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An equation that can be written as a1x1 + a2x2 + ... = b; a1, a2, etc. are real or complex numbers known in advance
1. All nonzero rows are above any all zero rows; 2. Each leading entry is in a column to the right of the previous leading entry; 3. All entries below a leading entry in its column are zeros
Reduced Echelon Form
Same as echelon form, except all leading entries are 1; each leading 1 is the only non-zero entry in its row; there is only one unique reduced echelon form for every matrix
the collection of all vectors in R^n that can be written as c1v1 + c2v2 + ... (where c1, c2, etc. are constants)
Ax = b
1. For each b in R^n, Ax = b has a solution; 2. Each b is a linear combination of A; 3. The columns of A span R^n; 4. A has a pivot position in each row
A position in the original matrix that corresponds to a leading 1 in a reduced echelon matrix
If only the trivial solution exists for a linear equation; the columns of A are independent if only the trivial solution exists
If non-zero weights that satisfy the equation exist; if there are more vectors than there are entries
Matrix multiplication warnings
1. AB != BA ; 2. If AB = AC, B does not necessarily equal C; 3. If AB = 0, it cannot be concluded that either A or B is equal to 0
Properties of transposition
1. (A^T)^T = A; 2. (A+B)^T = A^T + B^T; 3. (rA)^T = rA^T; 4. (AB)^T = B^TA^T
1. If A is invertible, (A^-1)^-1 = A; 2. (AB)^-1 = B^-1 * A^-1; 3. (A^T)^-1 = (A^-1)^T
Invertible Matrix Theorem (either all of them are true or all are false)
A is invertible; A is row equivalent to I; A has n pivot columns; Ax = 0 has only the trivial solution; The columns of A for a linearly independent set; The transformation x --> Ax is one to one; Ax = b has at least one solution for each b in R^n; The columns of A span R^n; x --> Ax maps R^n onto each R^m; there is an n x n matrix C such that CA = I; there is a matrix such that AD = I; A^T is invertible; The columns of A form a basis of R^n; Col A = R^n; dim Col A = n; rank A = n; Nul A = ; dim Nul A = 0
1. Ly = b; Ux = y; 2. Reduce A to echelon form; 3. Place values in L that, by the same steps, would reduce it to I
1. The zero vector is in H; 2. For u and v in H, u + v is also in H; 3. For u in H, cu is also in H (c is a constant)
A linearly independent set in H that spans H; the pivot columns of A form a basis for A's column space
A transformation that assigns a vector y in R^m for each x in R^n; there's a pivot in every column
a matrix product u^Tv or u . v where u and v are vectors; if U . V = 0, u and v are orthogonal
1. x is in W' if x is perpendicular to every vector that spans W; 2. W' is a subspace of R^n
A set of vectors where Ui . Uj = 0 (and i != j); if S is an orthogonal set, S is linearly independent and a basis of the subspace spanned by S
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