# Exercise with Linear Transformation

• Kernul
In summary, the matrix A has a rank of 3, which means that there is no basis for the endomorphism defined by f in terms of the vectors in the subspace of dimension 4 that is defined by the columns of A.
Kernul

## Homework Statement

Being ##f : \mathbb R^4\rightarrow\mathbb R^4## the endomorphism defined by:
$$f((x,y,z,t)) = (13x + y - 2z + 3t, 10y, 9z + 6t, 6z + 4t)$$

1) Determine the basis and dimension of ##Ker(f)## and ##Im(f)##. Complete the base chosen in ##Ker(f)## into a base of ##\mathbb R^4##;
2) Determine the basis and dimension of the subspaces of ##\mathbb R^4 Ker(f) \cap Im(f)## and ##Ker(f) + Im(f)##;
3) Determine the preimage of the vector ##\vec v = (0, 2h, 1, h - 4)##, with the variation of the real parameter ##h##;
4) Find, if they exist, the values of the ##h## parameter with ##[2\vec e_1 + \vec e_2, 3\vec e_3, \vec v, -\vec e_2 + \vec e_4]## being a base of ##\mathbb R^4##. ##ε = (\vec e_1, \vec e_2, \vec e_3, \vec e_4)## is the canonic base of ##\mathbb R^4##.
5) Having with ##A## the matrix ##M_{εε}(f)## associated to ##f## in respect to the canonic base, find out if ##A## is diagonalizable and, if so, determine a diagonalizing (?) matrix ##P## and the corresponding diagonal matrix ##D## to which ##A## is similar.
6) Find out if the matrix ##B = \begin{pmatrix} 3 & 0 & 0\\ 1 & 3 & 0\\ 1 & 0 & 1\end{pmatrix}## is diagonalizable and show a maximum independent system of eigenvectors of ##B##

## Homework Equations

Rank-Nullity Theorem
Operation on the vector subspaces (intersection and sum)
Canonic Base
Diagonalization
Eigenvectors and eigenvalues

## The Attempt at a Solution

So, I know how to do the first point of the exercise but I found myself stuck.
In order to find the dimension and basis of both ##Ker(f)## and ##Im(f)## I first have to find the rank of the matrix $$A = \begin{pmatrix} 13 & 1 & -2 & 3\\ 0 & 10 & 0 & 0\\ 0 & 0 & 9 & 6\\ 0 & 0 & 6 & 4 \end{pmatrix}$$
From what I see, the matrix' rank is 4. Knowing that ##ρ(A) = dim(Im(f))##, the base of ##Im(f)## is made of all the columns of ##A##. So: $$B_{Im(f)} = [A^1, A^2, A^3, A^4]$$
Now, knowing that ##dim(Ker(f)) = dim(A) - dim(Im(f))##, I have ##dim(Ker(f)) = 4 - 4 = 0##. This means that ##Ker(f)## has no basis, right? Then how do I complete the basis of ##Ker(f)##(as the exercise asks) if there is none?
For now I would like to concentrate on this first point and then ask other questions for the following points, if needed.

Kernul said:

## Homework Statement

Being ##f : \mathbb R^4\rightarrow\mathbb R^4## the endomorphism defined by:
$$f((x,y,z,t)) = (13x + y - 2z + 3t, 10y, 9z + 6t, 6z + 4t)$$

1) Determine the basis and dimension of ##Ker(f)## and ##Im(f)##. Complete the base chosen in ##Ker(f)## into a base of ##\mathbb R^4##;
2) Determine the basis and dimension of the subspaces of ##\mathbb R^4 Ker(f) \cap Im(f)## and ##Ker(f) + Im(f)##;
3) Determine the preimage of the vector ##\vec v = (0, 2h, 1, h - 4)##, with the variation of the real parameter ##h##;
4) Find, if they exist, the values of the ##h## parameter with ##[2\vec e_1 + \vec e_2, 3\vec e_3, \vec v, -\vec e_2 + \vec e_4]## being a base of ##\mathbb R^4##. ##ε = (\vec e_1, \vec e_2, \vec e_3, \vec e_4)## is the canonic base of ##\mathbb R^4##.
5) Having with ##A## the matrix ##M_{εε}(f)## associated to ##f## in respect to the canonic base, find out if ##A## is diagonalizable and, if so, determine a diagonalizing (?) matrix ##P## and the corresponding diagonal matrix ##D## to which ##A## is similar.
Item 6 below seems unrelated to this problem. Please start a new thread with that problem.
Kernul said:
6) Find out if the matrix ##B = \begin{pmatrix} 3 & 0 & 0\\ 1 & 3 & 0\\ 1 & 0 & 1\end{pmatrix}## is diagonalizable and show a maximum independent system of eigenvectors of ##B##

## Homework Equations

Rank-Nullity Theorem
Operation on the vector subspaces (intersection and sum)
Canonic Base
Diagonalization
Eigenvectors and eigenvalues

## The Attempt at a Solution

So, I know how to do the first point of the exercise but I found myself stuck.
In order to find the dimension and basis of both ##Ker(f)## and ##Im(f)## I first have to find the rank of the matrix $$A = \begin{pmatrix} 13 & 1 & -2 & 3\\ 0 & 10 & 0 & 0\\ 0 & 0 & 9 & 6\\ 0 & 0 & 6 & 4 \end{pmatrix}$$
From what I see, the matrix' rank is 4.
No.
It's fairly obvious (at least to me) that the rank of A is 3. To see this for yourself, row-reduce the matrix.
Kernul said:
Knowing that ##ρ(A) = dim(Im(f))##, the base of ##Im(f)## is made of all the columns of ##A##. So: $$B_{Im(f)} = [A^1, A^2, A^3, A^4]$$
Now, knowing that ##dim(Ker(f)) = dim(A) - dim(Im(f))##, I have ##dim(Ker(f)) = 4 - 4 = 0##. This means that ##Ker(f)## has no basis, right? Then how do I complete the basis of ##Ker(f)##(as the exercise asks) if there is none?
For now I would like to concentrate on this first point and then ask other questions for the following points, if needed.

Yeah, I know it wasn't part of the exercise but it was put there. I don't know why.

Mark44 said:
Item 6 below seems unrelated to this problem. Please start a new thread with that problem.
No.
It's fairly obvious (at least to me) that the rank of A is 3. To see this for yourself, row-reduce the matrix.
Oh yeah, I'm sorry. I tried that but it seems I didn't quite try enough...
Anyway, doing the calculations, I find myself with: ##B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}]##
In order to complete the base of ##Ker(f)## I'll put it together with the base of ##Im(f)##, so:$$B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}, \begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}]$$
I hope I'm right until now.

Now the second point.
I have to find the base and dimension of ##Ker(f) \cap Im(f)## and ##Ker(f)+Im(f)##, knowing that ##dim(Im(f)) = 3## and ##dim(Ker(f)) = 1##. At this point, do I have to find a vector that is in both of them? If yes, should I do something like this:$$\vec v = αA^1 + βA^2 +γA^3 + ωKer(f)$$

Kernul said:
Yeah, I know it wasn't part of the exercise but it was put there. I don't know why.Oh yeah, I'm sorry. I tried that but it seems I didn't quite try enough...
Anyway, doing the calculations, I find myself with: ##B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}]##
No, this isn't right. If you multiply A times the vector above, you don't get the zero vector. Check your work in row reducing A.
Kernul said:
In order to complete the base of ##Ker(f)## I'll put it together with the base of ##Im(f)##, so:$$B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}, \begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}]$$
I hope I'm right until now.
?
I don't know what you're doing here. As it turns out, dim(Ker(A)) = 1, so a basis for the kernel would consist of one vector.
Kernul said:
Now the second point.
I have to find the base and dimension of ##Ker(f) \cap Im(f)## and ##Ker(f)+Im(f)##, knowing that ##dim(Im(f)) = 3## and ##dim(Ker(f)) = 1##. At this point, do I have to find a vector that is in both of them?
For the first part of item 2, ##Ker(f) \cap Im(f)##, yes you need to find the vectors that are in both subspaces.
For the second part, ##Ker(f)+Im(f)##, how is '+' defined for two sets?
Kernul said:
If yes, should I do something like this:$$\vec v = αA^1 + βA^2 +γA^3 + ωKer(f)$$
I'm not sure you have the right vectors for your basis of Im(f).

Mark44 said:
No, this isn't right. If you multiply A times the vector above, you don't get the zero vector. Check your work in row reducing A.

Mark44 said:
I don't know what you're doing here. As it turns out, dim(Ker(A)) = 1, so a basis for the kernel would consist of one vector.
Kernul said:
Complete the base chosen in ##Ker(f)## into a base of ##\mathbb R^4##;
so I know that I have to put together both ##Ker(f)## and ##Im(f)## in order to have a basis of ##\mathbb R^4##. Or am I doing something wrong? Because it's something I learned recently.

Mark44 said:
For the first part of item 2, Ker(f)∩Im(f)Ker(f)∩Im(f)Ker(f) \cap Im(f), yes you need to find the vectors that are in both subspaces.
For the second part, Ker(f)+Im(f)Ker(f)+Im(f)Ker(f)+Im(f), how is '+' defined for two sets?
I know that the dimension of ##Ker(f) + Im(f) = dim(Ker(f)) + dim(Im(f)) - dim(Ker(f) \cap Im(f))## and that the sum is the subspace $$Ker(f) + Im(f) = {\vec v + \vec w : \vec v \in Ker(f), \vec w \in Im(f)}$$. I have to first find the dimension of the intersection in order to find the dimension of the sum.

Mark44 said:
I'm not sure you have the right vectors for your basis of Im(f).
Why? Now that I know that the dimension of ##Im(f)## is 3, the basis would be $$B_{Im(f)} = [A^1, A^2, A^3]$$ So these are the vectors I will have to use in order to find the intersection.

Kernul said:
OK, that's better. You could just as well have chosen ##\begin{bmatrix} 1 \\ 0 \\ 2 \\ -3 \end{bmatrix}## for your basis vector.
Kernul said:
The exercise asks me so I know that I have to put together both ##Ker(f)## and ##Im(f)## in order to have a basis of ##\mathbb R^4##. Or am I doing something wrong? Because it's something I learned recently.
For the four vectors, I would use the kernel basis vector you found, plus three linearly independent vectors that form a basis for the image of f.
Kernul said:
I know that the dimension of ##Ker(f) + Im(f) = dim(Ker(f)) + dim(Im(f)) - dim(Ker(f) \cap Im(f))## and that the sum is the subspace $$Ker(f) + Im(f) = {\vec v + \vec w : \vec v \in Ker(f), \vec w \in Im(f)}$$. I have to first find the dimension of the intersection in order to find the dimension of the sum.
Can any vector be in both the kernel and the image?
Kernul said:
Why? Now that I know that the dimension of ##Im(f)## is 3, the basis would be $$B_{Im(f)} = [A^1, A^2, A^3]$$ So these are the vectors I will have to use in order to find the intersection.
Writing them as ##A^1, A^2, A^3## is confusing, as it suggests you are raising them to powers. It's better to write them as ##A_1, A_2, A_3##, with subscripts, not exponents. Again, these three vectors should be a basis for Im(f).

Mark44 said:
For the four vectors, I would use the kernel basis vector you found, plus three linearly independent vectors that form a basis for the image of f.
Yes, that's exactly what I wrote.
Kernul said:
$$B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}, \begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}]$$
with ##\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}## being the kernel basis vector and ##\begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}## being the image basis vectors which are ##A^1, A^2, A^3## respectively.

Mark44 said:
Can any vector be in both the kernel and the image?
I guess no? I don't know how to check that. Do I have to calculate something?

Mark44 said:
Writing them as A1,A2,A3A1,A2,A3A^1, A^2, A^3 is confusing, as it suggests you are raising them to powers. It's better to write them as A1,A2,A3A1,A2,A3A_1, A_2, A_3, with subscripts, not exponents. Again, these three vectors should be a basis for Im(f).
I'm sorry but my lecturer taught us that when it is a column of a matrix we write ##A^x##, if it is a row of a matrix we write ##A_x##.
And yes, that's what I wrote:
Kernul said:
$$B_{Im(f)} = [A^1, A^2, A^3]$$

Mark44 said:
Can any vector be in both the kernel and the image?
Kernul said:
I guess no? I don't know how to check that. Do I have to calculate something?
How are kernel and image defined?

I would NOT write the first problem in terms of matrices. To find the kernel simply means finding x, y, z, and t such that
13x+y−2z+3t= 0,10y= 0, 9z+6t= 0, and 6z+4t= 0. From the second equation, y= 0 and then the other equations are easy.

Mark44 said:
How are kernel and image defined?
##Ker(f) = \{\vec v \in V : f(\vec v) = \vec 0\}##
##Im(f) = \{\vec w \in W : \exists \vec v \in V, f(\vec v) = \vec w\} = \{f(\vec v) : \vec v \in V\}##
So it is a vector ##\vec v = \vec 0##? The null vector?

And yeah, I know I wasn't bound to use the matrix. I actually put the first three rows in a system and assigned ##t = a##.

For any linear transformation, T, T(0)= 0 so the 0 vectors is always in the kernel- in fact the kernel is a subspace. But it is not necessarily the only vector in the kernel. As I said before, solve the equations 13x+y−2z+3t= 0,10y= 0, 9z+6t= 0, and 6z+4t= 0 for x, y, z, and t. Since you have only three equations in three unknowns you will have at least some of the variables in terms of others.

HallsofIvy said:
For any linear transformation, T, T(0)= 0 so the 0 vectors is always in the kernel- in fact the kernel is a subspace. But it is not necessarily the only vector in the kernel. As I said before, solve the equations 13x+y−2z+3t= 0,10y= 0, 9z+6t= 0, and 6z+4t= 0 for x, y, z, and t. Since you have only three equations in three unknowns you will have at least some of the variables in terms of others.
I have something like this:
\begin{array}{l}
13x + y - 2z + 3t = 0\\
10y = 0\\
9z + 6t = 0\\
6z + 4t = 0
\end{array}
\begin{array}{l}
x = + \frac{2z}{13} - \frac{3t}{13}\\
y = 0\\
z = - \frac{6t}{9} = - \frac{2t}{3}\\
t = - \frac{6z}{4} = - \frac{3z}{2}
\end{array}

So finish it! If x= 2z/13- 3t/13 and z= -2t/3, what is x in terms or 6 only?

I get this, which doesn't give me a number:
##\begin{array}{l}
x = - \frac{1z}{3}\\
y = 0\\
z = - \frac{2t}{3}\\
t = - \frac{3z}{2}
\end{array}##
And if I replace ##z## in ##t## as ##- \frac{2t}{3}## I get ##t = t##.

## 1. What is a linear transformation in the context of exercise?

A linear transformation is a mathematical operation that involves applying a linear equation to a set of input values to produce a set of output values. In the context of exercise, this can mean applying a linear equation to a person's physical movements or body measurements to track progress or make predictions about future performance.

## 2. How is a linear transformation used in exercise science?

In exercise science, a linear transformation can be used to analyze and interpret data related to physical activity and fitness. For example, a linear transformation can be applied to data from a person's workout routine to identify patterns, track progress, or predict future outcomes. It can also be used to create mathematical models for various aspects of exercise, such as energy expenditure or muscle growth.

## 3. What are some common examples of linear transformations in exercise?

Some common examples of linear transformations in exercise include calculating a person's body mass index (BMI), tracking changes in muscle mass over time, and predicting future performance based on past workout data. Linear transformations can also be used to analyze the relationship between different variables, such as heart rate and exercise intensity, or to create fitness testing protocols.

## 4. How do scientists determine the appropriate linear transformation to use for a specific exercise-related study?

The appropriate linear transformation for a specific exercise-related study will depend on the specific research question and the type of data being analyzed. Scientists will typically choose a linear transformation that best fits the data and allows for the most accurate analysis and interpretation of results. This may involve testing different equations or models and selecting the one that provides the best fit for the data.

## 5. What are some potential limitations of using linear transformations in exercise science?

While linear transformations can be a useful tool in exercise science, they also have some limitations. For example, they may not accurately represent complex relationships between variables or take into account individual differences in physical abilities or fitness levels. Additionally, linear transformations may not always be appropriate for non-linear data, and their results should be interpreted with caution and in conjunction with other methods and measures.

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