Linear Mapping - Linear Algebra
Card 1 of 208
Consider the mapping 
. Can f be an isomorphism?
(Hint: Think about dimension's role in isomorphism)
Consider the mapping
(Hint: Think about dimension's role in isomorphism)
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No, f, cannot be an isomorphism. This is because 
and 
have different dimension. Isomorphisms cannot exist between vector spaces of different dimension.
No, f, cannot be an isomorphism. This is because
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The last question showed us isomorphisms must be between vector spaces of the same dimension. This question now asks about homomorphisms.
Consider the mapping 
. Can f be a homomorphism?
The last question showed us isomorphisms must be between vector spaces of the same dimension. This question now asks about homomorphisms.
Consider the mapping
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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.
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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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 
. Is f an isomorphism?
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
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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)
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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Let f be a homomorphism from 
to 
. Can f be 1-to-1?
(Hint: look at the dimension of the domain and co-domain)
Let f be a homomorphism from
(Hint: look at the dimension of the domain and co-domain)
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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.
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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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)
Often we can get information about a mapping by simply knowing the dimension of the domain and codomain.
Let f be a mapping from
(Hint look at the dimension of the domain and codomain)
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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.
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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A mapping is said to be onto (sometimes called surjective) if it's image is the entire codomain.
Is the linear map 
such that 
onto?
A mapping is said to be onto (sometimes called surjective) if it's image is the entire codomain.
Is the linear map
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First we need to find the image of 
Any vector in the image of 
has the form 
.
Treat 
and 
like arbitrary constants. Then the image of 
is the subspace spanned by the set of vectors 
This subspace has a basis of 2 vectors. Therefore it has dimension of 
.
The image of 
does not span the whole codomain, 
, because the image does not have the same dimension as the codomain, 
.
Another way to do this problem is recognize that the domain,
, has dimension 
. Therefore, the linear map can have an image of 
with dimension of at most 2. But the codomain has dimension 
. Since 
, 
can't be onto.
First we need to find the image of
Treat
The image of
Another way to do this problem is recognize that the domain,
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A mapping is said to be onto (sometimes called surjective) if it's image is the entire codomain.
Is the linear map 
such that 
onto?
(This map is sometimes called a projection, specifically a projection onto the 
xy plane.)
A mapping is said to be onto (sometimes called surjective) if it's image is the entire codomain.
Is the linear map
(This map is sometimes called a projection, specifically a projection onto the
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This mapping is onto. For any given vector in the codomain, 
, there is a corresponding vector in the domain, 
.
This mapping is onto. For any given vector in the codomain,
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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 
. Also the vector space V has dimension 4 and the vector space W has dimension 8. What property of isomorphism can f NOT satisify.
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
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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.
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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Is the linear map 
such that 
onto?
(Note this is sometimes called the identity map because it maps every vector to itself)
Is the linear map
(Note this is sometimes called the identity map because it maps every vector to itself)
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The answer is yes, the mapping is onto. The image of 
is 
which is the codomain.
The answer is yes, the mapping is onto. The image of
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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
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
Tap to reveal 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
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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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
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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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.
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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Let f be a mapping such that 
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?
Let f be a mapping such that
Let f be defined such that
Is f a homomorphism?
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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.
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
Consider
This is the same thing as
Hence, f preserves vector addition because
Proof f preserves scalar multiplication
Let u be an arbitrary vector in
Consider
This is the same thing we get if we consider
Hence f preserves scalar multiplication because
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Let f be a mapping such that 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)
Let f be a mapping such that
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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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.
f is not onto. This is because not every vector in
We could also see this quicker by looking at the dimension of the domain and codomain. The domain
Finally, we know f preserves vector addition and scalar multiplication because it is a homomorphism.
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Let f be a mapping such that 
Let f be defined such that 
Is f 1-to-1 and onto?
Let f be a mapping such that
Let f be defined such that
Is f 1-to-1 and onto?
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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.
f is not 1-to-1 and it is not onto.
f is not onto because all of
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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Let f be a mapping such that 
Let f be defined such that 
Is f an isomorphism?
(Hint: Consider the zero vector)
Let f be a mapping such that
Let f be defined such that
Is f an isomorphism?
(Hint: Consider the zero vector)
Tap to reveal answer
f is 1-to-1 and onto but it is not a homomorphism. Therefore it is not an isomorphism. To see this consider f(0,0) = (0,5)
A homomorphism always takes the zero vector to the zero vector. This particular mapping does not. Thus it does not preserve structure ie not a homomorphism.
f is 1-to-1 and onto but it is not a homomorphism. Therefore it is not an isomorphism. To see this consider f(0,0) = (0,5)
A homomorphism always takes the zero vector to the zero vector. This particular mapping does not. Thus it does not preserve structure ie not a homomorphism.
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The null space (sometimes called the kernal) of a mapping is a subspace in the domain such that all vectors in the null space map to the zero vector.
Consider the mapping 
such that 
.
What is the the null space of 
?
The null space (sometimes called the kernal) of a mapping is a subspace in the domain such that all vectors in the null space map to the zero vector.
Consider the mapping
What is the the null space of
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Any vector in the null space satisfies 
.
Therefore we get the following equation:
Thus 
. Hence the null space is any vector in form 
where 
is any real number. Therefore, any point on the line 
gets mapped to the zero vector in 
Any vector in the null space satisfies
Therefore we get the following equation:
Thus
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This problem deals with the zero map. I.e the map that takes all vectors to the zero vector.
Consider the mapping 
such that 
.
What is the the null space of 
?
This problem deals with the zero map. I.e the map that takes all vectors to the zero vector.
Consider the mapping
What is the the null space of
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The zero map takes all vectors to the zero vector. Therefore, the entire domain of the map is the null space. The domain of this map is 
. Thus 
is the null space.
The zero map takes all vectors to the zero vector. Therefore, the entire domain of the map is the null space. The domain of this map is
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An important quantity for a linear map is the dimension of its image. This is called the rank.
Consider the mapping such that .
What is the the rank of 
?
An important quantity for a linear map is the dimension of its image. This is called the rank.
Consider the mapping
What is the the rank of
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To find the rank of 
, we first find the image of 
. The image of 
is any vector of form 
where a is any real number. This is a line in 
. Therefore the image of 
is a 1 dimensional subspace. Thus the answer is 1.
To find the rank of
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This problem deals with the zero map. I.e the map the takes all vectors to the zero vector.
Consider the mapping such that .
What is the the rank of 
?
This problem deals with the zero map. I.e the map the takes all vectors to the zero vector.
Consider the mapping
What is the the rank of
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The image of the the zero map is the zero vector. A single vector has dimension 
. Therefore the dimension of the image is zero. Hence the rank is zero.
The image of the the zero map is the zero vector. A single vector has dimension
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Consider the mapping 
such that 
.
What is the the rank of 
?
Consider the mapping
What is the the rank of
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The image is the space spanned by the vectors 
and 
. The image has a basis of 
vectors. Therefore the image has dimension 
. Thus the rank is 
.
The image is the space spanned by the vectors
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