av Chang Yeh-Yung för 11 årar sedan
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The orthogonal decomposition theorem
Let W be a subspace of R^n. Then each y ∈R^n can be written uniquely in the form y = y-hat + z where y-hat ∈W and z ∈W⊥ In fact, if {u1,...,up} is any orthogonal basis of W, then y-hat = (y·u1/u1·u1)u1+(y·u2/u2·u2)u2+...+(y·up/up·up)up and z = y - y-hat
The vector y-hat ∈ the Orthogonal decomposition theorem is called the orthogonal projection of y onto W and often is written as proj_[w]_y
The length of z, ||z||, is the distance between y and the vector space W.
Given a nonzero vector u ∈ R^n, we can decompose any vector y in R^n as y = y-hat + z where y-hat=αu for some scalar α and z is some vector orthogonal to u. The vector y-hat is called the orthogonal projection of y onto u, and the vector z is called the component of y orthogonal to u.
it can be shown that y-hat = (y·u/u·u)u
An orthogonal basis for a subspace W of R^n is a basis for W that is also an orthogonal set
Let {u1,...,up} be an orthogonal basis for a subspace W of R^n. For each y in W, the weights in the linear combination y=c1u1+...+cpup are given by cj=y·uj/uj·uj
We can apply the Gram-Schmidt process to produce an orthogonal basis for any nonzero subspace or R^n
A set of vectors {u1,...,up} ∈ R^n is said to be an orthogonal set if each pair of distinct vectors from the set is orthogonal, that is, ui·uj=0 when i is not equal to j
W=Span{[1,0]} W⊥=Span{[0,1]}
||u+v||^2=||u||^2+||v||^2
u·v=0
u^T * v
A vector whose length is 1 is called a unit vector. Given a vector v, the new vector u=v / ||v|| is a unit vector in the same direction as v
||av||=a||v||
u·v=(u^T)(v)
The Spectral Theorem for Symmetric Matrices
An nxn symmetric matrix has the following properties
A is orthogonally diagonalizable
The eigenspaces are mutually orthogonal
The dimension of the eigenspace for each eigenvalueλ equals the multiplicity of λ as a root of the characteristic equation
A has n real eigenvalues, counting multiplicities
A matrix A is said to be orthogonally diagonalizable if there are an othogonal matrix P (with P^-1 = P^T) and a Diagonalizable matrix D such that A = PDP^T=PDP^-1
An orthogonal matrix is a square matrix U such that U^-1 = U^T. Clearly the set of all column vectors from an orthogonal matrix forms an orthogonal set of unit vectors.
An nxn matrix A is orthogonally diagonalizable if and only if A isa symmetric matrix
If A is symmetric, then any two eigenvectors from different eigenspaces are orthogonal
Let A be an nxn matrix.
Find corresponding eigenvectors for eigenvalues of A by solving (A - λI)x=0
The dimension of the eigenspace is equal to the number of free variables in (A -λI)x=0
If the dimension of the eigenspace for each eigenvalue equals the multiplicity of that eigenvalue, then A is Diagonalizable
A=PDP^-1, where P is formed by the Eigenvectors and D is formed by the Eigenvalues corresponding the the eigenvectors
If the dimension of the eigenspace for each eigenvalue is not equal to the multiplicity of that eigenvalue, then A is NOT diagonalizable
All eigenvaectors and the zero vector from an eigenspace for each eigenvalue
Solve det(A - λI)=0 for λ to find all eigenvalues.
Check the algebraic multiplicity
Diagonalization Theorem
An nxn matrix is diagonalizable if and only if the sum of the dimensions of the distinct eigenspaces equals n, and this happens if and only if the dimension of the eigenspace for each eigenvalue equals the multiplicity of that eigenvalue
An nxn matrix with n distinct eigenvalues is diagonalizable
A=PDP^-1, with D a diagonal matrix, if and only if the columns of P are n Linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond , respectively to the eigenvectors in P.
A square matrix is said to be diagonalizable if A is similar to a diagonal matrix, that is, if A=PDP^-1 for some invertible matrix P and some Diagonal matrix D.
An nxn matrix A is diagonalizable if and only if A has n Linearly independent eigenvectors.
If v1, ..., vr are eigenvectors that correspond to distinct eigenvalues λ1, ..., λr of an nxn matrix A, then the set {v1,...,vr} is linearly independent
Let λ be an eigenvalue of a square matrix A. Then the set consisting of the zero vector and all the eigenvectors corresponding to λ iscalled the eigenspace of A corresponding to λ. Note that the eigenspace of A corresponding to λ is Nul (A - λI)
0 is an eigenvalue of a square matrix A if and only if A is not invertible
A scalar λ is an eigenvalue of an nxn matrix A if and only if λ satisfies the characteristic equation det(A - λI)=0
An eigenvector must be nonzero, but an eigenvalue may be zero
Let V be a p-dimensional vector space, p at least 1. Any Linearly independent set of exactly p elements in V is automatically a basis for V. Any set of exactly p elements that spans V is automatically a basis for V.
The mapping x |-> [x]B is the coordinate mapping (determined by B)
Let B = {b1, ..., bn} be a basis for a vector space V. Then the coordinate mapping x |-> [x]B is a one-to-one linear transformation from V onto R^n
Let B = {b1, ..., bn} be a basis for a vector space V. Then for each x ∈ V, there exists a unique set of scalars c1, ..., cn such that x=c1b1+...+cnbn
S is also a Linearly independent set that is as large as possible because adding any vector to S results in a linearly dependent set
Span
S is a spanning set that is as small as possible because deleting any vector from S results in a non-spanning set of V
vp is a Linear combination of v1,v2,...,v(p-1) if [ v1 v2 ... v(p-1)] x = vp
Span {v1,...,vp}=Span{v1...v(p-1)}
Let S = {v1, ..., vp} be a set in V, and let H = Span {v1, ..., vp}
If one of the vectors in S--say vk--is a linear combination of the remaining vectors in S, then the set formed from S by removing vk still spans H
If H is not {0} (the null set), some subset of S is a basis for H.
the subspace spanned by B coincides with H
H = Span {b1, ..., bp}
B is a Linearly independent set
the set of all u ∈ V such that T(u)=0.
T(cu)=cT(u) for all u ∈ V and scalars C
T(u+v) = T(u) + T(v) for all u,v ∈ V
Rank A + dimension Nul A = n
Number of Basic Variables in A
The number of pivot positions in matrix A
The Dimension of Col A
dim Col A
number of basic variables in the matrix equation Ax=0
Col A = Span {a1, ..., an}
Dimension of Nul A
Number of Free Variables in the matrix equation Ax=0
Nul A = { x : x is ∈ R^n and Ax=0}
The set of all solutions to a system Ax=0 of m homogenous Linear equations in n unknowns is a subspace of R^n
To check if a vector u is in the null space of A, simply solve the equation Au (not a matrix equation, but matrix A times a vector u). If the answer is 0, then u is in Nul A
Related to Eigenspace
H is closed under multiplication by scalars.
for each u ∈ H and each scalar c, the vector cu is in H
H is closed under vector addition
for each u and v ∈ H, the sum u+v is in H
The zero vector of V is in H
A and B are nxn matrices: det AB = (det A)(det B)
if A is an nxn matrix, then det A^T = det A
A square matrix A is invertible if and only if det A is not 0
A is a square matrix and B is the resultant matrix from one row operation
Scalar Multiplication
det(B) = k^(n) det A
interchange
det(B) = -det(A)
Replacement
det(B)=det(A)
Shortcut
If A is a triangular matrix, then det A is the product of the entries on its main diagonal
The determinant of an nxn matrix can be computed by cofactor expansion across any row or down any column
det A = a11C11+a12C12+...a1nC1n, where Cij = (-1)^(i+j) det Aij
Row Reduce [ U y ] to find x
Place entries in L such that the same sequence of row operations reduces L to I.
A Linear Transformation T : R^n -> R^n is said to be invertible if there exists a transformation S : R^n -> R^n such that S(T(x))=x for all x in R^n AND T(S(x))=x for all x in R^n
Let T : R^n -> R^n be a Linear transformation and let A be the standard matrix for T. Then T is invertible if and only if A is an invertible matrix. In that case, the linear transformation S given by S(x)=(A^-1)x is the unique function satisfying the previous conditions
S is the inverse transformation of T, written T^-1
Let A and B be nxn matrices. If AB=I, then A and B are both invertible with A=B^-1 and B=A^-1
A is an invertible, NxN matrix
Let C and D be NxN Matrices
CA=I; AD=I
C=D
A is row equivalent to I
A^T is invertible
(A^-1)^T is its inverse
Columns of A form a Basis for R^n
Rank A = n
Nul A = {0}
dim Nul A = 0
Col A = R^n
dim Col A = n
For all B ∈ R^N, Ax=b has the unique solution x=(A^-1)b
x |-> AX is ONE-TO-ONE
The Columns of A are Linearly Independent
The columns of A do not contain the Zero Vector
det A is not 0
0 is NOT an eigenvalue of A
At Most One Solution
x |-> Ax maps R^N ONTO R^N
A has N pivot columns
The Columns of A Span R^N
b is a Linear combination of the columns of A
N Pivot Positions
At least ONE solution
Row reduce the augmented matrix [ A I ]. If A is row equivalent to I, then [ A I ] is row equivalent to [ I A^-1]. Otherwise, A does not have an inverse.
The Product of invertible Matrices is Invertible and the inverse is the product of their inverses in the reverse order
(A^T)^-1 = (A^-1)^T
If A and B are nxn invertible matrices, then so is AB, and the inverse of AB is the product of the inverses of A and B in the reverse order. That is, (AB)^-1 = (B^-1)(A^-1)
If A is an invertible matrix, then A^-1 is invertible and (A^-1)^-1 = A
If A is an invertible nxn matrix, then for each b in R^n, the equation Ax=b has the unique solution x = (A^-1)b
A^-1 = (1/ad-bc) [d -b, -c a]
ad-bc = det A, which is the Determinant of A
Let A = [a b, c d]. If ad-bc is not equal to 0, then A is invertible
Non-singular Matrix
invertible Matrix
Singular Matrix
NOT invertible
(AB)^T = (B^T)(A^T)
for any scalar r, (rA)^T = rA^T
(A+B)^T = A^T + B^T
(A^T)^T=A
A is mxn, A^T is nxm
If AB=0, you cannot conclude that A=0 or B=0
If B=C, AB is not always equal to AC
AB is not always equal to BA
Assuming that T : R^n -> R^m is a linear transformation, there exists a unique matrix A, such that T(x)=Ax for all x in R^n
A is the mxn matrix whose jth column is the vector T(ej)
The Standard matrix for the Linear transformation T
A = [ T(e1) ... T(ej) ]
Identity Matrix
ej is the jth column of the identity matrix ∈ R^n for 1
has 1's on the diagonal and zeros elsewhere
nxn matrix denoted I
ONE-TO-ONE
The columns of a matrix A are Linearly independent
T(x)=0 has ONLY the trivial solution
A mapping T : R^n -> R^m is said to be one-to-one if each b in R^m is the image of at MOST one x in R^n
T(x1) is not equivalent to T(x2) when x1 is not equivalent to x2
ONTO
Columns of a matrix A span R^m
A mapping T : R^n -> R^m is said to be onto R^m if each b in R^m is the image of at LEAST one x in R^n
All Matrix transformations are Linear transformations
T(0)=0
T(cu+dv)=cT(u)+dT(v) for all vectors u,v and scalars c,d
T(cu)=cT(v) for all u and scalars c
T(u+v)=T(u)+T(v) for all u,v ∈ the domain of T
For x ∈ R^n, the vector T(x) in R^m is called the image of x (under the action of T). The set of all images T(x) is called the range of T
The set R^m
The CODOMAIN of T
The set R^n
DOMAIN of T
n > m
Set contains the zero vector
there exists at least one vector in S that is a linear combination of the others
Scalar multiples
A has a pivot position in every row
The columns of A span R^m
Each b ∈ R^m is a linear combination of the columns of A
For each b ∈ R^m, the equation Ax=b has a solution
If a Linear system is consistent then the solution set contains either
infinitely many solutions
at least one free variable
a unique solution
no free variables
For a given equation Ax=0, a nontrivial solution is a nonzero vector x that satisfies Ax=0
Happens when there is at least one free variable
Parametric Vector Form
The zero solution of Ax=0 is called the trivial solution.
A Linear system is homogenous if it can be written in the form Ax=0, where A is an mxn matrix and 0 is the zero vector in R^m.
The collection of all vectors that can be written in the form c1v1+c2v2+...+cpvp with c1, ..., cp scalars
Given the vectors v1, v2, v3, ..., vp ∈ R^n and given scalars c1,c2, ..., cp in R, the vector y defined by y=c1v1+c2v2+...+cpvp is called a linear combination of v1, v2, ..., vp with weights c1, c2, ..., cp
Variable not corresponding to a pivot column
Variable corresponding to a pivot column
Column of A containing a pivot position
a location in A that corresponds to a leading 1 in the reduced echelon form of A
Reduced echelon Form
Unique Row Reduced Matrix for each Eschelon matrix
Each leading 1 is the only nonzero entry in its column
The leading entry of each nonzero row is one
each leading entry of a row is in a column to the right of the leading entry of the row above it
all entries in a column below a leading entry are zeros
All nonzero rows are above any rows of all zeros
One or more solution