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# 数学代写|线性代数代写Linear algebra代考|What does this proposition mean?

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## 数学代写|线性代数代写Linear algebra代考|What does this proposition mean?

There is only one way of writing any vector as a linear combination of the basis vectors.

We have already proven this result for the standard basis in Proposition (2.18) of the last section.
Proof.
Let $\mathbf{u}$ be an arbitrary vector in $\mathbb{R}^n$. We are given that the vectors $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\right}$ form a basis, so they span $\mathbb{R}^n$ which means that we can write the vector $\mathbf{u}$ in $\mathbb{R}^n$ as a linear combination of $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\right}$. There exist scalars $k_1, k_2, k_3, \ldots$ and $k_n$ which satisfy
$$\mathbf{u}=k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+k_3 \mathbf{v}_3+\cdots+k_n \mathbf{v}_n$$
Suppose we can write this vector $\mathbf{u}$ as another linear combination of the basis vectors
$$\mathbf{u}=c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+c_3 \mathbf{v}_3+\cdots+c_n \mathbf{v}_n$$
where the $c$ ‘s are scalars.

What do we need to prove?
We need to prove that the two sets of scalars are equal: $k_1=c_1, k_2=c_2, \ldots$ and $k_n=c_n$. Equating the two linear combinations because both are equal to $\mathbf{u}$ gives
\begin{aligned} & k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+\cdots+k_n \mathbf{v}_n=c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+\cdots+c_n \mathbf{v}_n=\mathbf{u} \ & k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+\cdots+k_n \mathbf{v}_n-c_1 \mathbf{v}_1-c_2 \mathbf{v}_2-\cdots-c_n \mathbf{v}_n=\mathbf{u}-\mathbf{u}=\mathbf{O} \ & \left(k_1-c_1\right) \mathbf{v}_1+\left(k_2-c_2\right) \mathbf{v}_2+\cdots+\left(k_n-c_n\right) \mathbf{v}_n=\mathbf{O} \quad \text { [factorizing] } \end{aligned}
The basis vectors $\left{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$ are linearly independent, therefore all the scalars are equal to zero.

## 数学代写|线性代数代写Linear algebra代考|Why?

Because this is the definition of linear independence given in the last section (2.19):
Vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots$ and $\mathbf{v}_n$ in $\mathbb{R}^n$ are linearly independent $\Leftrightarrow$
$$m_1 \mathbf{v}_1+m_2 \mathbf{v}_2+m_3 \mathbf{v}_3+\cdots+m_n \mathbf{v}_n=\mathbf{O} \text { gives } m_1=m_2=m_3=\cdots=m_n=0$$
Applying this to the above derivation:
$$\left(k_1-c_1\right) \mathbf{v}_1+\left(k_2-c_2\right) \mathbf{v}_2+\cdots+\left(k_n-c_n\right) \mathbf{v}_n=\mathbf{O}$$
we have
$$\begin{gathered} k_1-c_1=0, k_2-c_2=0, k_3-c_3=0, \ldots \text { and } k_n-c_n=0 \ k_1=c_1, k_2=c_2, k_3=c_3, \ldots \text { and } k_n=c_n \end{gathered}$$
Hence any arbitrary vector $\mathbf{u}$ can be written uniquely as a linear combination of the basis vectors $\left{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$.

Next we prove a lemma. A lemma is a proposition or theorem, the proof of which is used as a stepping stone towards proving something of greater interest. However, there are many lemmas in mathematics which have become important results in themselves, such as Zorn’s lemma, Euclid’s lemma and Gauss’s lemma.

Lemma (2.30). Let $T=\left{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \ldots, \mathbf{w}_m\right}$ be a set of $m$ vectors that are linearly independent in $\mathbb{R}^n$ then $m \leq n$.

## 数学代写|线性代数代写Linear algebra代考|What does this proposition mean?

$$\mathbf{u}=k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+k_3 \mathbf{v}_3+\cdots+k_n \mathbf{v}_n$$

$$\mathbf{u}=c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+c_3 \mathbf{v}_3+\cdots+c_n \mathbf{v}_n$$

\begin{aligned} & k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+\cdots+k_n \mathbf{v}_n=c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+\cdots+c_n \mathbf{v}_n=\mathbf{u} \ & k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+\cdots+k_n \mathbf{v}_n-c_1 \mathbf{v}_1-c_2 \mathbf{v}_2-\cdots-c_n \mathbf{v}_n=\mathbf{u}-\mathbf{u}=\mathbf{O} \ & \left(k_1-c_1\right) \mathbf{v}_1+\left(k_2-c_2\right) \mathbf{v}_2+\cdots+\left(k_n-c_n\right) \mathbf{v}_n=\mathbf{O} \quad \text { [factorizing] } \end{aligned}

## 数学代写|线性代数代写Linear algebra代考|Why?

$\mathbb{R}^n$中的向量$\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots$和$\mathbf{v}_n$是线性无关的$\Leftrightarrow$
$$m_1 \mathbf{v}_1+m_2 \mathbf{v}_2+m_3 \mathbf{v}_3+\cdots+m_n \mathbf{v}_n=\mathbf{O} \text { gives } m_1=m_2=m_3=\cdots=m_n=0$$

$$\left(k_1-c_1\right) \mathbf{v}_1+\left(k_2-c_2\right) \mathbf{v}_2+\cdots+\left(k_n-c_n\right) \mathbf{v}_n=\mathbf{O}$$

$$\begin{gathered} k_1-c_1=0, k_2-c_2=0, k_3-c_3=0, \ldots \text { and } k_n-c_n=0 \ k_1=c_1, k_2=c_2, k_3=c_3, \ldots \text { and } k_n=c_n \end{gathered}$$

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