# 本次日本作业案例分享主要为线性代数代写的assignment

1. 矩阵的核

” ” #
” ” \$
a1,1×1 + a1,2×2 +… + a1,nxn =0
a2,1×1 + a2,2×2 +… + a2,nxn =0
………………
am,1×1 + am,2×2 +… + am,nxn =0

!” 0 是 Rm 的零向量，A 是系统的系数矩阵。

！” x = ！” 0
（即 x1 = x2 = … = xn = 0）总是一个解。

%
21
11
&
A的核是系统的解集：

2×1 + x2 =0
x1 + x2 =0

x1 = x2 =0。

!” x 是零向量。所以这里，A 的核只包含

！”0}。
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1. The kernel of a matrix
Deﬁnition 1.1. A system of linear equations is called homogeneous if the con-
stant term in each equation is equal to zero, that is, if the system can be written
as:
!
” ” #
” ” \$
a1,1×1 + a1,2×2 +… + a1,nxn =0
a2,1×1 + a2,2×2 +… + a2,nxn =0
… … … … …
am,1×1 + am,2×2 +… + am,nxn =0
A homogeneous system correponds to the matrix equation:
A!” x = !” 0
where
!” 0 is the zero vector of Rm and A is the coe!cient matrix of the system. The
last column of the augmented matrix B is zero. Such a system is always consistent,
because the zero vector, corresponding to
!” x = !” 0
(that is x1 = x2 = … = xn = 0) is always a solution.
Proposition 1.2. A homogeneous system of linear equations either has only one
solution, the zero vector of Rn, which is called the trivial solution, or it has
inﬁnitely many solutions.
Deﬁnition 1.3. Let A be a m # n matrix. The kernel or the nullspace of the
matrix A is the space of solutions of the homogeneous system A!” x = 0. It is denoted
by Ker(A).
Example 1.4. Set
A =
%
21
11
&
The kernel of A is the set of solutions of the system:

2×1 + x2 =0
x1 + x2 =0
We solve this system and we easily obtain:
x1 = x2 =0.
In other words,
!” x is the zero vector. So here, the kernel of A contains only the
zero vector:
Ker(A)= {
!” 0 }.
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