1))dt from 0 to 1

    I understand that for a transformation to be linear it must satisfy T(x+y) = T(x) + T(y) and T(cx) = cT(x), or closed under addition and scalar multiplication. But I am confused as to where to start on the problem above.

    The second problem I am having is finding the basis and dimensions of

    U = {all [a11 6a12 b21 5a22]}

    Once again I would know how to do this if it weren't for the a's and b.
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  2. #2
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    Re: Linear Algebra has me stumped

    Quote Originally Posted by zixxer View Post
    The first I am trying to decide of the transformation is linear or not. The transformation is
    T: F -> R, T(x) = Integral(x'(t)/(t2+1))dt from 0 to 1
    I understand that for a transformation to be linear it must satisfy T(x+y) = T(x) + T(y) and T(cx) = cT(x), or closed under addition and scalar multiplication. But I am confused as to where to start on the problem above.

    The second problem I am having is finding the basis and dimensions of
    U = {all [a11 6a12 b21 5a22]}
    Once again I would know how to do this if it weren't for the a's and b.

    For #1, you failed to give any clue as what F may be. I assume that R is the set of reals.

    For #2, I have no idea what you mean. Can you explain?
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  3. #3
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    Re: Linear Algebra has me stumped

    1) After looking to the beginning of the chapter in my book, it states "Let F, as before, be the vector space of all smooth functions" and yes R is the set of all reals.

    2) Sorry it seems that I left out some of this problem U = {all [a11 6a12 b21 5a22]} that U is the set of all 2x2 matrices A. A better way to visualize U is...

    U = |a 6a|
    ......|b 5a|

    where | are brackets.

    I hope this clears some stuff up?
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    Re: Linear Algebra has me stumped

    For number 1. It seems F would be the set of differentiable functions from R to R. This is the only way the question would make sense.
    The rules of integration and differentiation would tell us that we can pull out a constant, so T(cx(t))=cT(x(t)) is definitely satisfied. The rules of integration also give,
    T(x(t)+y(t))=Integral[((x(t)+y(t)))'/(t^2+1)]
    =Integral[(x'(t)+y'(t))/(t^2+1)] by the addition rule of differentiation
    =Integral[(x'(t)/(t^2+1)] + Integral[y'(t)/(t^2+1)] by the addition rule for integration
    =T(x(t)) + T(y(t)) by the definition of T.
    So it seems this is a linear transformation. As a side note, I feel that derivatives and /(t^2+1) were just a distraction
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    Re: Linear Algebra has me stumped

    Quote Originally Posted by zixxer View Post
    1) After looking to the beginning of the chapter in my book, it states "Let F, as before, be the vector space of all smooth functions" and yes R is the set of all reals.

    2) Sorry it seems that I left out some of this problem U = {all [a11 6a12 b21 5a22]} that U is the set of all 2x2 matrices A. A better way to visualize U is...
    U = |a 6a|
    ......|b 5a|
    For #1, recall that:
    (x(t)+y(t))'=x'(t)+y'(t)~\&~[cx(t)]'=c[x'(t)]

    AND \int_0^1 {\left( {f + g} \right)}  = \int_0^1 {\left( f \right)}  + \int_0^1 {\left( g \right)}~\&~\int_0^1 {\left( cf \right)}=c\int_0^1 {\left( f \right)}


    You show effort on #2.
    Last edited by Plato; April 6th 2013 at 11:07 AM.
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  6. #6
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    Re: Linear Algebra has me stumped

    So I have done some looking around and reading and for question 2 I found this...
    Linear Algebra has me stumped-image-2-.png

    So in my case according to this I would have

    a*[1,6;0,5] and b*[0,0;1,0]

    or

    [a,6a;b,5a] = a*[1,0;0,0] + 6a*[0,1;0,0] + b[0,0;1,0] + 5a[0,0;0,1]

    But is this my basis or do I have several more steps to go?
    Last edited by zixxer; April 6th 2013 at 01:25 PM.
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  7. #7
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    Re: Linear Algebra has me stumped

    Quote Originally Posted by zixxer View Post
    So I have done some looking around and reading and for question 2 I found this...
    Click image for larger version. 

Name:	image (2).png 
Views:	5 
Size:	34.6 KB 
ID:	27832

    So in my case according to this I would have

    a*[1,6;0,5] and b*[0,0;1,0]

    But is this my basis or do I have several more steps to go?

    Suppose that A = \left[ {\begin{array}{*{20}{c}}a&{6a} \\   b&{5b} \end{array}} \right]\;\& \;B = \left[ {\begin{array}{*{20}{c}}  c&{6c} \\   d&{5d} \end{array}} \right]

    Is it clear to you that \{A,B\}\subset U~?

    Is it true that (A+B)\in U~\&~(\forall t\in\mathbb{R})[tA\in U]~?
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  8. #8
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    Re: Linear Algebra has me stumped

    I am unsure what you are saying, I am not familiar with the notation that you are using.
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    Re: Linear Algebra has me stumped

    Quote Originally Posted by zixxer View Post
    I am unsure what you are saying, I am not familiar with the notation that you are using.

    Well then, there is no wonder that Linear Algebra has you stumped. You are simply not ready for it.
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  10. #10
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    Re: Linear Algebra has me stumped

    Your basis would be what you have written: [1,6;0,5] and [0,0;1,0].
    Every element of the set U = {[a,6a;b,5a] : a,b are real} can always be written as a linear combination of the two elements. Since it can be written uniquely in this way, it is a basis. Does that make sense? If W = {[a+b, b; 2a+b, 3a] : a,b are real}, a basis would be {[1,0; 2, 3], [1,1; 1,0]}. The comment you listed above is saying that {[1,0; 0,0], [0,1; 0,0], [0,0; 1,0], [0,0; 0,1]} is a basis for the space of all 2x2 matrices, but there are many more bases for this set.
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  11. #11
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    Re: Linear Algebra has me stumped

    Quote Originally Posted by Plato View Post
    Well then, there is no wonder that Linear Algebra has you stumped. You are simply not ready for it.
    I wouldn't say that's true. I have just personally never seen several of those symbols used in any of my classes.

    Quote Originally Posted by adambuck View Post
    Your basis would be what you have written: [1,6;0,5] and [0,0;1,0].
    Every element of the set U = {[a,6a;b,5a] : a,b are real} can always be written as a linear combination of the two elements. Since it can be written uniquely in this way, it is a basis. Does that make sense? If W = {[a+b, b; 2a+b, 3a] : a,b are real}, a basis would be {[1,0; 2, 3], [1,1; 1,0]}. The comment you listed above is saying that {[1,0; 0,0], [0,1; 0,0], [0,0; 1,0], [0,0; 0,1]} is a basis for the space of all 2x2 matrices, but there are many more bases for this set.
    So it is in fact my basis, I understand with those 2 matrices I can get my whole set of U, which is what a basis is. It just seems overly simple, or maybe I am thinking that it is harder than it is? I know that I thought the first question I had was harder then it really was.
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