Linear transformations and associated matrices

In this post, we discuss some facts about linear transformations. This latter play a key role in linear algebra. In particular, in infinite-dimensional spaces, this concept coincides with matrices calculus.

Definition: Let $E$ and $F$ be vector spaces on a field $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C}$). We say that an application $f: E\to F$ is linear if for any $u,v\in E$ and $\lambda\in\mathbb{K}$ we have \begin{align*} f(u+\lambda v)=f(u)+\lambda f(v).\end{align*}

The first remak is that all linear transformations $f$ satisfy $f(0)=0$. In fact, f(0)=f(0+0)=f(0)+f(0)=2f(0), which implies that $f(0)=0$.

Linear Transformations Worksheet

Exercise: Let $E=\mathcal{C}(\mathbb{R},\mathbb{R})$ be the vector space on $\mathbb{R}$ of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$. If $f$ and $g$ are real functions then the usual product of $f$ by $g$ is defined by $(fg)(x)=f(x)g(x)$ for any $x\in\mathbb{R}$. We denote by $id:\mathbb{R}\to \mathbb{R}$ the identity function ($id(x)=x$ for all $x\in \mathbb{R}$). Let us now define the application $\Phi:E\to E$ by \begin{align*} \Phi(f)=id\,f,\qquad \forall f\in E. \end{align*}

  • Prove that $\Phi$ is a linear transformation and determine the kernel of $\Phi$ denoted by $\ker(\Phi)$.
  • Does $\Phi$ is surjective?
  • Determine the image of $\Phi$ denoted by ${\rm Im}(\Phi)$.

Solution: 1) Let $f,g\in E$ and $\lambda\in \mathbb{R}$. For any $x\in \mathbb{R},$ we have \begin{align*} \Phi(f+\lambda g)(x)&=id(x)(f+\lambda g)(x)\cr &= x(f(x)+\lambda g(x))\cr &= xf(x)+\lambda (xg(x))\cr &=\Phi(f)(x)+\lambda \Phi(g)(x). \end{align*} Hence \begin{align*} \Phi(f+\lambda g)=\Phi(f)+\lambda \Phi(g). \end{align*} This means that $\Phi$ is linear.

Let $f\in E$. Then \begin{align*} f\in\ker(\Phi)&\;\Longleftrightarrow\;\Phi(f)=0\cr &\;\Longleftrightarrow\;xf(x)=0,\quad \forall x\in \mathbb{R}\cr &\;\Longrightarrow f(x)=0,\quad \forall x\in \mathbb{R}\setminus\{0\}. \end{align*} By taking the limit of $f$ at zero and by using the continuity of $f$ at zero we obtain $f(0)=0$. This prove that $f=0$, so that \begin{align*} \ker(\Phi)=\{0\}. \end{align*} which means that $\Phi$ is injective.

2) Let $g\in E$ such that $g(x)=1$ for all $x\in\mathbb{R}$. Assume that there exists $f\in E$ such that $\Phi(f)=g$. Then for all $x\in \mathbb{R},$ we have $xf(x)=1$. This is impossible since for $x=0$ we will obtain $0=1$!!. Then $\Phi$ is not surjective.

3) Let $g\in {\rm Im}(\Phi)$. Then there is $f\in E$ such that $g(x)=xf(x)$ for all $x\in \mathbb{R}$. In particular, $g(0)=0$. On the other have \begin{align*} \forall x\in \mathbb{R}\setminus\{0\},\quad \frac{g(x)-g(0)}{x-0}=f(x). \end{align*} As $f$ is continuous at $0,$ we deduce that $g$ is differentiable in $0$ and $g'(0)=f(0)$. This implies that \begin{align*} {\rm Im}(\Phi)\subset \{g\in E: g(0)=0\;\text{and}\; f\;\text{is differentiable in}\; 0\}. \end{align*} Conversely, let $g\in E$ such that $g(0)=0$ and $g$ is differentiable in $0$. Then the function \begin{align*} f(x)=\begin{cases}\frac{g(x)}{x},& x\neq 0,\cr g'(0),& x=0,\end{cases} \end{align*} is an element of $E$ and $g=\Phi(f)\in {\rm Im}(\Phi)$. Hence \begin{align*} {\rm Im}(\Phi)= \{g\in E: g(0)=0\;\text{and}\; f\;\text{is differentiable in}\; 0\}. \end{align*}

Exercise: Let $\mathbb{K}$ be a field and consider the $\mathbb{K}$-vector space $E=\mathbb{K}^n$ with $n\ge 0$. Let $a=(1,1,\cdots,1)\in E$ and denote \begin{align*} A&:=\mathbb{K}a=\{(\lambda,\lambda,\cdots,\lambda):\lambda\in \mathbb{K}\}\cr B&:=\{x=(x_1,x_2,\cdots,x_n)\in E: x_1+x_2+\cdots+x_n=0\}. \end{align*} Let the linear transformation \begin{align*} f:E\to \mathbb{K},\quad f(x_1,x_2,\cdots,x_n)=\sum_{i=1}^n x_i. \end{align*}

  • Determine $\ker(f)$. Deduce that $B$ is a subspace.
  • Let $x\in E$. Prove that there exists a unique scalar $\lambda\in\mathbb{K}$ such that $f(x)=\lambda f(a)$.
  • Justify that $A\cap B=\{0\}$.
  • Deduce that the sum $A+B$ is direct, that is, $E=A\oplus B$.

Solution: 1) $A$ is not empty because $a\in A$ (just take $\lambda=1$). Observe that $A={\rm span}(a)$ the vector space generated by $a\in E$. Then $A$ is a subspace of $E$.

The kernel of $f$ is, by definition, \begin{align*} \ker(f)&=\left\{(x_1,x_2,\cdots,x_n)\in E: f(x_1,x_2,\cdots,x_n)=0\right\}\cr & = =\left\{(x_1,x_2,\cdots,x_n)\in E: x_1+x_2+\cdots+x_n=0\right\}\cr &= B. \end{align*} On the other hand, \begin{align*} (x_1,x_2,\cdots,x_n)\in \ker(f)\;\Longleftrightarrow\; x_1=-x_2-x_3-\cdots-x_n. \end{align*} Then, we can write \begin{align*} (x_1,x_2,\cdots,x_n)&=(-x_2-x_3-\cdots-x_n,x_2,\cdots,x_n)\cr & = (-x_2,x_2,0,\cdots,0)+(-x_3,0,x_3,0,\cdots,0)\cr &=\hspace{2.5cm}+\cdots +(-x_n,0,\cdots,0,x_n)\cr &= x_2 v_2+x_3 v_3+\cdots+x_n v_n, \end{align*} where for each $i\in \{2,3,\cdots,n\}$, we denote $v_i:=(-1,0,\cdots,1,0,\cdots,0)$ where $1$ takes the $i^{\rm th}$ place. This is equivalent to \begin{align*} \ker(f)=B={\rm span}(v_2,v_3,\cdots,v_n). \end{align*} $B$ is then a subspace of $E$.

2) Let prove uniqueness of $\lambda$. Assume that there exist $(\lambda,\mu)\in\mathbb{K}^2$ such that $f(x)=\lambda f(a)=\mu f(a)$. As $f(a)=n,$ we then have $(\lambda-\mu) n=0$, so that $\lambda=\mu$. Now let us prove the existence. Let $x=(x_1,x_2,\cdots,x_n)\in X$. It suffice to prove that there exist $\lambda\in\mathbb{K}$ such that $x-\lambda a\in \ker(f)$, so that \begin{align*} \sum_{i=1}^n x_i-\lambda n=0. \end{align*} It suffice to take \begin{align*} \lambda=\frac{x_1+x_2+\cdots+x_n}{n}. \end{align*}

3) Let $x\in A\cap B$ Then we have there exist $\lambda\in \mathbb{K}$ such that \begin{align*} x=\lambda a\quad\text{and}\quad x_1+x_2+\cdots+x_n=0. \end{align*} As $f$ is linear then $f(x)=\lambda f(a)$. By the previous question we have \begin{align*} \lambda=\frac{x_1+x_2+\cdots+x_n}{n}=\frac{0}{n}=0. \end{align*} Hence $x=\lambda a=0\times a=0$, so that $A\cap B=\{0\}$.

4) Because of the previous question il suffice to show that $E=A+B$. According to the question $2$, there exists $\lambda\in\mathbb{K}$ such that $f(x)=\lambda f(a)$, so that $f(x-\lambda a)=0$. Then $x-\lambda a\in \ker(f)=B$. On the other hand, $\lambda a \in A$. Thus $x=\lambda a+(x-\lambda a)\in A+B$. This ends the proof.

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