# Linear Algebra has me stumped

• April 6th 2013, 10:33 AM
zixxer
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.
• April 6th 2013, 10:42 AM
Plato
Re: Linear Algebra has me stumped
Quote:

Originally Posted by zixxer
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?
• April 6th 2013, 10:52 AM
zixxer
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?
• April 6th 2013, 10:55 AM
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
• April 6th 2013, 11:03 AM
Plato
Re: Linear Algebra has me stumped
Quote:

Originally Posted by zixxer
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.
• April 6th 2013, 01:05 PM
zixxer
Re: Linear Algebra has me stumped
So I have done some looking around and reading and for question 2 I found this...
Attachment 27832

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?
• April 6th 2013, 01:31 PM
Plato
Re: Linear Algebra has me stumped
Quote:

Originally Posted by zixxer
So I have done some looking around and reading and for question 2 I found this...
Attachment 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 $(A+B)\in U~\&~(\forall t\in\mathbb{R})[tA\in U]~?$
• April 6th 2013, 01:56 PM
zixxer
Re: Linear Algebra has me stumped
I am unsure what you are saying, I am not familiar with the notation that you are using.
• April 6th 2013, 02:09 PM
Plato
Re: Linear Algebra has me stumped
Quote:

Originally Posted by zixxer
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.
• April 6th 2013, 02:45 PM
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.
• April 6th 2013, 02:59 PM
zixxer
Re: Linear Algebra has me stumped
Quote:

Originally Posted by Plato
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: