# Linear algebra - kernel and image question

## Homework Statement

$$f: K^{3} \rightarrow K^{4}$$ is a linear transformation of vector spaces:

$$K^{3} = \left\langle \vec{e}_{1}, \vec{e}_{2}, \vec{e}_{3} \right\rangle$$

and

$$K^{4} = \left\langle \vec{e}^{*}_{1}, \vec{e}^{*}_{2}, \vec{e}^{*}_{3}, \vec{e}^{*}_{4} \right \rangle$$

as well as:

$$f(\vec{e}_{1}) = \vec{e}^{*}_{1} - \vec{e}^{*}_{2} + \vec{e}^{*}_{3} - \vec{e}^{*}_{4}$$,

$$f(\vec{e}_{2}) = \vec{e}^{*}_{1} - 2 \vec{e}^{*}_{3}$$,

$$f(\vec{e}_{1}) = \vec{e}^{*}_{2} - 3 \vec{e}^{*}_{3} + \vec{e}^{*}_{4}$$.

Determine a matrix A so that for all $$x \in K^{3}$$ so that

$$f(x) = Ax$$
Determine the kernel and image of f.

## The Attempt at a Solution

well I assumed the following:

$$K^{3} = \left\langle \vec{e}_{1} \vec{e}_{2} \vec{e}_{3} \right\rangle $= \left[ {\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right]$$$

$$[tex]K^{4} = \left\langle \vec{e}^{*}_{1} \vec{e}^{*}_{2} \vec{e}^{*}_{3} \vec{e}^{*}_{4} \right \rangle $= \left[ {\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right]$$$
$$f(\vec{e}_{1}) $= \left[ {\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \\ \end{array} } \right]$$$,

$$f(\vec{e}_{2}) $= \left[ {\begin{array}{c} 1 \\ 0 \\ -2 \\ 0 \\ \end{array} } \right]$$$,

$$f(\vec{e}_{1}) $= \left[ {\begin{array}{c} 0 \\ 1 \\ -3 \\ 1 \\ \end{array} } \right]$$$.

$$f(x) $= \left[ {\begin{array}{ccc} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 2 & -3 \\ -1 & 0 & 1 \\ \end{array} } \right]$ $= Ax = A \left[ {\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right]$ = A = $\left[ {\begin{array}{ccc} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 2 & -3 \\ -1 & 0 & 1 \\ \end{array} } \right]$$$

so that's A, but I don't think it can be right for a start its not 4D.
I know how to get the kernel and image but I don't really know how else to start this problem

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so there are 2 questions in fact?

vela
Staff Emeritus
Homework Helper

## The Attempt at a Solution

well I assumed the following:

$$K^{3} = \left\langle \vec{e}_{1} \vec{e}_{2} \vec{e}_{3} \right\rangle $= \left[ {\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right]$$$

$$K^{4} = \left\langle \vec{e}^{*}_{1} \vec{e}^{*}_{2} \vec{e}^{*}_{3} \vec{e}^{*}_{4} \right \rangle $= \left[ {\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right]$$$
What is that supposed to mean? K3 and K4 are matrices?
$$f(x) $= \left[ {\begin{array}{ccc} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 2 & -3 \\ -1 & 0 & 1 \\ \end{array} } \right]$ $= Ax = A \left[ {\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right]$ = A = $\left[ {\begin{array}{ccc} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 2 & -3 \\ -1 & 0 & 1 \\ \end{array} } \right]$$$
You seriously need to clean up your notation. What you wrote doesn't make much sense.
so that's A, but I don't think it can be right for a start its not 4D.
I know how to get the kernel and image but I don't really know how else to start this problem
Your matrix A is correct (except for a typo in the second column). It's 4x3, so it represents a map from a three-dimensional vector space to a four-dimensional vector space.

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What is that supposed to mean? K3 and K4 are matrices?

K3 and k4 are vector spaces with bases:

$$K^{3} = \left\langle \vec{e}_{1} , \vec{e}_{2} , \vec{e}_{3} \right\rangle$$

$$K^{4} = \left\langle \vec{e}^{*}_{1} , \vec{e}^{*}_{2} , \vec{e}^{*}_{3} , \vec{e}^{*}_{4} \right \rangle$$

the elements within a basis are indeed vectors, therefore K3 and k4 can be represented as matrices. However, as i said the actual vectors I've chosen are assumed not given!

$$\vec{e}_{i}$$ is usually taken to be a unit vector in i-th dimension.

You seriously need to clean up your notation. What you wrote doesn't make much sense.

sorry, Latex is not fun. let me try again:

f(x) = Ax , where
$$$f(x) = \left( {\begin{array}{c} \vec{e}_{1} \\ \vec{e}_{2} \\ \vec{e}_{3} \\ \end{array} } \right)$$$

and
$$$x = \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array}$$$
therefore:

$$$A = \left[ {\begin{array}{ccc} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & -2 & -3 \\ -1 & 0 & 1 \\ \end{array} } \right]$$$

but i thought the dimension of a matrix was linked to the no. of pivots in this case the maximum is three.

thanks!

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HallsofIvy
Homework Helper
K3 and k4 are vector spaces with bases:

$$K^{3} = \left\langle \vec{e}_{1} , \vec{e}_{2} , \vec{e}_{3} \right\rangle$$

$$K^{4} = \left\langle \vec{e}^{*}_{1} , \vec{e}^{*}_{2} , \vec{e}^{*}_{3} , \vec{e}^{*}_{4} \right \rangle$$

the elements within a basis are indeed vectors, therefore K3 and k4 can be represented as matrices.
No, they cannot. Linear transformations from one vectors space to another are represented by matrices, or individual vectors (such as the basis vectors) as column matrices, not the vector spaces themselves.

However, as i said the actual vectors I've chosen are assumed not given!

$$\vec{e}_{i}$$ is usually taken to be a unit vector in i-th dimension.

sorry, Latex is not fun. let me try again:

f(x) = Ax , where
$$$f(x) = \left( {\begin{array}{c} \vec{e}_{1} \\ \vec{e}_{2} \\ \vec{e}_{3} \\ \end{array} } \right)$$$

and
$$$x = \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array}$$$
therefore:

$$$A = \left[ {\begin{array}{ccc} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & -2 & -3 \\ -1 & 0 & 1 \\ \end{array} } \right]$$$

but i thought the dimension of a matrix was linked to the no. of pivots in this case the maximum is three.

thanks!
What you have is a perfectly good matrix representing a linear transformation from a 3 dimensional space (it has 3 columns) to a 4 dimensional space (it has 4 rows).
The kernel is the subspace of K3 of vectors v such that Av= 0. That specifically means that they are of the form (x, y, z) such that x+ y= 0, -x+ z= 0, x- 2y- 3z= 0, and -x+ z= 0. Notice that the second and fourth equations are the same and both say z= x. Putting that into the first and third equations, gives you another equation for x and y so you can solve for y as a linear function of x and write any vector in the kernel as a multiple of x.

The image of f is the set of all vectors in K4 that are equal to f(v) for some v in K3. That is, (a, b, c, d) such that x+ y= a, -x+ z= b, x- 2y- 3z= c, and -x+ z= d have solutions. Again, the second and fourth equation both have "-x+ z" so we must have b= d. Putting z= b+ x into the third equation gives equations you can use to get conditions on b and c. Since the kernel is one dimensional and K3 is three dimensional, the image must have dimension 3- 1= 2.

No, they cannot. Linear transformations from one vectors space to another are represented by matrices, or individual vectors (such as the basis vectors) as column matrices, not the vector spaces themselves.

thanks for clearing that up :)

What you have is a perfectly good matrix representing a linear transformation from a 3 dimensional space (it has 3 columns) to a 4 dimensional space (it has 4 rows).

ok, so the four dimensions of K4 refer only to the space and our matrix A does not necessarily have the same number of dimensions as the resulting vector space ? thanks again