## 数学代写|线性代数代写linear algebra代考|MAST10022

2022年10月12日

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## 数学代写|线性代数代写linear algebra代考|APPLICATION: LU FACTORIZATION

Matrix factorizations or decompositions play an important role in numerical methods of computational linear algebra. They help in speeding up algorithms used in linear algebra such as solving linear systems, inverting a matrix or computing its determinant. One such decomposition is the LU factorization. In this section, we will describe this factorization as well as when and to what it applies. The LU factorization expresses a square matrix as a product of a unit lower triangular matrix times an upper triangular matrix, where a unit triangular matrix has ones on the diagonal.
Example 2.48 Below we express a matrix $A=L U$, where $L$ is unit lower triangular and $U$ is upper triangular. Shortly, we will see the algorithm for performing this factorization.
$$\left[\begin{array}{rrr} 1 & -3 & 1 \ 2 & -8 & -1 \ -3 & 1 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \ 2 & 1 & 0 \ -3 & 4 & 1 \end{array}\right]\left[\begin{array}{rrr} 1 & -3 & 1 \ 0 & -2 & -3 \ 0 & 0 & 16 \end{array}\right]$$
We will show that such a factorization is possible when one can row reduce a square matrix to an upper triangular matrix using only Type 3 elementary row operations, $a R_i+R_j$. To prove this we need a lemma the proof of which is left as an exercise. We also point out that this is not always possible. For example, for the matrix $\left[\begin{array}{ll}0 & 1 \ 1 & 0\end{array}\right]$ it is not possible (exercise).
Lemma $2.8$ The following statements about unit triangular matrices hold.

1. A finite product of unit lower (upper) triangular matrices is unit lower (upper) triangular.
2. The inverse of a unit lower (upper) triangular matrix is unit lower (upper) triangular.

Note first that the elementary matrix corresponding to an elementary row operation of the form $a R_i+R_j$ with $i<j$ is a unit lower triangular matrix (exercise). We can now prove the main result in this section.

## 数学代写|线性代数代写linear algebra代考|BASIS AND DIMENSION

This section introduces what we call a counting principle for comparing the relative sizes of vector spaces, called dimension. It will conform to our intuition of dimension for $\mathbb{R}^2$ (2-dimensional) and $\mathbb{R}^3$ (3-dimensional), but in addition it will assign dimension to many other vector spaces. Our focus will be on investigating vector spaces of finite dimension.

Definition $3.12$ A set of vectors $v_1, \ldots, v_n \in V$ a vector space is a basis for $V$ if

1. The vectors $v_1, \ldots, v_n$ span $V$, and
2. The vectors $v_1, \ldots, v_n$ are linearly independent.
Example 3.40 Take the earlier example. We have already verified that $[1,0,0]$, $[1,1,0],[1,1,1]$ span $\mathbb{R}^3$. Now we show they are linearly independent (and thus a basis for $\mathbb{R}^3$ ). Again, we use the technique from the Section 3.3. Putting the vectors in columns in a determinant,

$$\left|\begin{array}{lll} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{array}\right|=(1)(1)(1)=1 \neq 0$$
Example 3.41 The necessary generators of a particular span are a basis for that span. We shall formally prove this fact in Section 3.6. However, we illustrate this fact with one of the earlier examples. Although $[1,0,0],[1,1,0],[0,-2,0]$ do not span $\mathbb{R}^3$, from our calculations above, we know that $[1,0,0],[0,1,0]$ form a basis for $\operatorname{span}([1,0,0],[1,1,0],[0,-2,0])$, since $[1,0,0],[0,1,0]$ are the necessary generators of $\operatorname{span}([1,0,0],[1,1,0],[0,-2,0])$ and one can check that they are linearly independent.
We will not give too many examples at this point, because within this section we will shortcut the method of determining basis even further.

Definition 3.13 The following bases for their respective vector spaces are called standard bases:

1. The standard basis for $\mathbb{R}^n$ is the collection of vectors
$$e_1=[1,0, \ldots, 0], e_2=[0,1,0, \ldots, 0], \ldots, e_n=[0, \ldots, 0,1] .$$
Note that in $\mathbb{R}^2$ the notation for the standard basis is $\hat{\imath}, \hat{\jmath}$ and in $\mathbb{R}^3$ the notation is $\hat{\imath}, \hat{\jmath}, \hat{k}$
2. The standard basis for $P_n$ is the collection of vectors $1, x, x^2, \ldots, x^n$.
3. The standard basis for $P$ is the infinite collection of vectors $1, x, x^2, \ldots$.
4. The standard basis for $M_{m n}$ is the collection of vectors
$$\left{E_{i j} \mid 1 \leq i \leq m, 1 \leq j \leq n\right}$$
where each $E_{i j}$ is a matrix filled with zeros except that there is a 1 in the ijth entry.

# 线性代数代考

## 数学代写|线性代数代写线性代数代考|应用程序:LU FACTORIZATION

$$\left[\begin{array}{rrr} 1 & -3 & 1 \ 2 & -8 & -1 \ -3 & 1 & 1 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \ 2 & 1 & 0 \ -3 & 4 & 1 \end{array}\right]\left[\begin{array}{rrr} 1 & -3 & 1 \ 0 & -2 & -3 \ 0 & 0 & 16 \end{array}\right]$$

## 数学代写|线性代数代写线性代数代考|BASIS AND DIMENSION

1. 向量$v_1, \ldots, v_n$张成$V$，
2. 向量$v_1, \ldots, v_n$是线性无关的。
例3.40举前面的例子。我们已经验证了$[1,0,0]$, $[1,1,0],[1,1,1]$ span $\mathbb{R}^3$。现在我们证明它们是线性无关的(因此是$\mathbb{R}^3$的基)。我们再次使用3.3节中的技术。将向量放在行列式中的列中，

$$\left|\begin{array}{lll} 1 & 1 & 1 \ 0 & 1 & 1 \ 0 & 0 & 1 \end{array}\right|=(1)(1)(1)=1 \neq 0$$

$\mathbb{R}^n$的标准基是向量的集合
$$e_1=[1,0, \ldots, 0], e_2=[0,1,0, \ldots, 0], \ldots, e_n=[0, \ldots, 0,1] .$$

$$\left{E_{i j} \mid 1 \leq i \leq m, 1 \leq j \leq n\right}$$

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## MATLAB代写

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