## 线性代数代考_linear algebra代考_CALCULATION OF KERNEL AND IMAGE

This section is extremely important for the exercises as it provides us with practical methods for the calculation of bases for the kernel and the image of a given linear transformation.
We begin with the calculation of a basis of the kernel of a linear map.
Suppose we have a linear map $F: \mathbb{R}^n \longrightarrow \mathbb{R}^m$ and we want to determine a basis for the kernel. We endow $\mathbb{R}^n$ and $\mathbb{R}^m$ with the canonical bases; then, by Proposition $5.2 .2$, we have that $F(\mathbf{x})=A \mathbf{x}$, for a suitable matrix $A \in \mathrm{M}_{m, n}(\mathbb{R})$. By the definition of kernel, we have:
$$\text { Ker } F=\left{\mathbf{x} \in \mathbb{R}^n \mid A \mathbf{x}=0\right}$$
that is, the kernel of $F$ is the set of solutions of the homogeneous linear system associated with $A$.
Let us see a concrete example.
Example 5.7.1 Consider the linear map $F: \mathbb{R}^4 \longrightarrow \mathbb{R}^2$ defined by: $F\left(\mathbf{e}_1\right)=-\mathbf{e}_2$, $F\left(\mathbf{e}_2\right)=3 \mathbf{e}_1-4 \mathbf{e}_2, F\left(\mathbf{e}_3\right)=-\mathbf{e}_1, F\left(\mathbf{e}_4\right)=3 \mathbf{e}_1+\mathbf{e}_2$. We want to determine a basis for the kernel of $F$. We write the matrix $A$ associated with $F$ with respect to the canonical bases:
$$A=\left(\begin{array}{cccc} 0 & 3 & 3 & -1 \ -4 & -1 & 0 & 1 \end{array}\right) .$$
Therefore $F\left(x_1, x_2, x_3, x_4\right)=\left(3 x_2-x_3+3 x_4,-x_1-4 x_2+x_4\right)$, and Ker $F$ is the set of solutions of the homogeneous linear system:
$$\left{\begin{array}{l} 3 x_2-x_3+3 x_4=0 \ -x_1-4 x_2+x_4=0 \end{array}\right.$$
which is indeed associated with the matrix $A$.
To reduce $A$ in row echelon form, it is sufficient to exchange its two lines and we get:
$$A^{\prime}=\left(\begin{array}{cccc} -1 & -4 & 0 & 1 \ 0 & 3 & -1 & 3 \end{array}\right) \text {. }$$

## 线性代数代考_linear algebra代考_LINEAR SYSTEMS

In Chapter 5, we defined the row rank of a matrix (see Definition 5.7.3). We now want to deepen the study of this notion and have a clearer view of the link between matrices, linear systems and transformations.

Given a matrix $A \in \mathrm{M}{m, n}(\mathbb{R})$, we can read its rows as vectors of $\mathbb{R}^n$ and its columns as vectors of $\mathbb{R}^m$. It is therefore natural to introduce the following definition. Definition 6.2.1 We call column rank of a matrix $A \in \mathrm{M}{m, n}(\mathbb{R})$, the maximum number of linearly independent columns of $A$, i.e. the dimension of the subspace of $\mathbb{R}^m$ generated by the columns of $A$.

The following observation is already known and yet, given the its importance in the context that we are studying, we want to rexamine it.

Observation 6.2.2 If we write $A$ as the matrix associated with the linear transformation $L_A: \mathbb{R}^n \longrightarrow \mathbb{R}^m$ with respect to the canonical bases, then the column rank of $A$ is the dimension of the image of $L_A$. Indeed, the image is generated by the columns of the matrix $A$.

Although in general the row vectors and column vectors of a matrix $A \in \mathrm{M}_{m, n}(\mathbb{R})$ are elements of different vector spaces, the row and column rank of $A$ always coincide. This number is simply called $\operatorname{rank}$ of $A$, denoted by $\operatorname{rk}(A)$.

Proposition 6.2.3 If $A \in \mathrm{M}_{m, n}(\mathbb{R})$, then the row rank of $A$ is equal to the column rank of $A$.

# 线性代数代考

## 线性代数代考_linear algebra代考_CALCULATION OF KERNEL AND IMAGE

$\backslash$ text ${$ Ker $} F=\backslash$ left $\left{\right.$ mathbf ${x} \backslash$ in $\backslash \operatorname{mathbb}{R}^{\wedge} n \backslash$ mid $A \backslash$ mathbf ${x}=0 \backslash$ right $}$

$F\left(\mathbf{e}_2\right)=3 \mathbf{e}_1-4 \mathbf{e}_2, F\left(\mathbf{e}_3\right)=-\mathbf{e}_1, F\left(\mathbf{e}_4\right)=3 \mathbf{e}_1+\mathbf{e}_2$. 我们想确定内核的基础 $F$. 我们写矩阵 $A$ 有 关联 $F$ 关于规范基础:

$\$ \$$Veft {$$
3 x_2-x_3+3 x_4=0-x_1-4 x_2+x_4=0
$$正确的。 whichisindeedassociatedwiththematrix \ A \$$. Toreduce $\$ A \$$inrowechelon form, itissufficienttoe \mathrm{A}^{\wedge}{ \prime }=\backslash \operatorname{left}($$
\begin{array}{lllllllll}
-1 & -4 & 0 & 1 & 0 & 3 & -1 & 3
\end{array}
$$、右) \文本 {0} \ \$$

## 线性代数代考_linear algebra代考_LINEAR SYSTEMS

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