linear algebra final

If {v1, v2, v3} is a linearly independent set in R5 , then Span{v1, v2, v3} has dimension 3.

T → the dimension of a subspace depends if a set of vectors are spanning and independent

Every spanning set of a vector space is a basis.

F → there also must be linear independency

If A is a square matrix and det(A) = 0, then the columns of A are linearly dependent.

T → det(A) = 0 only when matrix is singular, which means dependent

If A and B are row equivalent matrices, then they have the same eigenvalues.

F → matrix row operations affect the eigenvaluees

If {v1, v2} is an orthogonal set of nonzero vectors, then {v1, v2} is linearly independent.

T → orthogonal nonzero vectors are always linearly independent

If u · v = 0, then one of the vectors must be the zero vector.

F → (1,0) * (0,1) = 0, yet none of them are zero

The distance between two vectors u and v is ∥u + v∥.

F → it is a minus sign

If W is a subspace of R n , then (W⊥) ⊥ = W.

T → Complement of complement returns subspace

Every subspace of R 3 has dimension 0, 1, 2, or 3.

T → subspace cannot exceed value 3

If A is the standard matrix of a linear transformation T, then Ker(T) = Null(A)

T → With standard matrix A, T(x) = Ax. so ker(T) {x: Ax = 0} = Null(A)

If det(A) doesnt = 0, then the linear transformation T(x) = Ax is one-to-one and onto.

T → if det not 0, then it is invertible. invertible matrixes are one-to-one and onto

(l) Every square matrix has at least one real eigenvalue.

F → Only complex eigenvalues for square matrices

A real matrix with a non-real eigenvalue must also have its complex conjugate as an eigenvalue.

T → if λ=a+bi  an eigenvalue, then  λ‾=a−bi is also an eigenvalue.

Eigenvectors corresponding to distinct eigenvalues are linearly independent.

T → they scale at different rates and never cancel out to 0

If 0 is an eigenvalue of A, then A is invertible.

F → Det is 0, so not invertible

(p) If A = PDP −1 and D is diagonal, then the columns of P are eigenvectors of A.

T → The equation A = PDP^{-1} can be rewritten as \(AP = PD\).

(q) A matrix can have repeated eigenvalues and still be diagonalizable.

T → A matrix is diagonalizable if for each eigenvalue, geometric multiplicity = algebraic multiplicity.

If λ is an eigenvalue of A, then λ^2 is an eigenvalue of A^2 .

T → See google doc Pic

F → Testing

If r≤2, then: nullity(A)=4−r ≥ 4−2=2

nullity(A)=4−r≥4−2=2

If T : V → W is a linear transformation of finite dimensional vector spaces and rank(T) = 3 then ker(T) has a basis with three elements.

F → ker(T) also depends on dim(V), not just rank

If T : R 7 → R 3 is a surjective linear transformation, then nullity(T) = 3

7-3 = 4, not 3