# A couple linear algebra questions (basis and linear transformation

by Roo2
Tags: algebra, basis, couple, linear, transformation
 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.
 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.
 Sci Advisor HW Helper PF Gold P: 2,606 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 $$V = \{f| f:\mathbb{R}\rightarrow\mathbb{R}\}$$ 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 $$(f_1, f_2,\ldots)$$ 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 $$\text{ker}(T) \subset V$$ is still defined by the real definition of kernel: $$\text{ker}(T) = \{ v\in V | T(v) =0\}.$$ 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.
P: 41

## A couple linear algebra questions (basis and linear transformation

 Quote by lkh1986 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.
Alright, I'll give it a shot.

x = -y -2z

{(y,z) | -y-2z, y, z}

{(y,z) | y(-1, 1, 0) + z(-2, 0, 1)}

Basis: {(-1, 1, 0), (-2, 0, 1)}

Is that correct?
P: 100
 Quote by Roo2 Alright, I'll give it a shot. x = -y -2z {(y,z) | -y-2z, y, z} {(y,z) | y(-1, 1, 0) + z(-2, 0, 1)} Basis: {(-1, 1, 0), (-2, 0, 1)} Is that correct?
{(x,y,z) | x + y + 2z = 0}
={(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 =-x-2z}
={(x,-x-2z,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]
Math
Emeritus
Thanks
PF Gold
P: 38,706
 Quote by Roo2 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?
Wouldn't you think x+ y+ 2z= 0 is relevant?

 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?
Interesting. In all the years I have been working with linear algebra, I have never even heard of the "basis" of a matrix! The only basis I know ,defined for linear algebra, is the basis of a vector space or subspace which is what is asked for here.

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.

 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?)
Only in the sense that the solution is 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?[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).

 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.
If $\lambda$ is an eigenvalue of linear transformation T, then there exist non-zero vectors, v, such that $Tv= \lambda v$. That means, in particular, that $T(v- \lambda v)= (T- \lambda I)v= 0$ so the space of all eigenvalues of T, corresponding to eigenvalue $\lambda$ is the kernel of the linear transformation $T- \lambda I$. By subtracting $\lambda$ from each of the diagonal values, you have given the matrix for $T- \lambda I$. Yes, you are looking for the kernal of that matrix. You must have $x_2= x_3= 0$ but have no condition on $x_1$ so it can be anything. Yes, (t, 0, 0), where t can be any number, is an eigenvector.

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