D Linear Algebra: Determinants, Inverses, Rank
... If the determinant |A| of a n × n square matrix A ≡ An is zero, then the matrix is said to be singular. This means that at least one row and one column are linearly dependent on the others. If this row and column are removed, we are left with another matrix, say An−1 , to which we can apply the same ...
... If the determinant |A| of a n × n square matrix A ≡ An is zero, then the matrix is said to be singular. This means that at least one row and one column are linearly dependent on the others. If this row and column are removed, we are left with another matrix, say An−1 , to which we can apply the same ...
LINEAR DEPENDENCE AND RANK
... Another theorem of linear algebra provides a way to evaluate whether a consistent set of equations has an infinite number of solutions: a consistent system of n equations in n unknowns has a unique solution if the rank of the coefficient matrix is equal to its order, where r(A) = n. In the example h ...
... Another theorem of linear algebra provides a way to evaluate whether a consistent set of equations has an infinite number of solutions: a consistent system of n equations in n unknowns has a unique solution if the rank of the coefficient matrix is equal to its order, where r(A) = n. In the example h ...
Math 480 Notes on Orthogonality The word orthogonal is a synonym
... We now consider in detail the question of why every subspace of Rn has a basis. Theorem 3. If S is a subspace of Rn , then S has a basis containing at most n elements. Equivalently, dim(S) 6 n. Proof. First, recall that every set of n + 1 (or more) vectors in Rn is linearly dependent, since they for ...
... We now consider in detail the question of why every subspace of Rn has a basis. Theorem 3. If S is a subspace of Rn , then S has a basis containing at most n elements. Equivalently, dim(S) 6 n. Proof. First, recall that every set of n + 1 (or more) vectors in Rn is linearly dependent, since they for ...
APPM 2360 17 October, 2013 Worksheet #7 1. Consider the space
... fails to satisfy second condition of subspace definition. Therefore W is not subspace. (b) YES Let u = ax3 + bx2 + cx + d ∈ W and v = a1 x3 + b1 x2 + c1 x + d1 ∈ W then i. u + v = (a + a1 )x3 + (b + b1 )x2 + (c + c1 )x + (d + d1 ) ∈ W ii. ∀λ ∈ R we get λu = λax3 + λbx2 + λcx + λd ∈ W Thus W is subsp ...
... fails to satisfy second condition of subspace definition. Therefore W is not subspace. (b) YES Let u = ax3 + bx2 + cx + d ∈ W and v = a1 x3 + b1 x2 + c1 x + d1 ∈ W then i. u + v = (a + a1 )x3 + (b + b1 )x2 + (c + c1 )x + (d + d1 ) ∈ W ii. ∀λ ∈ R we get λu = λax3 + λbx2 + λcx + λd ∈ W Thus W is subsp ...
