Linear Mapping - Linear Algebra

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Question

A linear map $f:R^4 \rightarrow R^2$ has a null space that is spanned by the vectors,

and . is this function 1-to-1?

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Answer

This linear mapping is not 1-to-1. We know this because the null space is spanned by two vectors. Therefore the null space has dimension . A linear map is 1-to-1 only if it has a null space with dimension zero. This linear map doesn't have a null space of dimension zero, therefore it is not 1-to-1.

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Question

A linear map $f:R^4 \rightarrow R^3$ has a rank of 3. Is the linear map, , 1-to-1?

(Hint- Use the formula where

is the dimension of the domain

is the dimension of the null space

is the rank of the linear map )

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Answer

The answer is no because the dimension of the null space is not zero. This comes from the equation

We know that the domain is which has dimension of . Therefore

Also from the problem statement .

Plugging these into the equation gives

Since the dimension of the null space is and not , then the function can't be 1-to-1.

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Question

A linear map $f:R^4 \rightarrow R^5$ has a rank of 4. Is the linear map, , 1-to-1?

(Hint- Use the formula where

is the dimension of the domain

is the dimension of the null space

is the rank of the linear map )

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Answer

The answer is yes because the dimension of the domain and the rank are equal. This implies that the dimension of the null space is zero.

This comes from the equation

We know that the domain is which has dimension of . Therefore

Also from the problem statement .

Plugging these into the equation gives

Since the dimension of the null space is then the function is 1-to-1.

Notice that whenever , then is always zero. Thus whenever , the linear mapping is 1-to-1.

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Question

$C_{ \mathbb{R}}$ refers to the set of all functions with domain $\mathbb{R}$ and range a subset of $\mathbb{R}$ .

Define the transformation $T: C_{ \mathbb{R}} \rightarrow \mathbb{R}$ to be

$T(f) = \sum_{j=1}^{100} f'(j)$

True or false: is a linear transformation.

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Answer

For to be a linear transformation, it must hold that

and

for all in the domain of and and for all scalar .

Let $f, g \in C_{ \mathbb{R}}$ .

$T(f) = \sum_{j=1}^{100} f'(j)$ and $T(g) = \sum_{j=1}^{100} g'(j)$ , so

$T(f)+ T(g)= \sum_{j=1}^{100} f'(j)+ \sum_{j=1}^{100} g'(j)$

By the sum rule for finite sequences,

$T(f)+ T(g)= \sum_{j=1}^{100}[ f'(j) + g'(j)]$

By the derivative sum rule,

$T(f)+ T(g)= \sum_{j=1}^{100}( f '+g')(j)$

$T(f)+ T(g)= \sum_{j=1}^{100}( f +g)'(j)$

The first condition is met.

Let $f \in C_{ \mathbb{R}}$ and be a scalar.

$T(f) = \sum_{j=1}^{100} f'(j)$

$c T(f) = c\sum_{j=1}^{100} f'(j)$

By the scalar product rule for finite sequences,

$c T(f) = \sum_{j=1}^{100}c f'(j)$

By the scalar product rule for derivatives,

$c T(f) = \sum_{j=1}^{100}(c f)'(j)$

The second condition is met.

is a linear transformation.

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Question

$C_{[0, 2\pi]}$ refers to the set of all functions that are continuous on $[0,2\pi]$ .

Define the linear mapping $T: C_{[0, 2\pi]}\rightarrow \mathbb{R}$ as follows:

$T(f)= \int_{0}^{2 \pi} f(x) \textup{ d}x$

True or false: $f(x) = \cos x + \sin x$ is in the kernel of .

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Answer

The kernel of a linear transformation is the subset of the domain of that maps into the zero of its range. It follows that $f \in \ker (T)$ if and only if

$T(f)= \int_{0}^{2 \pi} f(x) \textup{ d}x = 0$

To determine whether this is true or false, evaluate the integral:

$T(f)= \int_{0}^{2 \pi} (\cos x + \sin x) \textup{ d}x = 0$

$=\ \left.\begin{matrix} \sin x - \cos x \ ; \end{matrix}\right|\ \begin{matrix} 2 \pi \ 0 \end{matrix}$

$= (\sin 2\pi - \cos 2 \pi) - (\sin 0 - \cos 0)$

Therefore, $f \in \ker (T)$ .

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Question

$C_{[0, \pi]}$ refers to the set of all functions that are continuous on $[0, \pi]$ .

Define the linear mapping $T: C_{[0, \pi]}\rightarrow \mathbb{R}$ as follows:

$T(f)= \int_{0}^{ \pi} f(x) \textup{ d}x$

True or false: $f(x) = \cos x + \sin x$ is in the kernel of .

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Answer

The kernel of a linear transformation is the subset of the domain of that maps into the zero of its range, so, by definition, $f \in \ker (T)$ if and only if

$T(f)= \int_{0}^{ \pi} f(x) \textup{ d}x = 0$

To determine whether this is true or false, evaluate the integral:

$T(f)= \int_{0}^{ \pi} (\cos x + \sin x) \textup{ d}x = 0$

$=\ \left.\begin{matrix} \sin x - \cos x \ ; \end{matrix}\right|\ \begin{matrix} \pi \ 0 \end{matrix}$

$= (\sin \pi - \cos \pi) - (\sin 0 - \cos 0 )$

$T(f) \ne 0$ , so $f \notin \ker (T)$ .

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Question

True or false: If $T: V \rightarrow W$ is a linear mapping, and is a vector space, then is a subspace of .

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Answer

For example, if is the space of all vectors in $\mathbb{R}^2$ of the form , and is the space of all vectors in $\mathbb{R}^2$ the form , then $T: (a_1,a_2) \rightarrow (a_1,0)$ is a linear mapping, but is not a subset of , let alone a subspace of .

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Question

True or false: The identity mapping $I:V \rightarrow V$ , is also considered a linear mapping, regardless of the vector space .

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Answer

Verifying the conditions for a linear mapping, we have

$I[a\mathbf{x}+b]=a\mathbf{x}+b = aI[\mathbf{x}]+b.$

Hence the identity mapping is closed under vector addition and scalar multiplication, and is therefore a linear mapping.

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Question

$P_{3}$ is the set of all real polynomials with degree 3 or less.

Define the linear mapping $T: P_{3}\rightarrow \mathbb{R}$ as follows:

$T(P(x)) \rightarrow P(1)$

Is this linear mapping one-to-one and onto?

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Answer

$T: P_{3}\rightarrow \mathbb{R}$ is a linear mapping of a four-dimensional vector space into a one-dimensional vector space; it cannot be one-to-one.

is onto if, for every element in its codomain, which here is $\mathbb{R}$ , there exists at least one in the domain $P_{3}$ so that .

For each , we can choose the constant polynomial , so each has at least one domain element that maps into it.

is onto but not one-to-one.

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Question

Isomorphism is an important concept in linear algebra. To be able to tell if a mapping is isomorphic, it is important to be able to know what an isomorphism is.

Let f be a mapping between vector spaces V and W. Then a mapping f is an isomorphism if it is

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Answer

An isomorphism is homomorphism (preserves vector addition and scalar multiplcation) that is bijective (both onto and 1-to-1). Therefore an isomorphism is a mapping that is

onto

1-to-1

Preserves vector addition

Preserves scalar multiplcation

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Question

That last question dealt with isomorphism. This question is meant to point out the difference between isomorphism and homomorphisms.

A homomorphism is a mapping between vector spaces that

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Answer

By definition a homomorphism is a mapping that preserves vector addition and scalar multiplication.

Compare this to the previous problem. An isomorphism is a homomorphism that is also 1-to-1 and onto. Therefore isomorphism is just a special homomorphism. In other words, every isomorphism is a homomorphism, but not all homomorphisms are an isomorphisms.

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Question

Consider the mapping $f:R^2\rightarrow R^3$ . Can f be an isomorphism?

(Hint: Think about dimension's role in isomorphism)

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Answer

No, f, cannot be an isomorphism. This is because and have different dimension. Isomorphisms cannot exist between vector spaces of different dimension.

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Question

The last question showed us isomorphisms must be between vector spaces of the same dimension. This question now asks about homomorphisms.

Consider the mapping $f:R^2\rightarrow R^3$ . Can f be a homomorphism?

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Answer

The answer is yes. There is no restriction on dimension for homomorphism like there is for isomorphism. Therefore f could be a homomorphism, but it is not guaranteed.

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Question

In the previous question, we said an isomorphism cannot be between vector spaces of different dimension. But are all homomorphisms between vector spaces of the same dimension an isomorphism?

Consider the homomorphism $f:R^2\rightarrow R^2$ . Is f an isomorphism?

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Answer

The answer is not enough information. The reason is that it could be an isomorphism because it is between vector spaces of the same dimension, but that doesn't mean it is.

For example:

Consider the zero mapping f(x,y)= (0,0).

This mapping is not onto or 1-to-1 because all elements go to the zero vector. Therefore it is not an isomorphism even though it is a mapping between spaces with the same dimension.

Another example:

Consider the identity mapping f(x,y) = (x,y)

This is an isomorphism. It clearly preserves structure and is both onto and 1-to-1.

Thus f could be an isomorphism (example identity map) or it could NOT be an isomorphism ( Example the zero mapping)

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Question

Let f be a homomorphism from to . Can f be 1-to-1?

(Hint: look at the dimension of the domain and co-domain)

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Answer

No, f can not be 1-to-1. The reason is because the domain has dimension 3 but the co-domain has dimension of 2. A mapping can not be 1-to-1 when the the dimension of the domain is greater than the dimension of the co-domain.

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Question

Often we can get information about a mapping by simply knowing the dimension of the domain and codomain.

Let f be a mapping from to . Can f be onto?

(Hint look at the dimension of the domain and codomain)

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Answer

No, f cannot be onto. The reason is because the dimension of the domain (2) is less than the dimension of the codomain(3).

For a function to be onto, the dimension of the domain must be less than or equal to the dimension of the codomain.

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Question

The previous two problems showed how the dimension of the domain and codomain can be used to predict if it is possible for the mapping to be 1-to-1 or onto. Now we'll apply that knowledge to isomorphism.

Let f be a mapping such that $f:V\rightarrow W$ . Also the vector space V has dimension 4 and the vector space W has dimension 8. What property of isomorphism can f NOT satisify.

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Answer

f cannot be onto. The reason is because the domain, V, has a dimension less than the dimension of the codomain, W.

f can be 1-to-1 since the dimension of V is less-than-or-equal to the dimension of W. However, just because f can be 1-to-1 based off its dimension does not mean it is guaranteed.

f preserves both vector addition and scalar multiplication because it was stated to be a homomorphism in the problem statemenet. The definition of a homomorphism is a mapping that preserves both vector addition and scalar multiplication.

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Question

Let f be a mapping such that $f:R^2 \rightarrow P^2$ where is the vector space of polynomials up to the term. (ie polynomials of the form )

Let f be defined such that

Is f a homomorphism?

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Answer

f is a homomorphism because it preserves both vector addition and scalar multiplication.

To show this we need to prove both statements

Proof f preserves vector addition

Let u and v be arbitrary vectors in with the form and

Consider . Applying the definition of f we get

This is the same thing as

Hence, f preserves vector addition because

Proof f preserves scalar multiplication

Let u be an arbitrary vector in with the form and let k be an arbitrary real constant.

Consider

This is the same thing we get if we consider

Hence f preserves scalar multiplication because for all vectors u and scalars k.

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Question

Let f be a mapping such that $f:R^2 \rightarrow P^2$ where is the vector space of polynomials up to the term. (ie polynomials of the form )

Let f be defined such that

Is f an isomorphism?

(Hint: The last problem we showed this particular f is a homomorphism)

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Answer

f is not onto. This is because not every vector in is in the image of f. For example, the vector is not in the image of f. Hence, f is not onto.

We could also see this quicker by looking at the dimension of the domain and codomain. The domain has dimension 2 and the codomain has dimension 3. A mapping can't be onto and have a domain with a lower dimension than the codomain.

Finally, we know f preserves vector addition and scalar multiplication because it is a homomorphism.

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Question

Let f be a mapping such that $f:R^2 \rightarrow R^2$

Let f be defined such that

Is f 1-to-1 and onto?

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Answer

f is not 1-to-1 and it is not onto.

f is not onto because all of is not in the image of f. For example, the vector (1,1) is not in the image of f.

f is not 1-to-1. For example, the vector (1,1) and (1,0) both go to the same vector.

Ie f(1,1) = f(1,0). Therefore f is not 1-to-1.

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