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# 数学代写|线性代数代写Linear algebra代考|MATH311 Homogeneous linear systems

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## 数学代写|线性代数代写Linear algebra代考|Homogeneous linear systems

We begin with a definition.
Definition 5.1: A linear system of the form $\mathrm{Ax}=\mathbf{0}$ is called a homogeneous linear system.

A homogeneous system $\mathbf{A x}=\mathbf{0}$ always has at least one solution, namely, the zero solution because $\mathbf{A} \mathbf{0}=\mathbf{0}$. A homogeneous system is therefore always consistent. The zero solution $\mathbf{x}=\mathbf{0}$ is called the trivial solution and any non-zero solution is called a nontrivial solution. From the existence and uniqueness theorem (Theorem 2.5), we know that a consistent linear system will have either one solution or infinitely many solutions. Therefore, a homogeneous linear system has nontrivial solutions if and only if its solution set has at least one parameter.

Recall that the number of parameters in the solution set is $d=n-r$, where $r$ is the rank of the coefficient matrix $\mathbf{A}$ and $n$ is the number of unknowns.
Example 5.2. Does the linear homogeneous system have any nontrivial solutions?
\begin{aligned} 3 x_1+x_2-9 x_3 & =0 \ x_1+x_2-5 x_3 & =0 \ 2 x_1+x_2-7 x_3 & =0 \end{aligned}
Solution. The linear system will have a nontrivial solution if the solution set has at least one free parameter. Form the augmented matrix:
$$\left[\begin{array}{llll} 3 & 1 & -9 & 0 \ 1 & 1 & -5 & 0 \ 2 & 1 & -7 & 0 \end{array}\right]$$

## 数学代写|线性代数代写Linear algebra代考|Nonhomogeneous systems

As we have seen, a homogeneous system $\mathbf{A x}=\mathbf{0}$ is always consistent. However, if $\mathbf{b}$ is nonzero, then the nonhomogeneous linear system $\mathbf{A x}=\mathbf{b}$ may or may not have a solution. A natural question arises: What is the relationship between the solution set of the homogeneous system $\mathbf{A x}=\mathbf{0}$ and that of the nonhomogeneous system $\mathbf{A} \mathbf{x}=\mathbf{b}$ when it is consistent? To answer this question, suppose that $\mathbf{p}$ is a solution to the nonhomogeneous system $\mathbf{A} \mathbf{x}=\mathbf{b}$, that is, $\mathbf{A p}=\mathbf{b}$. And suppose that $\mathbf{v}$ is a solution to the homogeneous system $\mathbf{A x}=\mathbf{0}$, that is, $\mathbf{A v}=\mathbf{0}$. Now let $\mathbf{q}=\mathbf{p}+\mathbf{v}$. Then
\begin{aligned} \mathbf{A q} & =\mathbf{A}(\mathbf{p}+\mathbf{v}) \ & =\mathbf{A p}+\mathbf{A} \mathbf{v} \ & =\mathbf{b}+\mathbf{0} \ & =\mathbf{b} \end{aligned}
Therefore, $\mathbf{A q}=\mathbf{b}$. In other words, $\mathbf{q}=\mathbf{p}+\mathbf{v}$ is also a solution of $\mathbf{A} \mathbf{x}=\mathbf{b}$. We have therefore proved the following theorem.

Theorem 5.5: Suppose that the linear system $\mathbf{A x}=\mathbf{b}$ is consistent and let $\mathbf{p}$ be a solution. Then any other solution $\mathbf{q}$ of the system $\mathbf{A x}=\mathbf{b}$ can be written in the form $\mathbf{q}=\mathbf{p}+\mathbf{v}$, for some vector $\mathbf{v}$ that is a solution to the homogeneous system $\mathbf{A x}=\mathbf{0}$.

Another way of stating Theorem $5.5$ is the following: If the linear system $\mathbf{A x}=\mathbf{b}$ is consistent and has solutions $\mathbf{p}$ and $\mathbf{q}$, then the vector $\mathbf{v}=\mathbf{q}-\mathbf{p}$ is a solution to the homogeneous system $\mathbf{A x}=\mathbf{0}$. The proof is a simple computation:
$$\mathbf{A v}=\mathbf{A}(\mathbf{q}-\mathbf{p})=\mathbf{A q}-\mathbf{A p}=\mathbf{b}-\mathbf{b}=\mathbf{0}$$
More generally, any solution of $\mathbf{A} \mathbf{x}=\mathbf{b}$ can be written in the form
$$\mathbf{q}=\mathbf{p}+t_1 \mathbf{v}_1+t_2 \mathbf{v}_2+\cdots+t_p \mathbf{v}_d$$
where $\mathbf{p}$ is one particular solution of $\mathbf{A x}=\mathbf{b}$ and the vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_d$ span the solution set of the homogeneous system $\mathbf{A x}=\mathbf{0}$.

## 数学代写|线性代数代写Linear algebra代考|Homogeneous linear systems

$$3 x_1+x_2-9 x_3=0 x_1+x_2-5 x_3 \quad=02 x_1+x_2-7 x_3=0$$

## 数学代写|线性代数代写Linear algebra代考|Nonhomogeneous systems

$$\mathbf{A q}=\mathbf{A}(\mathbf{p}+\mathbf{v}) \quad=\mathbf{A p}+\mathbf{A} \mathbf{v}=\mathbf{b}+\mathbf{0} \quad=\mathbf{b}$$

$$\mathbf{A} \mathbf{v}=\mathbf{A}(\mathbf{q}-\mathbf{p})=\mathbf{A} \mathbf{q}-\mathbf{A} \mathbf{p}=\mathbf{b}-\mathbf{b}=\mathbf{0}$$

$$\mathbf{q}=\mathbf{p}+t_1 \mathbf{v}_1+t_2 \mathbf{v}_2+\cdots+t_p \mathbf{v}_d$$

## MATLAB代写

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