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A couple linear algebra questions (basis and linear transformation 
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#1
Oct3110, 12:35 AM

P: 41

Hi guys, I have two practice problems with no solutions that i was not able to figure out. If anyone could help I'd appreciate it.
Question 1 1. The problem statement, all variables and given/known data Find the basis of {(x,y,z)  x + y + 2z = 0} 2. Relevant equations None? 3. The attempt at a solution I can find the basis of a matrix but I'm not sure how to go about finding the matrix of a homogeneous equation. Can I just turn it into a 1x3 matrix, give it the coefficients (1,1,2) and call that my basis? Question 2 1. The problem statement, all variables and given/known data Let T(y) = y'' + y. Find ker(T) 2. Relevant equations ker(T) = nullspace(matrix(T)) solution to y + y'' = 0 is asin(x) + bcos(x) (Relevant?) 3. The attempt at a solution I'm not even sure how to begin this. I know what the kernel of a linear transformation is but I am not sure how to turn y + y'' into a matrix. Could I do as follows? y y' y'' 1 0 1 [1, 0, 1  0] = REF [1, 0, 1  0] y = y'' Would that be the solution? Finally, I have a quick question that doesn't really warrant the entire template. I'm to calculate an eigenvector and after plugging one of the eigenvalues I found into the matrix, I got 0 2 0 0 1 3 0 0 1 Row reduction gives 0 1 3 0 0 1 0 0 0 Setting the matrix equal to 0: x3 = 0 x2 = 3(x3) = 0 x1 = t? In this case, is my eigenvector (t,0, 0)? Sorry for all the questions; the exam is on Monday and the prof didn't give out any answers to his study materials. I just got around to these questions so I would really like to figure them out before the exam. Thanks for any help or advice. 


#2
Oct3110, 01:09 AM

P: 100

For question 1 with the condition x + y + 2z = 0, try to express one of the variables in other two variables. For example, x = y  2z. Then, replace x with y  2z in {(x,y,z)x,y,z are real}. Lastly, try to "factor" or decompose the vector into several vectors to get the basis.



#3
Oct3110, 01:53 AM

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For the second question, you should really go back to the definitions of vector spaces and linear transformations. It is often helpful, but by no means necessary to use a matrix to represent a linear transformation.
In question 2, the vector space in question is the set [tex]V = \{f f:\mathbb{R}\rightarrow\mathbb{R}\}[/tex] of differentiable functions f from the reals to the reals. This space is infinite dimensional, but we still know how to add functions and multiply them by scalars. However we can't really write them down in a list [tex](f_1, f_2,\ldots)[/tex] unless we impose further constraints that would let us define a basis. Furthermore, the linear operator T is a differential operator, and since we don't have a basis yet for the space on which it acts, we can't write a matrix down for it. However [tex]\text{ker}(T) \subset V[/tex] is still defined by the real definition of kernel: [tex]\text{ker}(T) = \{ v\in V  T(v) =0\}.[/tex] In fact, as you showed, it is spanned by only two functions, so we can write the basis down for it. However, as you saw, you didn't need to refer to any matrix or nullspace to compute it. For your last question, I'm baffled by what you mean by "plugging one of the eigenvalues I found into the matrix." The eigenvector equation is M v = a v. You should take the most general parametrization of v = (x,y,z) and solve the set of equations that result. If the matrix you wrote down is M a I, then you probably have the right value. 


#4
Oct3110, 10:38 AM

P: 41

A couple linear algebra questions (basis and linear transformation
x = y 2z {(y,z)  y2z, y, z} {(y,z)  y(1, 1, 0) + z(2, 0, 1)} Basis: {(1, 1, 0), (2, 0, 1)} Is that correct? 


#5
Oct3110, 12:04 PM

P: 100

={(x,y,z)  x = y 2z} ={(y 2z,y,z)  y,z are real} ={y(1,1,0) + z(2,0,1)  y, z are real} =L{(1, 1, 0), (2, 0, 1)} Basis: {(1, 1, 0), (2, 0, 1)} Yup, same answer obtained. :) Or, if you have chosen to express y in terms of x and z, you will get: {(x,y,z)  x + y + 2z = 0} ={(x,y,z)  y =x2z} ={(x,x2z,z)  x,z are real} ={x(1,1,0) + z(0,2,1)  x, z are real} =L{(1, 1, 0), (0, 2, 1)} Basis: {(1, 1, 0), (0, 2, 1)} [I think this answer is also acceptable] 


#6
Nov210, 09:05 AM

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P: 39,346

From x+y+ 2z= 0, x= y 2z. In particular, (x, y, z)= (y 2z, y, z)= (y, y, 0)+ (2z, 0, z)= y(1, 1, 0)+ z(2, 0, 1). Now what do you think a basis would be? Notice that we could as easily have said y= x 2z so that (x, y, z)= (x, x 2z, z)= (x, x, 0)+ (0, 2z, z)= x(1, 1, 0)+ z(0, 2, 1) giving another answer to the same question. Or, z= x/2 y/2 so that (x, y, z)= (x, y, x/2 y/2)= (x, 0, x/2)+ (0, y, y/2)= x(1, 0, 1/2)+ y(1, 0, 1/2) giving a third basis. Of course, there exist an infinite number of "bases" for any vector space. 3. The attempt at a solution I'm not even sure how to begin this. I know what the kernel of a linear transformation is but I am not sure how to turn y + y'' into a matrix. Could I do as follows? y y' y'' 1 0 1 [1, 0, 1  0] = REF [1, 0, 1  0] y = y'' Would that be the solution?[quote] Why do you keep talking about matrices? Linear transformations are more general, more fundamental, and more important than matrices. The kernel of a linear transformation is the set of all vectors that it maps to 0. Here, since T involves differentiation, the vector space is the space of all infinitely differentiable functions with function addition as vector addition. T(y)= y+ y"= 0 is satisified, as you say, by asin(x)+ bcos(x). The kernel of T is exactly the set of all functions of the form f(x)= a sin(x)+ b cos(x). It is the subspace spanned by sin(x) and cos(x). 


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