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# 数学代写|数学分析作业代写Mathematical Analysis代考|Linear Functionals and Operators

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## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Linear Functionals and Operators

A particularly important set of linear transformations is that from a vector space $U$ to the base field $\mathbb{K}$.
Definition. A linear mapping from a vector space $U$ to the base field $\mathbb{K}$ is called a linear functional on $U$.
The following are examples of linear functionals.
Example 12. Define $\lambda: \mathbb{P} \rightarrow \mathbb{R}$ by $\lambda(f)=\int_0^1 f(x) d x$. (The base field is $\mathbb{R}$ and the polynomials have real coefficients.)
Example 13. Define $\lambda: \mathbb{K}{n \times n} \rightarrow \mathbb{K}$ by $\lambda(A)=\sum{i=1}^n a_{i i}$. Here $A=\left(a_{i j}\right)$ is an $n \times n$ matrix. The quantity $\sum_{i=1}^n a_{i i}$ is called the trace of $A$, often written $\operatorname{tr}(A)$.

Theorem 3.4.12. Let $M$ be a subspace of a vector space $U$. The following are equivalent:
(a) $M=\operatorname{Ker}(\lambda)$ for some nonzero linear functional $\lambda$ on $U$.
(b) M has a one-dimensional complement.
Proof. (a) implies (b). Let $x \in U$ be such that $\lambda(x) \neq 0$. By replacing $x$ with $x / \lambda(x)$, we may assume that $\lambda(x)=1$. For $y \in U$, let $w=y-\lambda(y) x$. Then $\lambda(w)=\lambda(y)-$ $\lambda(\lambda(y) x)=\lambda(y)-\lambda(y) \lambda(x)=0$. This shows that $w \in \operatorname{Ker}(\lambda)=$; hence $y=$ $w+\lambda(y) x \in M+\operatorname{Span}({x})$, and $U=M+\operatorname{Span}({x})$. Next we show that $M \cap$ $\operatorname{Span}({x})={0}$. This will complete the proof. If $y \in M \cap \operatorname{Span}({x})$, then $y=a x$ for some $a \in \mathbb{K}$, and $\lambda(y)=0$. But $\lambda(y)=a \lambda(x)=a$. Thus $a=0$, and $y=0$.
Conversely, suppose that $U=M \oplus \operatorname{Span}({x})$ for some nonzero $x \in U$. Let $S_1$ be a basis for $M$, and let $S=S_1 \cup{x}$. Then $S$ is a basis for $U$. Define $\lambda: S \rightarrow \mathbb{K}$ by $\lambda(x)=1$, and $\lambda(u)=0$ for all $u \in S_1$. Finally, extend $\lambda$ to a linear functional, which we also denote by $\lambda$, on $U$ according to theorem 3.4.4. The reader can easily verify that $\operatorname{Ker}(\lambda)=M$.

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Matrix Representation and Diagonalization

A careful reading of example 2 in section 3.4 reveals that the set of linear mappings from $\mathbb{K}^n$ to $\mathbb{K}^m$ is in one-to-one correspondence with the set of $m \times n$ matrices. This section generalizes this result. Suppose $U$ and $V$ are finite-dimensional vector spaces and that $\left{u_1, \ldots, n_n\right}$ and $\left{v_1, \ldots, v_m\right}$ are bases for $U$ and $V$, respectively. Theorem 3.4.4 states that a linear mapping $T: U \rightarrow V$ is uniquely determined by the vectors $T\left(u_1\right), \ldots, T\left(u_n\right)$. Since each of the vectors $T\left(u_j\right)$ can be uniquely written as a linear combination of $\left{v_1, \ldots, v_m\right}$ with coefficients in $\mathbb{K}$, the set of coefficients determines $T$ uniquely. This observation is the basis for the opening definition of this section. The information in this section is standard, and we assume familiarity with its contents.
Matrix Representations of Linear Mappings
Let $U$ and $V$ be finite-dimensional vector spaces, and let $n=\operatorname{dim}(U), m=\operatorname{dim}(V)$. Fix a pair of bases $B=\left{u_1, \ldots, u_n\right}$ and $C=\left{v_1, \ldots, v_m\right}$ for $U$ and $V$, respectively. If $T \in \operatorname{Hom}(U, V)$, then, for every $1 \leq j \leq n, T\left(u_j\right)$ can be written as a linear combination of $C$, say, $T\left(u_j\right)=\sum_{i=1}^m a_{i j} v_i$.

Definition. Given the construction in the previous paragraph, the matrix $A=\left(a_{i j}\right)$ is called the matrix of $T$ relative to the base pair $(B, C)$.

The matrix representing a linear mapping is totally dependent on the base pair $(B, C)$ and is even sensitive to the permutation of the elements in each basis. Thus the bases $B$ and $C$ are assumed to be ordered.

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Linear Functionals and Operators

(a) $M=\operatorname{Ker}(\lambda)$对于$U$上的非零线性泛函$\lambda$。
(b) M具有一维补。

## MATLAB代写

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