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The null space of an m x n matrix A is a subspace of R^n. Equivalently, the set of all solutions to a system Ax = 0 of m homogeneous linear equations in n unknowns is a subspace of R^n. |
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The pivot columns of a matrix A form a basis for the column space of A. |
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Theorem 2.14 (Rank Theorem) |
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If a matrix A has n columns, then rankA + dim NulA = n. |
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Theorem 2.15 (The Basis Theorem) |
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Let H be a p-dimensional subspace of R^n. Any linearly independent set of exactly p elements in H is automatically a basis for H. Also, any set of p elements of H that spans H is automatically a basis for H. |
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The eigenvalues of a triangular matrix are the entries on its main diagonal. |
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If v1,...,vr are eigenvectors that correspond to distinct eigenvalues &1,...,&r of an n x n matrix A, then the set {v1,...,vr} is linearly independent. |
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If n x n matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities). |
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Theorem 5.5 (The Diagonalization Theorem) |
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An n x n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. In fact, 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 digaonal entries of D are eigenvalues of A that correspond, repectively, to the eigenvectors of P. |
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An n x n matrix with n distinct eigenvalues is diagonalizable. |
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Let A be a real 2 x 2 matrix with a complex eigenvalue & = a-bi (b=/0) and an associated eigenvector v in C^2. Then, A = PCP^(-1) where P = [Re v Im v] and C = {a, -b, b, a}. |
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Theorem 6.2 (Pythagorean Theorem) |
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Two vectors u and v are orthogonal if and only if ||u + v||^2 = ||u||^2 + ||v||^2. |
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Let A be an m x n matrix. The orthogonal complement of the row space of A is the nullspace of A and the orthogonal complement of the column space of A is the nullspace of A^(perp): (Row A)^(perp) = NulA and (Col A_^(perp) = Nul A^(perp). |
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If S = {u1,...,up} is an orthogonal set of nonzero vectors in R^n, then S is linearly independent and hence is a basis for the subspace spanned by S. |
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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 dot uj)/(uj dot uj) (j = 1,...,p). |
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An m x n matrix U has orthonormal columns if an donly if (U^T)U = 1 |
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Le U be an m x n matrix with orthonormal columns, and let x and y be in R^n. Then: (a) ||Ux|| = ||x|| (b) (Ux) dot (Uy) = x dot y (c) (Ux) dot (Uy) = 0 if and only if x dot y = 0. |
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