《线性代数导论》笔记 第3章 Vector Spaces and Subspaces
本文为《Introduction to Linear Algebra》的读书笔记
3.0 个人心得
矩阵A做消元(elimination)，就是假设Ax=0是有解的，也就是假设矩阵A的列向量是线性相关的，然后求这个系数。
消元其实就是后面的等式，减去前面几个等式，来达到 化简后面几个等式的效果。
3.1 Spaces of Vectors
In the definition of a vector space, vector addition x + y and scalar multiplication ex must obey the following eight rules:
3.2 The Nullspace of A: Solving Ax=0 and Rx=0
该小节笔记 摘自Gilbert的讲课视频《7. Solving Ax=0 Pivot Variables, Special Solutions》
\(Ax = 0\) 的所有\(x\)解构成了\(A\)的nullspace，记作\(N(A)\) 。
这里我们对矩阵\(A\)通过行操作进行消元，不会改变\(x\)的解。
例如A可以转化为U，这里U矩阵又被称为echelon form. (echelon这里是阶梯形状的意思)
这里\(U_{11}\) 和\(U_{23}\) 是pivot，\(A\)的秩(rank)等于pivot的个数。
我感觉这里pivot本身的定义就不太清楚。 A的阶还是记作消元后，不为0的行的个数。
另外rank翻译成秩，我感觉也有很大的问题。在英文中rank可以表示等级，阶级，行列 等多重意思。翻译后则语义全无。
pivot所在的列\(A_{1}\) \(A_{3}\) 被称为pivot column , 其他列被称为 free column（虽然我觉得叫dependent column 可能会更好一点，因为它们是依赖 pivot column的）
求\(Ax=0\)的解，通过消元就转化为 求\(Ux=0\)的解。
也就是求
为了获得\(Ux=0\)的解，对于 free column对应的x分量，我们可以赋值为1或者0。
- 为了获得一个特殊解，我们需要将一个free column对应的x分量设置为1，然后其他的free column对应的分量设置为0。比如 \(x_{2}=1\) , \(x_{4}=0\) 。 然后回替法，我们可以求出pivot column对应的 \(x_{3}=0\) , \(x_{1}=-2\) 。 x=(-2, 1, 0, 0)
- 接着，我们按照上面的步骤，获取其他 free column的特殊解。对于当前案例，也就是设置\(x_{4}=1\) ， \(x_{2}=0\) 。 我们可以得到 x=(2, 0, -2, 1)
- 最终我们可以得到A的nullspace , 它们是 每个 free column取1时的特殊解的 线性组合
free column的个数 等于列的个数减去 pivot列的个数， n-r = 4-2
reduced row echelon form
上面的U矩阵，也就是echelon form，可以进一步转化为简化行阶梯形式(reduced row echelon form)，记作R，或者是rref(A)，也就是reduced row echelon form的缩写。
- ==Produce zeros above the pivots. ==Use pivot rows to eliminate upward in R
- Produce ones in the pivots. Divide the whole pivot row by its pivot.
这里我们将所有pivot column提取出来，也就是第1列和第3列提取出来，可以得到\(I\) 单位矩阵(identity matrix)，再将原矩阵剩下的free column加进去，可以得到。
\(rref\) 列整理后 通常是下面这样的格式， \(I\) 是单位矩阵，也就是pivot column的部分， \(F\) 是free column的部分
这里我们将 \(Rx=0\) 的特殊解放入矩阵\(N\) 中， 就变成求 \(RN=0\) 中\(N\)的值
观察可得(根据block multiplication )，
A and U and R also have r independent columns (the pivot columns). Section 3.4
says what it means for rows or columns to be independent.
A third definition of rank, at the top level of linear algebra, will deal with spaces of
vectors. The rank \(r\) is the "dimension" of the column space. It is also the dimension of
the row space. The great thing is that n - r is the dimension of the nullspace.
这里 rank of the column space 为什么等于 rank of the row space?
Rank-Nullity Theorem(阶-无效力 定理)： 对于 \(m×n\) 的矩阵\(A\)，存在
\(n = rank(A) + nullity(A)\)
n = the number of columns of A. 矩阵A的列的个数
Rank = the dimension of the column space (the number of linearly independent columns) Pivot Column的个数
Nullity = the dimension of the nullspace (the number of linearly independent solutions to Ax=0) Free Column的个数
Elimination: The Big Picture
I mention that reducing \([A\space I]\) to \([R\space E]\) will tell you even more about \(A\) —in fact
virtually everything (including \(EA = R\)). The matrix \(E\) keeps a record, otherwise lost,
of the elimination from \(A\) to \(R\). When \(A\) is square and invertible, \(R\) is \(I\) and \(E\) is \(A^{-1}\)
Why row rank equals column rank
Step1: Start with a matrix
The row rank is the number of linearly independent rows.
The column rank is the number of linearly independent columns.
At first glance, there is no reason these numbers should be the same.
Step2: Perform row operations
Use Gaussian elimination to reduce the matrix to row echelon form.
下面是重点， 行操作不会修改 行或列的线性独立性
For any matrix, row operations:
- do not change the linear dependence among the rows, so the row rank stays the same.
- do not change the linear dependence relationships among the columns (although the columns themselves change), so the number of independent columns also stays the same.
Eventually, you obtain a matrix like
Step 3: Count the pivots
Notice there are 2 pivot positions.
These pivots tell us two things:
- There are 2 nonzero rows, so the row rank is 2.
- There are 2 pivot columns, so the column rank is also 2.
Thus,
==row rank= number of pivots = column rank
Why do pivots count both?
Each pivot corresponds to:
- one new independent row, and
- one new independent column
So every pivot increases both the row rank and the column rank by exactly one.
3.3 The Complete Solution to \(Ax=b\)
3.4 Independence, Basis and Dimension
Every vector in the space is a unique combination of the basis vectors.
basis vectors: independent vectors that "span the space".
Linear Independence
DEFINITION The columns of \(A\) are linearly independent when the only solution to \(Ax=0\) is \(x=0\). **No other combination Ax of the columns gives the zero vector
DEFINITION The sequence of vectors \(v_{1}, ..., v_{n}\) is linearly independent if the only combination that gives the zero vector is \(0v_{1}+0v_{2}+...+0v_{n}\)
Another way to describe linear dependence is this: "One vector is a combination of the other vectors."
Vectors that Span a Subspace
DEFINITION A set of vectors spans a space if their linear combinations fill the space.
DEFINITION A basis for a vector space is a sequence of vectors with two properties:
- The basis vectors are linearly independent
- they span the space
There is one and only one way to write v as a combination of the basis vectors.
Reason: Suppose \(v = a_{1}v_{1} +· ··+a_{n}v_{n}\) and also \(v = b_{1}v_{1} +...+b_{n}v_{n}\). By subtraction
\((a_{1} - b_{1})v_{1}+···+(a_{n} - b_{n} )v_{n}\) is the zero vector. From the independence of the \(v\)'s, each
\(a_{i} - b_{i} = 0\). Hence \(a_{i} = b_{i}\) , and there are not two ways to produce \(v\).
n×n可逆矩阵的列是 \(R^{n}\) 的一个基(basis)。 \(R^{n}\) 有无穷多个基。
The vectors \(v_{1}, ... , v_{n}\) are a basis for \(R^{n}\) exactly when they are the columns of an n by n invertible matrix. Thus \(R^{n}\) has infinitely many different bases.
The pivot columns of A are a basis for its column space. The pivot rows of A are a basis
for its row space. So are the pivot rows of its echelon form R
任意的基(basis)里面的向量的个数，就是space的维度。
There are many choices for the basis vectors, but the number of basis vectors doesn't change.
If \(v_{1}, ... , v_{m}\) and \(w_{1}, ... , w_{n}\) are both bases for the same vector space, then m = n . p171
DEFINITION The dimension of a space is the number of vectors in every basis
We never say "the rank of a space" or "the dimension of a basis" or "the basis of a matrix". Those terms have no meaning. It is the dimension of the column space that equals the rank of the matrix.
3.5 Dimensions of the Four Subspaces
The dimension of a subspace is the number of vectors in a basis. We count pivots or we count basis vectors.
The row space and column space have the same dimension \(r\). This number \(r\)
is the rank of the matrix.
\(N(A)\) and \(N(A^T)\) have dimensions n - r and m - r, to make up the full n and m.
==If A is m by n of rank r, its left nullspace has dimension m - r. ==
\(C(A)\) 的基向量(basis vectors) 是A的pivot columns，维度是A的rank，也就是r
\(N(A)\)的基向量 是Ax=0的free column的special solutions，维度是n-r。（因为列向量除了pivot columns 就是free columns）
Why is this a "left nullspace"? The reason is that \(R^Ty = 0\) can be transposed to
\(y^TR = 0^T\) . Now \(y^T\) is a row vector to the left of R.
In \(R^n\) the row space and nullspace have dimensions r and n - r (adding to n).
In \(R^m\) the column space and left nullspace have dimensions r and m - r (total m).
Every rank one matrix is one column times one row \(A=uv^{T}\)
浙公网安备 33010602011771号