Linear Algebra - Linear Mapping

A mapping is said to be onto (sometimes called surjective) if it's image is the entire codomain.

Is the linear map $f:R^2 \rightarrow R^1$ such that onto?

Yes

Not enough information

Explanation

Yes, is onto because any vector in the codomain, , is the image of a vector from the domain.

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

False

True

Explanation

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 .

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?

Yes

No because vector addition is not preserved

No, because scalar multiplication is not preserved

No, because both scalar multiplication and vector addition is not preserved

Explanation

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.

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.

Onto

t-to-1

Preserve vector addition

Preserve scalar multiplication

Explanation

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.

A mapping is said to be onto (sometimes called surjective) if it's image is the entire codomain.

Is the linear map $f:R^2 \rightarrow R^2$ such that onto?
(Note this is called the zero mapping)

Yes

Not enough information

Explanation

No, is not onto. This is because the image of is only the zero vector, not all of

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

True

False

Explanation

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.

This problem deals with the zero map. I.e the map that takes all vectors to the zero vector.

Consider the mapping $f:R^2\rightarrow R^3$ such that .

What is the the null space of ?

The line

The vector

Explanation

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.

is the set of all polynomials of finite degree in .

Define a linear mapping $T:P \rightarrow P$ as follows:

$T(p(x)) =\frac{\mathrm{d} ^2}{\mathrm{d} x^2}p(x)$ .

True or false: is a one-to-one and onto linear mapping.

False: is onto but not one-to-one.

True

False: is one-to-one but not onto.

False: is neither one-to-one nor onto.

Explanation

The domain and the codomain are both of infinite dimension, so it is possible for be one-to-one, onto, both, or neither.

is one-to-one if and only if

implies .

Let and

Then

$T(p)= \frac{\mathrm{d} ^2}{\mathrm{d} x^2}p(x)= 0$ and $T(q)= \frac{\mathrm{d} ^2}{\mathrm{d} x^2}q(x)= 0$ .

Since

, but , is not one-to-one.

Now let $t(x)= \sum_{n= 0}^{\infty} a_{n}x^{n}$ , where finitely many $a_{n}$ are nonzero.If

$p (x)= \sum_{n= 0}^{\infty} \frac{a_{n}}{(n+2)(n+1)}x^{n+2}$ ,

then

$T(p)= \frac{\mathrm{d} ^2}{\mathrm{d} x^2} \left ( \sum_{n= 0}^{\infty} \frac{a_{n}}{(n+2)(n+1)}x^{n+2} \right )$

$= \sum_{n= 0}^{\infty} a_{n}x^{n} = t(x)$

is therefore onto.

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?

Not enough information

Yes

Explanation

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)

Consider the mapping $f:R^2\rightarrow R^3$ such that .