A permutation matrix
A row or column exchange matrix may be a matrix multiplication means to achieve additional
First-line exchange:
$\underbrace{\left[\begin{array}{ccc}{1} & {1} & {1} \\ {2} & {2} & {2} \\ {3} & {3} & {3}\end{array}\right]}_{A} \stackrel{S_{12}}{\rightarrow} \underbrace{\left[\begin{array}{ccc}{2} & {2} & {2} \\ {1} & {1} & {1} \\ {3} & {3} & {3}\end{array}\right]}_{A_{2}}$
That is the first row and the second row of the matrix were interchanged
$ $ A_2 for the first row, corresponding to 0 out of the first row, a second row from $ A $, the third row 0 and add
$ $ A_2 to the second row, and come up with an equivalent of the first row from the $ A $, 0 second row, the third row 0 and add
For the third row $ $ A_2, 0 corresponds to the first row come from the $ A $, 0 second row, a third row was added and
Before we said before , the row vector by a matrix, the matrix is equal to a linear combination of the results of each line
The above $ P_ {12} $ referred to as row permutation matrix. It can be seen row permutation matrix is a unit matrix rearrangement, it is a characteristic: $ \ mathrm {P} ^ {- 1} = \ mathrm {P} ^ {\ mathrm {T}} $
Exchange column and row exchange is similar, but the vector into a column vector becomes the right (left-right column is what I mean)
$\underbrace{\left[\begin{array}{lll}{1} & {2} & {3} \\ {1} & {2} & {3} \\ {1} & {2} & {3}\end{array}\right]}_{A} \stackrel{C_{12}}{\longrightarrow} \underbrace{\left[\begin{array}{ccc}{2} & {1} & {3} \\ {2} & {1} & {3} \\ {2} & {1} & {3}\end{array}\right]}_{A_{2}}$
I.e. first and second columns of the matrix interchanged
For the first column A_2 $ $, equivalent to come from the first column 0 in $ A $, by adding up a second column, the third column 0
For the second column A_2 $ $, corresponding to come up with a first row, second column 0, 0 from the third column is carried out adding $ A $ and
For the third column A_2 $ $, corresponding to the first column out of 0, 0 second column, a third column from $ A $ performed and added
We have previously talked about: matrix by a column vector to obtain a column vector - that is, the original matrix must have a linear combination of the columns
The above obtained $ C_ {12} $ called a column permutation matrix. Note that the results column permutation matrix is constructed according to the column
Summary: Unit permutation matrix is rearranged row or column matrix, the permutation matrix may be used in the process of elimination to prevent the main element 0
Second, the transpose matrix
I.e., obtain the original matrix rows into columns, rows become columns, with $ A ^ T $ represents:
$\left[\begin{array}{ll}{1} & {1} \\ {4} & {5} \\ {0} & {3}\end{array}\right]$的转置是$\left[\begin{array}{lll}{1} & {4} & {0} \\ {1} & {5} & {3}\end{array}\right]$
Note: a matrix transpose matrix multiplication can be symmetric matrix, as demonstrated
$ (R ^ TR) ^ T = R ^ T (R ^ T) ^ T = R ^ TR $, so a $ R ^ TR $ matrix transpose is equal to itself, then he is symmetrical
Third, the vector space
There are two basic vector arithmetic: adding $ v_a + v_b $ sum by $ num * v $
$ R ^ 2 $ is a two-dimensional vector space, $ R ^ 3 $ is three-dimensional vector space, is the vector space of all three-dimensional vector consisting of empathy $ R ^ n $ is the vector space of all $ n-$ dimensional vector composed of
All vector space vector must contain 0
We proceed from the $ R ^ 2 $, take the first quadrant of the two-dimensional coordinate system, then the region is a vector space? Obviously not, because the vector in the region multiplied by a negative number, then ran out of space in the region, and not in the region, so the first quadrant is obviously not a vector space
Well, since the first quadrant is not a vector space, the presence of other vector space in $ R ^ 2 $ vector space it ( in fact, also known as the quantum space )? There are of course, such as $ R ^ 2 $ through the origin of the vector space of a straight line
Since the straight line through the origin, so that a vector of any arbitrary straight line on the vector or on the straight line multiplied by the number of addition, the results still fall on the straight line, the straight line so that $ R ^ 2 $ subspace of the vector space
But note that not all straight lines are $ R ^ 2 $ subspace of vector space , such as a straight line, but if after the origin, then the number of vectors on the straight line vector multiplication result is 0 0, 0 vector is not on the line , so that the origin is not a straight line but $ R ^ 2 $ subspace of the vector space
Since spoke $ R ^ 2 $ vector space, then the subspace, what does?
a) it is $ R ^ 2 $ itself, i.e., the entire $ R ^ 2 $ vector space
b) a straight line: to be exact straight line through 0:00
c) Point: 0:00 i.e., typically referred to as the Z $ $
Similarly, $ R ^ 3 $ vector space, then the subspace, what does?
a）其$R^3$本身，即整个$R^3$向量空间
b）平面：确切的说是穿过0点的平面
c）直线：确切的说是穿过0点的直线
d）点：即0点，通常记为$Z$
下面我们来看看矩阵是如何构成子空间的，也就是我们从一个矩阵构造出子空间：
1）通过列向量来构造，如
$A=\left[\begin{array}{ll}{1} & {3} \\ {2} & {3} \\ {4} & {1}\end{array}\right]$各列属于$R^3$
我们想用其各列来构成$R^3$的子空间：方法就是矩阵各列的线性组合，记作$C(A)$，C代表列空间
针对上面的矩阵$A$，其列空间就是穿过各列和原点的平面
