Review of Linear Algebra

Transpose and Inverse

$$(AB{)}^T=B^T{A}^T$$

https://blog.csdn.net/u011240016/article/details/52914262

$$(AB{)}^{-1}=B^{-1}{A}^{-1}$$

$$(A^T{)}^{-1}=(A^{-1}{)}^T$$

$$A{A}^{-1}=I,(A^{-1}{)}^T(A^T)=I$$

Permutation matrix

$$P^{-1}=P^T$$

$$P^{T}P=I$$

Trace

• cyclic property
  $$tr(ABCD)=tr(BCDA)=tr(CDAB)=tr(DABC)$$

• • derivation
    

https://blog.csdn.net/keineahnung2345/article/details/120481487

Vector space Vector space

• Vector space is closed to linear operations , That is, the vector obtained by linear operation of the vector in space is still in space .
• Linear operation is ： Add and multiply , The linear combination of vectors can be obtained through linear operation .
• All vector spaces must contain zero vectors , Because any vector number multiplies 0 Or add the inverse vector to get the zero vector , Because vector space is closed to linear operations , So zero vector must belong to vector space .

Subspace Subspace

• A vector space contained in a vector space is called a subspace of the original vector space .
• Must contain zero vector , Because vector space is closed to linear operations .
• Pay attention to distinguish between “ Subspace ” and “ A subset of ”. all Elements that are all in the original space can be called subsets , But only a subset that is closed to linear operations can become a subspace .
• Any subspace S and T The intersection of is a subspace , Can pass S and T Itself is closed to linear combination to prove .

Column space Column Space

• Given matrix A, Its column vector belongs to $R^3$ Space , These column vectors and their linear combinations form $R^3$ A subspace in space , Matrix A Column space of C(A).
• Ex: A = [ [1, 3], [2, 3], [4, 1] ], be A The column space of is $R^3$ In the space , Including vectors $[1,2,4{]}^T$ and $[3,3,1{]}^T$, And pass through the plane of the origin , Space contains all linear combinations of two vectors .
• $Ax=b$, Only b In the matrix A Column space of C(A) In time ,x That's why ; If there is a solution , Just explain b Can be A Is represented by a linear combination of column vectors of .
• The rank of a matrix r = The number of columns of matrix principal elements = The dimension of the column space

Zero space Null space

• matrix A Zero space of N(A) It refers to satisfaction Ax=0 The set of all solutions of .
• For the given matrix A( Upper figure ), Its column vector contains 4 Weight , So column space is space $R^4$ The subspace of ,x To contain 3 A vector of components , So the matrix A The zero space of is $R^3$ The subspace of . about m*n matrix , The column space is $R^m$ The subspace of , Zero space is $R^n$ Subspace of space .
• In this example, the matrix A Zero space of N(A) Include for $[1,1,-1{]}^T$ Set of any multiples of , Because it's easy to see the first column of vectors (1) And the second column vector (1) Add and subtract the third column of vectors (-1) zero . This zero space is $R^3$ A straight line in .
• In order to verify Ax=0 The solution set of is a vector space , We can test whether it is closed to linear operations . if v and w For the elements in the solution set , Then there are : $A(v+w)=Av+Aw=0+0=0,A(cv)=cAv=0$, Therefore, it is proved that N(A) Is, indeed, $R^n$ A subspace of space .
• If the equation becomes the following form , Then its solution set cannot form a subspace . The zero vector is not in this set . Solution set is space $R^3$ Neiguo $[1,0,0{]}^T,[0,-1,1{]}^T$ A plane of , But it doesn't cross the origin .
• The dimension of a null space = Number of free columns = n-r

Be careful Column Space and Null space The difference between

• For column spaces , It is a space formed by linear combination of column vectors .
• Zero space starts from equations , By making x Subspace obtained by satisfying specific conditions .

Linearly independent independent

• if $c_1{x}_1+c_2{x}_2+......+c_n{x}_n=0$ Only in $c_1=c_2=......=c_n=0$ It was founded when , It's called a vector $x_1,x_2\cdots x_n$ yes Linearly independent Of . If these vectors form a matrix as column vectors A, Then the equation $Ax=0$ There is only zero solution x=0, Or matrix A The zero space of has only zero vectors .
• In other words , If there is a non-zero vector $c$, bring $Ac=0$, Then this matrix $A$ The column vector Linear correlation .
• matrix A by $m\ast n$ matrix , among $m<n$ ( therefore $Ax=b$ in There are more unknowns than equations ). be A There is at least one free variable in , that $Ax=0$ There must be a non-zero solution .A The column vectors of can be linearly combined to get the zero vector , therefore A The column vector of is linearly related .
• If matrix A The column vector of is linearly independent , be A All columns are primary columns （pivot column）, There is no free column (free column), The rank of the matrix is n. if A The column vector of is linearly correlated , Then the rank of the matrix is less than n, And there are free columns .

Zhang Cheng space Spanning a space

• When a space is composed of vectors $v_1,v_2,\dots v_k$ When composed of all linear combinations of , We call These vectors form this space . For example, the column vector of a matrix becomes the column space of the matrix .
• If the vector $v_1,v_2,\dots v_k$ Zhang Cheng space $S$, be $S$ Is the smallest space containing these vectors .

Basis and dimension Basis &Dimension

• The basis of vector space is a set of vectors with the following two properties $v_1,v_2,\dots v_k$:
  • $v_1,v_2,\dots v_k$ Linearly independent
  • $v_1,v_2,\dots v_k$ Zhang Cheng the vector space
• The basis of space tells us all the information of space .
• If the $R^n$ In the space n A matrix composed of column vectors is a reversible matrix , Then these vectors can form $R^n$ A set of bases in space .
• Every set of bases in a space has the same number of vectors , This number is the dimension of space (dimension). therefore $R^n$ Each set of bases of space contains n Vector .

Rank Rank

• “ Rank ”： That's ok ( Column ) vector The number of vectors of the largest linear independent vector group in
• Elementary row transformation does not change the linear correlation of column vectors

Orthogonal vector Orthonormal vectors

A vector that satisfies the following conditions $q_1,q_2,\dots q_n$ by Orthonormal :

$$q_i^T{q}_j=0,ifi\ne j$$
$$q_i^T{q}_j=1,ifi=j$$