# A couple linear algebra questions (basis and linear transformation

• Roo2
In summary: What "basis" to those two vectors form?Question 2Homework Statement Let T(y) = y'' + y. Find ker(T)Homework Equationsker(T) = nullspace(matrix(T))solution to y + y'' = 0 is asin(x) + bcos(x) (Relevant?)The Attempt at a SolutionI'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,
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

## Homework Statement

Find the basis of {(x,y,z) | x + y + 2z = 0}

None?

## 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

## Homework Statement

Let T(y) = y'' + y. Find ker(T)

## Homework Equations

ker(T) = nullspace(matrix(T))
solution to y + y'' = 0 is asin(x) + bcos(x) (Relevant?)

## 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.

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.

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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 $$V = \{f| f:\mathbb{R}\rightarrow\mathbb{R}\}$$ 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.

lkh1986 said:
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?

Roo2 said:
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)}

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]

Roo2 said:
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

## Homework Statement

Find the basis of {(x,y,z) | x + y + 2z = 0}

## Homework Equations

None?
Wouldn't you think x+ y+ 2z= 0 is relevant?

## 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

## Homework Statement

Let T(y) = y'' + y. Find ker(T)

## Homework 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!

## 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?
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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## 1. What is a basis in linear algebra?

A basis is a set of linearly independent vectors that span a vector space. This means that any vector in the vector space can be written as a linear combination of the basis vectors. In other words, the basis provides a way to represent any vector in the vector space.

## 2. How do you determine if a set of vectors is linearly independent?

A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the others. This can be determined by setting up a system of equations and checking if there is a unique solution, or by using the determinant of the matrix formed by the vectors.

## 3. What is a linear transformation?

A linear transformation is a function between two vector spaces that preserves vector addition and scalar multiplication. In other words, the output of a linear transformation is always a linear combination of its inputs.

## 4. How do you determine if a transformation is linear?

To determine if a transformation is linear, you can check if it preserves vector addition and scalar multiplication. This means that if you apply the transformation to the sum of two vectors, it should be equal to the sum of the transformation applied to each individual vector. Similarly, if you apply the transformation to a vector multiplied by a scalar, it should be equal to the scalar multiplied by the transformation applied to the original vector.

## 5. How do you find the matrix representation of a linear transformation?

To find the matrix representation of a linear transformation, you can apply the transformation to the standard basis vectors of the vector space. The resulting vectors will form the columns of the matrix representation. This matrix can then be used to perform the transformation on any vector by multiplying it with the vector.

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