Operations and Properties - Linear Algebra
Card 0 of 2140
is an involutory matrix.
True, false, or indeterminate: 0 is an eigenvalue of
.
is an involutory matrix.
True, false, or indeterminate: 0 is an eigenvalue of .
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An eigenvalue of an involutory matrix must be either 1 or
. This can be seen as follows:
Let
be an eigenvalue of involutory matrix
. Then for some eigenvector
,

Premultiply both sides by
:


By definition, an involutory matrix has
as its square, so



By transitivity,

Thus,
, or 
It follows that
. The statement is false.
An eigenvalue of an involutory matrix must be either 1 or . This can be seen as follows:
Let be an eigenvalue of involutory matrix
. Then for some eigenvector
,
Premultiply both sides by :
By definition, an involutory matrix has as its square, so
By transitivity,
Thus, , or
It follows that . The statement is false.
What is the dimension of the space spanned by the following vectors:



What is the dimension of the space spanned by the following vectors:
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Since there are three linearly independent vectors, they span a 3 dimensional space.
Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.
Since there are three linearly independent vectors, they span a 3 dimensional space.
Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.
In a 5 dimensional vector space, what is the maximum number of vectors you can have in a linearly dependent set?
In a 5 dimensional vector space, what is the maximum number of vectors you can have in a linearly dependent set?
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Linearly dependent sets have no limit to the number of vectors they can have.
Linearly dependent sets have no limit to the number of vectors they can have.
Consider the following set of three vectors:

where 
Is the set linearly independent?
Consider the following set of three vectors:
where
Is the set linearly independent?
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Since
can be written as a linear combination of of
and
then the set cannot be linearly independent.
Since can be written as a linear combination of of
and
then the set cannot be linearly independent.
Does the following row reduced echelon form of a matrix represent a linearly independent set?

Does the following row reduced echelon form of a matrix represent a linearly independent set?
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The set is linearly dependent because there is a row of all zeros.
Notice that having columns of all zeros does not tell if the set is linearly independent or not.
The set is linearly dependent because there is a row of all zeros.
Notice that having columns of all zeros does not tell if the set is linearly independent or not.
Does the following row reduced echelon form of a matrix represent a linearly independent set?

Does the following row reduced echelon form of a matrix represent a linearly independent set?
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The set must be linearly independent because there are no rows of all zeros. There are columns of all zeros, but columns do not tell us if the set is linearly independent or not.
The set must be linearly independent because there are no rows of all zeros. There are columns of all zeros, but columns do not tell us if the set is linearly independent or not.
Consider the mapping
. Can f be an isomorphism?
(Hint: Think about dimension's role in isomorphism)
Consider the mapping . Can f be an isomorphism?
(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 and
have different dimension. Isomorphisms cannot exist between vector spaces of different dimension.
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 . Can f be a homomorphism?
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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.
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 . Is f an isomorphism?
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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)
Find the Eigen Values for Matrix
.

Find the Eigen Values for Matrix .
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The first step into solving for eigenvalues, is adding in a
along the main diagonal.

Now the next step to take the determinant.


Now lets FOIL, and solve for
.


Now lets use the quadratic equation to solve for
.



So our eigen values are


The first step into solving for eigenvalues, is adding in a along the main diagonal.
Now the next step to take the determinant.
Now lets FOIL, and solve for .
Now lets use the quadratic equation to solve for .
So our eigen values are
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 to
. Can f be 1-to-1?
(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.
Find the eigenvalues for the matrix 
Find the eigenvalues for the matrix
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The eigenvalues,
, for the matrix are values for which the determinant of
is equal to zero. First, find the determinant:

Now set the determinant equal to zero and solve this quadratic:
this can be factored:

The eigenvalues are 5 and 1.
The eigenvalues, , for the matrix are values for which the determinant of
is equal to zero. First, find the determinant:
Now set the determinant equal to zero and solve this quadratic:
this can be factored:
The eigenvalues are 5 and 1.
Which is an eigenvector for
,
or 
Which is an eigenvector for ,
or
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To determine if something is an eignevector, multiply times A:

Since this is equivalent to
,
is an eigenvector (and 5 is an eigenvalue).

This cannot be re-written as
times a scalar, so this is not an eigenvector.
To determine if something is an eignevector, multiply times A:
Since this is equivalent to ,
is an eigenvector (and 5 is an eigenvalue).
This cannot be re-written as times a scalar, so this is not an eigenvector.
Find the eigenvalues for the matrix 
Find the eigenvalues for the matrix
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The eigenvalues are scalar quantities,
, where the determinant of
is equal to zero.
First, find an expression for the determinant:


Now set this equal to zero, and solve:
this can be factored (or solved in another way)

The eigenvalues are -5 and 3.
The eigenvalues are scalar quantities, , where the determinant of
is equal to zero.
First, find an expression for the determinant:
Now set this equal to zero, and solve:
this can be factored (or solved in another way)
The eigenvalues are -5 and 3.
Which is an eigenvector for
,
or
?
Which is an eigenvector for ,
or
?
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To determine if something is an eigenvector, multiply by the matrix A:

This is equivalent to
so this is an eigenvector.

This is equivalent to
so this is also an eigenvector.
To determine if something is an eigenvector, multiply by the matrix A:
This is equivalent to so this is an eigenvector.
This is equivalent to so this is also an eigenvector.
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 to
. Can f be onto?
(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.
Determine the eigenvalues for the matrix 
Determine the eigenvalues for the matrix
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The eigenvalues are scalar quantities
where the determinant of
is equal to zero. First, write an expression for the determinant:


this can be solved by factoring:

The solutions are -2 and -7
The eigenvalues are scalar quantities where the determinant of
is equal to zero. First, write an expression for the determinant:
this can be solved by factoring:
The solutions are -2 and -7
Which is an eigenvector for the matrix
,
or 
Which is an eigenvector for the matrix ,
or
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To determine if a vector is an eigenvector, multiply with A:
. This cannot be expressed as an integer times
, so
is not an eigenvector
This can be expressed as
, so
is an eigenvector.
To determine if a vector is an eigenvector, multiply with A:
. This cannot be expressed as an integer times
, so
is not an eigenvector
This can be expressed as
, so
is an eigenvector.
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