Properties of linear maps – Serlo

We consider some properties of linear maps between vector spaces.

Overview

A linear map, also called vector space homomorphism, preserves the structure of the vector space. This is shown in the following properties of a linear mapping $f:V\to W$:

• The zero vector is mapped to the zero vector: $f(0)=0$.
• Inverses are mapped to inverses: $f(-v)=-f(v)$.
• Linear combinations are mapped to linear combinations.
• Compositions of linear maps are again linear
• Images of subspaces are subspaces
• The image of a span is the span of the individual image vectors: $f(\operatorname {span} (M))=\operatorname {span} (f(M))$ ($M\subseteq V$ is supposed to be an arbitrary set)

Zero vector is mapped to the zero vector

The zero vector / origin has a central meaning in our view of vector spaces. And indeed, the origin is sent to the origin by any linear map. Mathematically, "origin" means the neutral element $0$ of addition.

And the $0\to 0$-property can be shown as a mathematical theorem:

Theorem (Zero vector is mapped to the zero vector)

Any linear map $f:V\to W$ between two $K$-vector spaces maps the neutral element of $V$ to the neutral element in $W$. Formally, this means $f(0_{V})=0_{W}$.

How to get to the proof? (Zero vector is mapped to the zero vector)

We first start with the vector $0_{V}$, which is the neutral element of addition of the vector space $V$. Thus, it does not change a vector to which it is added. Therefore we have that in particular $0_{V}+_{V}0_{V}=0_{V}$.

We need the additivity of a linear map. Thus, $f(0_{V}+0_{V})=f(0_{V})+f(0_{V})$ follows.

Using these two properties, we get $f(0_{V})=f(0_{V})+f(0_{V})$. This equation is satisfied only of $0_{W}$. We can subtract $f(0_{V})$ on both sides and get $f(0_{V})=0_{W}$.

Proof (Zero vector is mapped to the zero vector)

We have that

${\begin{aligned}&f(0_{V})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ 0_{V}=0_{V}+_{{}_{V}}0_{V}\right.}\\[0.3em]=&f(0_{V}+_{{}_{V}}0_{V})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}f\right.}\\[0.3em]=&f(0_{V})+_{{}_{W}}f(0_{V}).\\[0.3em]\end{aligned}}$

So we have that

$f(0_{V})=f(0_{V})+_{{}_{W}}f(0_{V}).$

Now, let us add $-f(0_{V})$ on both sides:

${\begin{aligned}0_{W}=&f(0_{V})+_{{}_{W}}(-f(0_{V}))\\[0.3em]=&f(0_{V})+_{{}_{W}}f(0_{V})+_{{}_{W}}(-f(0_{V}))\\[0.3em]=&f(0_{V})+_{{}_{W}}0_{W}\\[0.3em]=&f(0_{V}).\end{aligned}}$

Hence, we have that $0_{W}=f(0_{V})$.

Inverses are mapped to inverses

Another important structure of vector space is that there is an additive inverse $-v$ to every element $v$. We now want to show that inverses are preserved by linear maps.

Theorem (Inverses are mapped to inverses)

Each linear map sends the inverse of an element to the inverse of the image of the element. Or within one formula, for all $v$ in $V$ we have that $f(-v)=-f(v)$.

How to get to the proof? (Inverses are mapped to inverses)

Our goal is to show that $f(-v)=-f(v)$. We already know that in a vector space $V$ the inverse of any vector $v$ is given by $(-1)\cdot _{{}_{V}}v$. Likewise we have that for any vector $w\in W$ that $-w=(-1)\cdot _{{}_{W}}w$. This simplifies the statement to be shown to $f((-1)\cdot _{{}_{V}}v)=(-1)\cdot _{{}_{W}}f(v)$.

We can transform the expression $f((-1)\cdot _{{}_{V}}v)$ into $(-1)\cdot _{{}_{W}}f(v)$ using the homogeneity of the linear map $f$.

Proof (Inverses are mapped to inverses)

Let $v$ be any element of the vector space $V$.

${\begin{aligned}&f(-v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ -v=(-1)\cdot _{{}_{V}}v\right.}\\[0.3em]=&f((-1)\cdot _{{}_{V}}v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{homogeneity of }}f\right.}\\[0.3em]=&(-1)\cdot _{{}_{W}}f(v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ (-1)\cdot _{{}_{W}}f(v)=-f(v)\right.}\\[0.3em]=&-f(v)\end{aligned}}$

Thus we have that indeed $f(-v)=-f(v)$.

above, we have used that $-v=(-1)\cdot _{{}_{V}}v$ for $v\in V$ and $-w=(-1)\cdot _{{}_{W}}w$ for all $w\in W$. This holds in every vector space. You can find the proof here (link missing).

Alternative proof (Inverses are mapped to inverses)

Let $v$ again be any element of the vector space $V$. Our goal is to show that $-f(v)=f(-v)$ holds. Let's start with a statement that we know is true: $f(v)+(-f(v))=0_{V}$.

The addition of an element with its inverse always gives $0$. So we have that $f(v)+(-f(v))=0_{V}$ . Now, we show that also $f(v)+f(-v)=0_{W}$ holds. Since we are working with maps, we should use their properties. For instance, additivity:

${\begin{aligned}&f(v)+_{{}_{W}}f(-v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}f\right.}\\[0.3em]=&f(v+_{{}_{V}}(-v))\\[0.3em]=&f(0_{V})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{previous theorem}}\right.}\\[0.3em]=&0_{W}\end{aligned}}$

thus, $f(v)+_{{}_{W}}f(-v)=0_{W}$. It follows that $f(-v)$ is the additive inverse of $f(v)$ with respect to $+_{{}_{W}}$. Or, in other words, $f(-v)=-f(v)$.

The statement of this theorem also holds in any abelian group. However, scalar multiplication does not exist there. Hence, the alternative (2nd) version of the proof must be used in this case.

Linear combinations are mapped to linear combinations

Linear mappings preserve the structure of a linear combination and thus map linear combinations in the domain of definition to their corresponding linear combinations in the range of values:

Theorem (Linear combinations are mapped to linear combinations)

A map $f:V\to W$ between two $K$-vector spaces $V$ and $W$ is a linear map if and only if it preserves linear combinations. That is, every linear map sends the linear combination of elements to the linear combination of the images of the elements. Put in a formula, this means that for finitely many $v_{1},\dots ,v_{n}\in V$ and $\lambda _{1},\dots ,\lambda _{n}\in K$ we have that:

$f\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)$

How to get to the proof? (Linear combinations are mapped to linear combinations)

We want to show that for all $v_{i}\in V$ and $\lambda _{i}\in K$ we have that: $f{\bigg (}\sum _{i=1}^{n}\lambda _{i}\cdot _{V}v_{i}{\bigg )}=\sum _{i=1}^{n}\lambda _{i}\cdot _{W}f(v_{i})\iff f$ is a linear map.

We know from the definition of the linear map that additivity and homogeneity hold, and we make use of then.

For the direction "left to right" within the proof we choose two linear combinations in such a way that we get the two properties by inserting them into the above formula.

For the "right to left" direction we know that $f$ is a linear map. We can show by induction that above formula holds true for all elements. This way, we reduce the linear combination to single addition and scalar multiplication, to which we can apply additivity and homogeneity.

Proof (Linear combinations are mapped to linear combinations)

Proof step: $\left(\forall v_{1},\dots ,v_{n}\in V\,\forall \lambda _{1},\dots ,\lambda _{n}\in K:\,f\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)\right)\implies f$ is a linear map.

Let $v,v_{1},v_{2}\in V$ and $\lambda \in K$. The terms $v_{1}+v_{2}$ and $\lambda \cdot v$ are two linear combinations in $V$. If we plug them into the formula ${\textstyle f\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)}$ , we obtain

${\begin{aligned}f(v_{1}+v_{2})&=f(v_{1})+f(v_{2})\\[0.5em]f(\lambda \cdot v)&=\lambda \cdot f(w)\end{aligned}}$

So $f$ is by definition a linear map.

Proof step: $f$ is a linear map $\implies \left(\forall v_{1},\dots ,v_{n}\in V\,\forall \lambda _{1},\dots ,\lambda _{n}\in K:\,f\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)\right)$.

Let $f$ be a linear map. We prove this equation by induction in $n$:

Theorem whose validity shall be proven for the $n\in \mathbb {N}$:

$\forall v_{1},\dots ,v_{n}\in V\,\forall \lambda _{1},\dots ,\lambda _{n}\in K:\,f\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)$

1. Base case:

We start with the induction with $n=1$ and find that the property of homogeneity is sufficient:

${\begin{aligned}&f\left(\lambda _{1}\cdot _{{}_{V}}v_{1}\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{homogeneity of }}f\right.}\\[0.3em]=&\lambda _{1}\cdot _{{}_{W}}f\left(v_{1}\right)\end{aligned}}$

1. inductive step:

2a. inductive hypothesis:

$\forall v_{1},\dots ,v_{n}\in V\,\forall \lambda _{1},\dots ,\lambda _{n}\in K:\,f\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)$

2b. induction theorem:

$\forall v_{1},\dots ,v_{n},v_{n+1}\in V\,\forall \lambda _{1},\dots ,\lambda _{n},\lambda _{n+1}\in K:\,f\left(\sum _{i=1}^{n+1}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)=\sum _{i=1}^{n+1}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)$

2b. proof of induction step:

Let $v_{1},\dots ,v_{n+1}\in V$ and $\lambda _{1},\dots ,\lambda _{n+1}\in K$. Then

${\begin{aligned}&f\left(\sum _{i=1}^{n+1}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{split the sum}}\right.}\\[0.3em]=&f\left(\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)+_{{}_{V}}\left(\lambda _{n+1}\cdot _{{}_{V}}v_{n+1}\right)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}f\right.}\\[0.3em]=&f\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)+_{{}_{W}}f\left(\lambda _{n+1}\cdot _{{}_{V}}v_{n+1}\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{homogeneity of }}f\right.}\\[0.3em]=&f\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{V}}v_{i}\right)+_{{}_{W}}\left(\lambda _{n+1}\cdot _{{}_{W}}f\left(v_{n+1}\right)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{induction assumption}}\right.}\\[0.3em]=&\left(\sum _{i=1}^{n}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)\right)+_{{}_{W}}\left(\lambda _{n+1}\cdot _{{}_{W}}f\left(v_{n+1}\right)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{join the sum}}\right.}\\[0.3em]=&\sum _{i=1}^{n+1}\lambda _{i}\cdot _{{}_{W}}f\left(v_{i}\right)\end{aligned}}$

Compositions of linear maps are again linear

Let us take two linear maps $f:V_{1}\to V_{2}$ and $g:V_{2}\to V_{3}$. Both are compatible with the vector space structure and preserve linear combinations. This preservation should also hold for the consecutive execution of both maps $g\circ f:V_{1}\to V_{3}$ with $(g\circ f)(v)=g(f(v))$. This is mathematically established by the following theorem:

Theorem (Composition of linear maps)

Let $f:V_{1}\to V_{2}$ and $g:V_{2}\to V_{3}$ be two linear maps between the $K$-vector spaces $V_{1}$, $V_{2}$ and $V_{3}$. Then the composition $g\circ f:V_{1}\to V_{3}$ of these two maps with $(g\circ f)(v)=g(f(v))$ for $v\in V_{1}$ is also a linear map.

How to get to the proof? (Composition of linear maps)

We know that the composition (missing) of two maps is again a well-defined map. So we just need to show that $g\circ f$ is linear. To do this, we need to prove that $(g\circ f)$ satisfies additivity and homogeneity.

• For all $v_{1},\,v_{2}\in V_{1}$ we have that $(g\circ f)(v_{1}+v_{2})=(g\circ f)(v_{1})+(g\circ f)(v_{2})$ and
• For all $\lambda \in K$ and $v\in V$ we have that $(g\circ f)(\lambda \cdot v)$.

To prove this, we exploit the additivity and homogeneity of the individual maps $f$ and $g$.

Proof (Composition of linear maps)

Let first $v_{1},v_{2}\in V_{1}$ be any two vectors. We have that

${\begin{aligned}&(g\circ f)(v_{1}+v_{2})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}g\circ f\right.}\\[0.3em]=&g(f(v_{1}+v_{2}))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}f\right.}\\[0.3em]=&g(f(v_{1})+(v_{2}))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}g\right.}\\[0.3em]=&g(f(v_{1}))+g(f(v_{2}))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}g\circ f\right.}\\[0.3em]&(g\circ f)(v_{1})+(g\circ f)(v_{2})\end{aligned}}$

For the proof of homogeneity we choose any $\lambda \in K$ and any $v\in V_{1}$:

${\begin{aligned}&(g\circ f)(\lambda \cdot v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}g\circ f\right.}\\[0.3em]=&g(f(\lambda \cdot v))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{homogeneity of }}f\right.}\\[0.3em]=&g(\lambda \cdot f(v))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{homogeneity of }}g\right.}\\[0.3em]=&\lambda \cdot g(f(v))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}g\circ f\right.}\\[0.3em]&\lambda \cdot (g\circ f)(v)\end{aligned}}$

Subspaces are mapped to subspaces

That linear maps preserve the vector space structure can also be seen in the following property: The images of subspaces of a linear map are again subspaces.

Theorem (Subspaces are mapped to subspaces)

Let $f:V\to W$ be a linear map between two $K$-vector spaces $V$ and $W$. Then the image $f(U)=\{f(v):v\in U\}$ of every subspace $U\subseteq V$ is again a subspace of $W$.

Proof (Subspaces are mapped to subspaces)

Let $U$ be a subspace of $V$. The image $f(U)=\{f(v):v\in U\}$ is the set of all function values of arguments from $U$ and thus a subset of the range of values $W$. To show that $f(U)$ is a subspace, the following criteria must be shown:

1. $f(U)\neq \emptyset$
2. For all $w_{1},w_{2}\in f(U)$ we have that $w_{1}+w_{2}\in f(U)$.
3. For all $w\in f(U)$ and for all $\lambda \in K$ we have that $\lambda \cdot w\in f(U)$.

Proof step: $f(U)\neq \emptyset$

Since $U$ is a subspace of $V$ we have $0_{V}\in U$. With $f(0_{V})$ , we have at least one element in $f(U)$ and so $f(U)\neq \emptyset$.

Proof step: For all $w_{1},w_{2}\in f(U)$ we have that $w_{1}+w_{2}\in f(U)$

Let us take any two vectors $w_{1},w_{2}\in f(U)$. Because these vectors lie in the image, there are at least two vectors $v_{1},v_{2}\in U$ with $f(v_{1})=w_{1}$ and $f(v_{2})=w_{2}$. Now

$w_{1}+w_{2}=f(v_{1})+f(v_{2})=f(v_{1}+v_{2})$

Thus $w_{1}+w_{2}$ is the image of $v_{1}+v_{2}$ (the vector $v_{1}+v_{2}$ is mapped to $w_{1}+w_{2}$). Because $U$ is a subspace of $V$, we have that $v_{1}+v_{2}\in U$ and thus $w_{1}+w_{2}$ lies in $f(U)$.

Proof step: For all $w\in f(U)$ and for all $\lambda \in K$ we have that $\lambda \cdot w\in f(U)$

Let $\lambda \in K$ and $w\in f(U)$. Since $w$ lies within the image of $U$ , we can find a $v\in V$ with $f(v)=w$. Now

$\lambda \cdot w=\lambda \cdot f(v)=f(\lambda \cdot v)$

Thus $\lambda \cdot w$ is the image of $\lambda \cdot v$ (the vector $\lambda \cdot v$ is mapped to $\lambda \cdot w$). Because $U$ is a subspace of $V$, we have that $\lambda \cdot v\in U$ and thus $\lambda \cdot w$ lies in $f(U)$.

Hint

The above theorem also proves that the image $f(V)$ of a linear map $f:V\to W$ is always a vector space. This follows from the fact that the vector space $V$ is also a subspace of itself. According to the theorem above $f(V)=\{f(v):v\in V\}$ is a subspace of $W$.

Spans are mapped to spans

Now suppose we have a subset $M\subseteq V$. For this subset, it does not matter whether we first calculate the span and then apply the map or vice versa. This is content of the following theorem:

Theorem (Spans are mapped to spans)

Let $M\subseteq V$ be any subset (not necessarily a subspace!) of the vector space $V$. Then we have that for the span of $M$:

$\operatorname {span} (f(M))=f(\operatorname {span} (M))$

How to get to the proof? (Spans are mapped to spans)

Since we want to show the equality of two sets, we must show that the sets are contained in the other. Once we have shown this, it follows that the two sets are equal.

To show that $f(\operatorname {span} (M))\subseteq \operatorname {span} (f(M))$, we first choose an arbitrary vector $v\in \operatorname {span} (M)$. Because this is in the span, it can be written as a linear combination of elements from the set $M$:

${\begin{aligned}&v=\lambda _{1}\cdot m_{1}+\cdots +\lambda _{n}\cdot m_{n}\\[0.3em]\end{aligned}}$

where $\lambda _{1},\ldots ,\lambda _{n}\in K$.

We then apply the linear map to both $v$ and the linear combination of elements from $M$. This yields the following expression:

${\begin{aligned}&f(v)=f(\lambda _{1}\cdot m_{1}+\cdots +\lambda _{n}\cdot m_{n})\\[0.3em]\end{aligned}}$

Then, using the properties of linear maps, we reshape the expression:

${\begin{aligned}&f(v)=\lambda _{1}\cdot f(m_{1})+\cdots +\lambda _{n}\cdot f(m_{n})\\[0.3em]\end{aligned}}$

Since the right-hand side is contained in $\operatorname {span} (f(M))$, we have that $f(v)\in \operatorname {span} (f(M))$.

Similarly, we show that $\operatorname {span} (f(M))\subseteq f(\operatorname {span} (M))$ . Now we need to prove that for any $w\in \operatorname {span} (f(M))$ there exists a vector $v\in \operatorname {span} (M)$ with $f(v)=w$. We know

$w=\lambda _{1}\cdot f(m_{1})+\cdots +\lambda _{n}\cdot f(m_{n})$

with $\lambda _{1},\ldots ,\lambda _{n}\in K$. Using the linearity of $f$, we can transform the expression to:

$w=f(\lambda _{1}\cdot m_{1}+\cdots +\lambda _{n}\cdot m_{n})$

But the right hand side is in $f(\operatorname {span} (M))$ and thus we have that the same holds for $w$.

Proof (Spans are mapped to spans)

Proof step: $f(\operatorname {span} (M))\subseteq \operatorname {span} (f(M))$

We first take any vector $v\in \operatorname {span} (M)$, for which we have that: $f(v)=w\in W$.

Since we know that the vector is in the span of $M$, there are coefficients $\lambda _{1},\cdots ,\lambda _{n}\in K$ and vectors $m_{1},\cdots ,m_{n}\in M$, such that:

${\begin{aligned}&v=\lambda _{1}\cdot m_{1}+\cdots +\lambda _{n}\cdot m_{n}\\[0.3em]\end{aligned}}$

If we now apply our linear map to this expression, we get:

${\begin{aligned}&f(v)=f(\lambda _{1}\cdot m_{1}+\cdots +\lambda _{n}\cdot m_{n})\\[0.3em]\end{aligned}}$

We now further transform the right-hand side of this expression using the properties of linear maps:

${\begin{aligned}&f(\lambda _{1}\cdot m_{1}+\cdots +\lambda _{n}\cdot m_{n})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}f\right.}\\[0.3em]=&f(\lambda _{1}\cdot m_{1})+\cdots +f(\lambda _{n}\cdot m_{n})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{homogeneity of }}f\right.}\\[0.3em]=&\lambda _{1}\cdot f(m_{1})+\cdots +\lambda _{n}\cdot f(m_{n})\\[0.3em]\end{aligned}}$

On the right-hand side we now have $\operatorname {span} (f(M))$.

So we have shown that $w\in \operatorname {span} (f(M))$ and thus $f(\operatorname {span} (M))\subseteq \operatorname {span} (f(M))$.

Proof step: $\operatorname {span} (f(M))\subseteq f(\operatorname {span} (M))$

We again choose an arbitrary vector $w\in \operatorname {span} (f(M))$. Since it is in the span of the map, we can write it as:

${\begin{aligned}&w=\lambda _{1}\cdot f(m_{1})+\cdots +\lambda _{n}\cdot f(m_{n})\\[0.3em]\end{aligned}}$

We can now start transforming the expression in the same way as in the previous proof step:

${\begin{aligned}w=&\lambda _{1}\cdot f(m_{1})+\cdots +\lambda _{n}\cdot f(m_{n})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{homogeneity of }}f\right.}\\[0.3em]=&f(\lambda _{1}\cdot m_{1})+\cdots +f(\lambda _{n}\cdot m_{n})\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{additivity of }}f\right.}\\[0.3em]=&f(\lambda _{1}\cdot m_{1}+\cdots +\lambda _{n}\cdot m_{n})\\[0.3em]\end{aligned}}$

On the right-hand side there is now a vector in $f(\operatorname {span} (M))$. We have thus shown that $w\in f(\operatorname {span} (M))$ holds.

Moreover, we have shown the equality of the sets $f(\operatorname {span} (M))$ and $\operatorname {span} (f(M))$, as both are contained within each other.