Operations and Properties - Linear Algebra

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.

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.

Linearly dependent sets have no limit to the number of vectors they can have.

Since can be written as a linear combination of of and then the set cannot be linearly independent.

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 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.

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

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 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 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

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.

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.

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.

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.

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.

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.

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

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.