Similarly, we can show closure under scalar multiplication. \end{proof}
injective
A function is injective if implies that
Injectivity is equivalent to null space equals
Let . Then is injective iff .
\begin{proof}
Consider . We already know since it's a subspace of and thus must contain the zero-vector. We'll show the other way now, so suppose . Then:
and because is injective, then that implies that , completing the proof for this direction.
Consider . Suppose are arbitrary, and that . Then:
thus which equals , so then so so then is injective. \end{proof}
Range
For the range of is the subset of consisting of those vectors that are of the form for some :
Some examples include:
If is the zero map, ie for all , then .
The differentiation operator has the range of .
The range is a subspace
If then the is a subspace of .
\begin{proof}
as .
Since is linear then
A similar argument is made for closure under s. mult.
So is a subspace of . \end{proof}
surjective
A function is surjective if its range equals .
For instance, is surjective since the range of is the entire polynomial space.
Fundamental Theorem of Linear Maps
Fundamental Theorem of Linear maps
Suppose is finite-dimensional and . Then is finite dimensional and:
Suppose are finite-dimensional vector spaces where . Then no linear maps from to is injective. Use the lemma above to show this which implies that so then there's other basis vectors that aren't zero in this space, so is not injective via Chapter 3 - Linear Maps#^ba1b7d.
If instead then no linear map from to is surjective as using the above theorem implies that so then so then is not surjective.
Some other Notes
We can use this idea to show when and when not a system of linear equations has a solution. There's a lot more information on relationships to systems of linear equations, which you can find at [[Year3/Winter2024/MATH306-LinearAlgebraII/2015_Book_LinearAlgebraDoneRight.pdf#page=65]] and onward. Some lemmas:
Summary of Lemmas
Homogeneous sytem of linear equations with more variables than equations has nonzero solutions.
An inhomogeneous system of linear equations with more equations than variables has no solution for some choice of the constant terms.
3.C: Matrices
Representing matrices
matrix,
Let denote positive integers. An matrix is a rectangular array of elements of with rows and columns:
where denotes the entry in row and column of .
matrix of a linear map,
Suppose and is a basis of and is a basis of . The matrix of with respect to these bases is the matrix whose entries are defined by:
If the bases are not clear from the context, then the notation is used.
You can use the following to help construct the matrix:
Thus:
Addition and Scalar Multiplication of Matrices
We've done matrix addition before and scalar multiplication:
As a result, then:
Matrix sum of linear maps
Suppose . Then .
and:
The matrix of a scalar times a linear map
Suppose and . Then .
As such, we define this as a new vector space . As a lemma:
If then denotes the matrix consisting of column of .
As such, then multiplication of left by via is:
for ) and .
Column of matrix product equals matrix times column
Suppose and ,. Then:
For example, consider:
We generalize the results from the examples above to give:
Linear Combinations of Columns
Suppose is an matrix and is an matrix. Then:
In other words, is a linear combination of the columns of , with the scalars that multiply the columns coming from .
3.D: Invertibility and Isomorphic Vector Spaces
Invertible Linear Maps
invertible, inverse
A linear map is called invertible if there exists a linear map such that equals the identity map on and equals the identity map on .
A linear map satisfying and is called the inverse of .
Note that the inverse is unique:
\begin{proof}
Suppose is invertible and are inverses of . Then:
Thus . \end{proof}
Since is unique, we call it to denote it as 's inverse.
Invertibility equals bijectivity
A linear map is invertible iff it is bijective (injective and surjective).
A proof is at [[Year3/Winter2024/MATH306-LinearAlgebraII/2015_Book_LinearAlgebraDoneRight.pdf#page=81]]. Some examples of non-invertible linear maps include:
The multiplication of from to itself is not invertible because it is not surjective (1 is not in the range).
The backward shift linear map from to itself is not invertible because it is not injective ( is in the null space).
Isomorphic Vector Spaces
isomorphism, isomorphic
An isomorphism is an invertible linear map. Two vector spaces are isomorphic if there is an isomorphism from one vector space onto the other one.
We are essentially relabeling all the to some new label . Hence, this is why certain spaces are so similar to one another, like and . You might wonder if dimension has anything to do with it.
Dimension shows vector spaces as isomorphic
Two finite-dimensional vector spaces over are isomorphic iff they have the same dimension.
\begin{proof}
First, suppose and are isomorphic finite-dimensional vector spaces. Thus there exists an isomorphism . Because is invertible then and So then:
For the other way, suppose are finite-dimensional vector spaces with the same dimension. Then and are bases for respectively. Let be defined by:
Then is a well-defined linear map because is a basis for via Chapter 3 - Linear Maps#^9a80a8. Also, is surjective since spans . Furthermore since is LI, so is injective. Because is both injective and surjective, it's an isomorphism, so are isomorphic as desired. \end{proof}
Thus any vector space where is isomorphic to . If is a basis for and is a basis for then for each we have a matrix . So in other words, once bases have been set for then is a function from to , so since via Chapter 3 - Linear Maps#^bfb9a4, then is invertible.
and are isomorphic
Suppose is a basis of and is a basis of . Then is an isomorphism between and .
\begin{proof}
We already noted that is linear. We need to prove that is injective and surjective.
With injectivity, if and then for all and because is a basis of this implies , showing as injective via Chapter 3 - Linear Maps#^ba1b7d.
For surjectivity, suppose . Let be the linear map from to such that:
A linear map from a vector space to itself is called an operator. The notation denotes the set of all operators on . In other words, .
Removing from this may suggest that we can remove one condition from bijectivity of , but recall from our earlier examples Chapter 3 - Linear Maps#^153ab0. But both of these are still not invertible despite being themselves operators. This is actually because both are infinite-dimensional vector spaces. But when they are finite dimensional, we get a remarkable result:
Injectivity is equivalent to surjectivity in finite dimensions
Suppose is finite-dimensional and . Then the following are equivalent:
is invertible
is injective
is surjective
\begin{proof}
Clearly if (i) holds then we get (ii, iii) for free. Suppose instead (ii) just holds. Then is injective so then so we just have the zero vector in the null space, so via the FTOLM:
so then implying surjectivity. So we get (iii), which gives us (i).
Instead if (iii) holds then is surjective, so so then via the FTOLM we have so then the null space is just the zero vector, showing injectivity. Hence is invertible. \end{proof}