Exercises Linear Maps – "Math for Non-Geeks"

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We have compiled some tasks on linear maps here. The proof structures can help you to solve other similar tasks. As a reminder, here is the definition of a linear map:

Definition (Linear map)

Let $f\colon {V}\to {W}$ be a mapping between the two vector spaces ${V}$ and ${W}$. We call $f$ a linear map from ${V}$ to ${W}$ if the following two properties are satisfied:

Showing linearity of a mapping

Linear maps from $\mathbb {R} ^{n}$ to $\mathbb {R} ^{m}$

We consider an example for a linear map of $\mathbb {R} ^{2}$ to $\mathbb {R} ^{2}$:

$f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2}$ with $f{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}\\x_{1}-5x_{2}\end{pmatrix}}$

Important special cases

Math for Non-Geeks: Template:Aufgabe

Linear maps between function spaces

Math for Non-Geeks: Template:Aufgabe

Exercises: Isomorphisms

Math for Non-Geeks: Template:Aufgabe

Exercises: Images

Exercise (Image of a matrix)

1. Consider the matrix $(1,2)\in \mathbb {R} ^{1\times 2}$ and the mapping $f\colon \mathbb {R} ^{2}\to \mathbb {R} ,x\mapsto (1,2)x$ induced by it. What is the image $\operatorname {im} (f)$?
2. Now let $A=(a_{1},\ldots ,a_{m})\in K^{n\times m}$ be any matrix over a field $K$, where $a_{1},\ldots ,a_{m}\in K^{n}$ denote the columns of $A$. Consider the mapping $f_{A}\colon K^{m}\to K^{n},x\mapsto Ax$ induced by $A$. Show that $\operatorname {im} (f_{A})=\operatorname {span} \{a_{1},\ldots ,a_{m}\}$ holds. So the image of a matrix is the span of its columns.

Solution (Image of a matrix)

Solution sub-exercise 1:

We know that the image $\operatorname {im} (f)$ of the linear map $f$ is a subspace of $\mathbb {R}$. Since the $\mathbb {R}$-vector space $\mathbb {R}$ has dimension $1$, a subspace can only have dimension $0$ or $1$. In the first case the subspace is the null vector space, in the second case it is already all of $\mathbb {R}$. So $\mathbb {R}$ has only the two subspaces $\{0\}$ and $\mathbb {R}$. Since $(1,2)(1,0)^{T}=1\neq 0$ holds, we have that $\operatorname {im} (f)\neq \{0\}$. Thus, $\operatorname {im} (f)=\mathbb {R}$.

Solution sub-exercise 2:

Proof step: "$\subseteq$"

Let $y\in \operatorname {im} (f_{A})$. Then, there is some $x=(x_{1},\ldots ,x_{m})^{T}\in K^{m}$ with $Ax=y$. We can write $x$ as $x=\sum _{i=1}^{m}x_{i}e_{i}$. Plugging this into the equation $Ax=y$, we get.

${\begin{aligned}y&=Ax\\&{\color {OliveGreen}\left\downarrow \ x=\sum _{i=1}^{m}x_{i}e_{i}\right.}\\[0.3em]&=A\left(\sum _{i=1}^{m}x_{i}e_{i}\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{application of }}A{\text{ is linear}}\right.}\\[0.3em]&=\sum _{i=1}^{m}x_{i}Ae_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ Ae_{i}=a_{i}{\text{, the }}i{\text{-th column of }}A\right.}\\[0.3em]&=\sum _{i=1}^{m}x_{i}a_{i}.\end{aligned}}$

Since $\sum _{i=1}^{m}x_{i}a_{i}\in \operatorname {span} \{a_{1},\ldots ,a_{n}\}$, we obtain $y\in \operatorname {span} \{a_{1},\ldots ,a_{n}\}$.

Proof step: "$\supseteq$"

Let $y=\sum _{i=1}^{m}y_{i}\cdot a_{i}\in \operatorname {span} (f_{A})$ with $y_{i}\in K$ for $i=1,\ldots ,m$. We want to find $x\in K^{m}$ with $Ax=y$. So let us define $x:=\sum _{i=1}^{m}y_{i}e_{i}$. The same calculation as in the first step of the proof then shows

$Ax=A\left(\sum _{i=1}^{m}y_{i}e_{i}\right)=\sum _{i=1}^{m}y_{i}Ae_{i}=\sum _{i=1}^{m}y_{i}a_{i}=y.$

Exercises: Kernel

Check your understanding: Can you visualize $\ker(f)$ in the plane? What does the image of $f$ look like? How do the kernel and the image relate to each other?

The kernel of f

We have already seen that $\ker(f)=\left\{{\begin{pmatrix}x\\x\end{pmatrix}}\mid x\in \mathbb {R} \right\}=\operatorname {span} \left({\begin{pmatrix}1\\1\end{pmatrix}}\right)$

Now we determine the image of $f$ by applying $f$ to the canonical basis. ${\begin{aligned}f{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}-3\\1\end{pmatrix}}\\f{\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}3\\-1\end{pmatrix}}\end{aligned}}$ So $\operatorname {im} (f)=\operatorname {span} (f((1,0)^{T}),f((0,1)^{T}))$ holds. We see that the two vectors are linearly dependent. That is, we can generate the image with only one vector: $\operatorname {im} (f)=\operatorname {span} ((-3,1)^{T})$.

• 
  The image of f
• 
  Image and kernel of f together

In our example, the image and the kernel of the linear map $f$ are straight lines through the origin. The two straight lines intersect only at the zero and together span the whole $\mathbb {R} ^{2}$.

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