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Example Questions
Example Question #502 : Operations And Properties
Let be a function such that . Is it possible for to have a rank of and a null space with dimension ?
no
not enough information
yes
no
The rank and the dimension of the null space must add up to the dimension of the domain from the formula
where is the dimension of the domain,
is the dimension of the null space,
and is the rank.
For this problem the dimension of the domain () is . So
From the problem statement, we want to consider and But if we plug this into the formula we get
Hence the equation for dimension doesn't hold.
Example Question #21 : Linear Mapping
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?
Not enough information
No
Yes
Yes
Yes, is onto because any vector in the codomain, , is the image of a vector from the domain.
Example Question #22 : Linear Mapping
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.)
Not enough information
No
Yes
Yes
This mapping is onto. For any given vector in the codomain, , there is a corresponding vector in the domain, .
Example Question #23 : Linear Mapping
Is the linear map such that onto?
(Note this is sometimes called the identity map because it maps every vector to itself)
Not enough information
No
Yes
Yes
The answer is yes, the mapping is onto. The image of is which is the codomain.
Example Question #506 : Operations And Properties
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?
No
Yes
Not enough information
No
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.
Example Question #24 : Linear Mapping
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?
(Note this is called the zero mapping)
No
Yes
Not enough information
No
No, is not onto. This is because the image of is only the zero vector, not all of
Example Question #25 : Linear Mapping
A mapping is said to 1-to-1 (sometimes called injective) if no two vectors in the domain go to the same vector in the image of the mapping.
Is the linear map such that 1-to1?
Yes
Not enough information
No
Yes
The linear map, , is 1-to-1 because no vector in the codomain is the output of two different vectors in the domain.
Example Question #26 : Linear Mapping
There are relationships between the dimension of the null space (sometimes called kernal) and if a function is 1-to-1
The dimension of a linear map's null space is zero. Is the linear map 1-to-1?
Yes
No
Not enough information
Yes
A linear map is always 1-to-1 if the null space has dimension zero.
The converse of this statement is also true. A linear map is 1-to-1 if the null space has dimension 0.
Example Question #26 : Linear Mapping
A mapping is said to 1-to-1 (sometimes called injective) if no two vectors in the domain go to the same vector in the image of the mapping.
Is the linear map such that 1-to1?
(Note this is called the zero mapping)
No
Not enough information
Yes
No
The linear map, is not 1-to-1 because more than one vector goes to the zero vector. In fact, all vectors go to the same vector for the zero mapping!
For example, the vector and both go to the same vector. Thus is not 1-to-1
Example Question #27 : Linear Mapping
A linear map has a null space that is spanned by the vectors,
and . is this function 1-to-1?
Not enough information
No
Yes
No
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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