Please enable JavaScript.

Coggle requires JavaScript to display documents.

MATH, 1, 2, Minors and cofactors, If A is an n x n matrix, then the number…

- - - - Notes Row Reduction
      - Youtube Row Reduction
  - - - Triangular matrix
      - For 2x2 and 3x3 matrix only
- - - - Using Row Operations to Find A-1
        
        Inversion Algorithm:
        To find the inverse of an invertible matrix A, find a sequence of elementary row operations that reduces A to the identity and then perform that same sequence of operations on In
        to obtain A-1.
        
        Solution of a Linear System using A-1
- - - - Definition 1
        
        u + (v+w) = (u + v ) + w
        
        u + v = v+ u
        
        If u and v are objects in V , then u + v is in V
        
        if k is any scalar and u is any object in V, then ku is in V.
        
        there exists an object in v, called the zero vector, that is denoted by 0 and has the property that 0 + u = u +0 for all u in v .
        
        7.1u= u
        
        For each u in V , there is an object -u in V called a negative of u ,such that u + (-u) = (-u) + u = 0
        
        k(u + v ) = ku + kv
        
        (k + m)u = ku + mu
        
        k(mu) = (km)(u)
      - steps to show that a Set with Two Operations is a vector space
        
        Step 1 : Identify the set V of objects that will become vectors.
        
        Step 2 : Identify the addition and scalar multiplication operation on V
        
        step 3 : Verify Axions 1 and 6 : that is adding two vectors in v produces a Vector in V , and multiplying a vector in V by a scalar by a scalar also produces a vector in V .
        
        Step 4 : Confirm that Axions 2,3,4,5,7,8,9 and 10.
      - Some Properties of vectors
        
        0u + 0u = (0 +0)u [axion] 8 = 0
        
        Let V be a vector space, u a vector in V, and k a scalar ; then
        (1) 0u = 0
        (2) k0 = 0
        (3) (-1)u = 0
        (4) if ku = 0 , then k = 0 or u = 0
        
        By Axion 5 the vector 0u has a negative , -0u , adding this negative to both side above yields
        [ 0u + 0u ] + (-0) = 0u + ( -0u)
        or
        0u+ [0u +(-0u) = 0u +(-0u) [ axion 3]
        0u + 0 =0 [axion 5 ]
        0u = 0 [ axion 4 ]
        
        to prove that (-1)u = -u we must show that u + (-1)u = 0 , the proof is as follows :
        u + (-1)u = 1u +(-1)u [ Axion 10 ]
        = (1+ (-1)u [ Axion 8 ]
        = 0u [ property of number ]
        = 0 [ Part(a) of this theorem ]
  - - - If W1,W2...., Wr are subspaces of a vector space V , then the intersection of these subspaces is also a subspace of V .
      - Let W be the intersection of the subspaces W1,W2,.....Wr. This set is not empty because each of these subspaces contains the zero vector of V, and hence so does their intersection. Thus, it remains to show that W is closed under addition and scalar multiplication.
      - To prove closure under addition, let u and v be vectors in W. Since W is the intersection of W1,W2,...,Wr, it follows that u and v also lie in each of these subspaces. Moreover, since these subspaces are closed under addition and scalar multiplication , they also all contain the vectors u + v and ku for every scalar k, and hence so does their intersection W. this proves that W is closed under addition and scalar multiplication.
    - - The solution set of a homogenous system Ax = 0 of m equations in n unknowns is a subspaces of Rn.
      - Let W be the solution set of the system . The set W is not empty because it contains at least the trivial solution x = 0 .
      - it follows from these equations and the distributive property of matrix multiplication that
        A(x1 + x2) = Ax1 + Ax2 = 0+0=0
        so W is closed under addition. Similarly, if k is any scalar then
        A ( kx1)= kAx1 = k0=0
        so W is also closed under scalar multiplication
    - - If A is an m x n matrix , then the kernel of the matrix transformation TA : Rn Rm is a subspace of Rn .
    - - If A is an m x n matrix , then the kernel of the matrix transformation TA : Rn Rm is a subspaces of Rn.
  - - - (a) A set with finitely many vectors that contain 0 is linearly dependent.
        (b) A set with exactly two vectors is linearly independent if and only if neither vector is a scalar multiple of the other.
      - For any vector v1,v2,...,vr, the set S ={ v1,v2} is linearly dependent since the equation
        0v1 + 0v2 =0
        expresses 0 as a linear combination of the vectors in S with coefficients that are not all zero.
      - A nonempty set S = {v1,v2 } in a vector space V is linearly independent if and only if the coefficients satisfying the vector equation
        k1v1 + k2v2 + ....+ = 0
        are k1 = 0 , k2 =0 ,...,kr = 0
      - If S = {v1,v2 } is a of two or more vectors in a vector space V, then S is said to be a linearly independent is said to be linearly dependent. If S has only one vector , we will agree that it is linearly independent if and only if that vector is nonzero .
    - - if f1 = f1(x) , f2=f2(x) are functions that are n -1 times differentiable on the interval (-∞,∞),then the determinant
        
        is called the Wronskian of F1,F2
      - if the function f1,f2 .., fn have n-1 continuous derivatives on the interval (-∞,∞) , and if the Wronskian of these functions is not identically zero on (-∞,∞), then these function form a linearly independant set of vectors in C(n-1)(-∞,∞)
  - - - Uniqueness of Basis Representation If S = {v1,v2...vn} is a basis for a vector spaces v , then every vector v in V can be expressed in the form v = c1v1 +c2v2...cnvn in exactly one way .
      - If S = {v1,v2,..Vn} is an ordered basis for a vector space V,
        V= c1v1 +c2v2+....+cnvn
        is the expression for a vector v in term of the basis S, then the scalar c1,C2, are called the coordinates of v relative to the basis S . the vector (c1,c2,..,cn) in Rn constructed is called the coordinate vector of v relative to S ; it is denoted by (v)s=(c1,c2,cn)
    - - If S = {v1,V2 } is a set of vectors in a finite-dimensional vector space V, then S is called a basis for V if
        (a) S span V
        (B) S is linearly independent.
- - - - If A is an n × n matrix, the following statements are equivalent
      - • λ is an eigenvalue of A
      - • The system of equations (λI-A)x=0 has nontrivial solutions
      - • There is a nonzero vector x such that Ax= λx
      - • λ is a solution of the characteristic equation det(λI-A)=0
  - - - If B is similar to A, then A is similar to B, since we can express B as B=Q-1AQ by taking Q=P-1. We usually say that A and B are similar matrices. If B=P-1AP, then A and B have the same determinant det(B)=det(P-1AP)=det(P-1)det(A)det(P) = (1/det(P)) det(A)det(P) = det(A) .Any property that is shared by all similar matrices is called a similarity invariant or is said to be invariant under similarity.
        
        A square matrix A is said to be diagonalizable if it is similar to some diagonal matrix; that is, if there exists an invertible matrix P such that P-1AP is diagonal. In this case the matrix P is said to diagonalize A. Theorem 5.2.1: If A is an n × n matrix, the following statements are equivalent. (a) A is diagonalizable (b) A has n linearly independent eigenvectors.
      - The preceding theorem guarantees that an n x n matrix A with n linearly independent eigenvectors is diagonalizable, and the proof provides the following method for diagonalizing A. Step 1. Find n linear independent eigenvectors of A, say, p1, p2, …, pn. Step 2. From the matrix P having p1, p2, …, pn as its column vectors. Step 3. The matrix P-1AP will then be diagonal with λ1, λ 2, …, λ n as its successive diagonal entries, where i is the eigenvalue corresponding to pi, for i = 1, 2, …, n. If there is total of n such vectors, then A is diagonalizable. Otherwise, A is not diagonalizable.
        
        If a matrix has all distinct eigenvalues, then it is diagonalizable. Matrices with repeated eigenvalues might be nondiagonalizable.
        for example has repeated eigenvalues (1) but is diagonalizable since any nonzero vector in R3 is an eigenvector of I3, and so, we can find three linearly independent eigenvectors.
  - - - Except for the use of complex scalars, the notions of linear combination, linear independence, subspace, spanning, basis, and dimension carry over without change to Cn. If A is an n x n matrix with complex entries, then the complex roots of the characteristic equation are called complex eigenvalues of A. is a complex eigenvalue of A iff there exists a nonzero vector x in Cn such that . x is called a complex eigenvector of A corresponding .
        
        The reduced row echelon form must have a row of zeros
        
        because it has nontrivial solutions.
        
         Each row must be a scalar multiple of the other, and
        
        hence the first row can be made into a row of zeros by
        
        adding a suitable multiple of the second row to it.
        
         Accordingly, we can simply set the entries in the first row
        
        to zero, then interchange the rows, and then multiply the
        
        new first row by to obtain the reduced row echelon
        
        form
        
         A general solution of the system is
- - - - Theorem 8.1.1
        If T : V -> W is a linear transformation, then:
        
        T(-v) = -T(v) for all v in V
        
        T(u-v ) = T(u) - T(v) for all u and v in V
        
        T(0) = 0
      - Theorem 8.1.2
        Let T : V -> W be a linear transformation, where V is finite V dimensional. If S = {v1,v2,...., vn} is a basis for V, then the image of any vector v in V can be expressed as
        
        T(v) = c1T(v1) + c2T(v2) + ... + cnT(vn)
        where c1,c2,,,,cn are the coefficients required to express v as a linear combination of the vectors in S.
    - - Theorem 8.1.3
        If T : V -> W is a linear transformation, then :
        a) The kernel of T is a subspace of V
        b) The range of T is a subspace of W
      - Theorem 8.1.4 Dimension Theorem for Linear Transformations
        If T : V -> W is a linear transformation from an n-dimensional vector space V to a vector space W, then
        Rank (T) + Nullity(T) = n
- - - - properties of symmetric matrices
        (a) The eigenvalues of A are all real numbers.
        (b) Eigenvectors from different eigenspaces are orthogonal
      - Orthogonally Diagonalizing an n × n Symmetric Matrix
        Step 1. Find a basis for each eigenspace of A.
        Step 2. Apply the Gram–Schmidt process to each of these bases to obtain an orthonormal basis for each eigenspace.
        Step 3. Form the matrix P whose columns are the vectors constructed in Step 2. This matrix will orthogonally diagonalize A, and the eigenvalues on the diagonal of D = PTAP will be in the same order as their corresponding eigenvectors in P.