MATH 3350 T/F Practice Q's
A subspace S ≠ {0} can have a finite number of vectors.
F
An eigenvalue λ must be nonzero, but an eigenvalue u can be equal to the zero vector.
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Cramer's rule can be used to find the solution to any system that has the same number of equations as unknowns.
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Every matrix A has a determinant.
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If A and B are 2 x 2 matrices, then det(A - B) = det(A) - det(B).
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If A and B are invertible n × n matrices, then the inverse of A+B is A⁻¹+B⁻¹.
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If A and B are n x n matrices, then det(A + B) = det(A) + det(B).
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If A is a 3 x 3 matrix and det(A) = 0, then rank(A) = 0.
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If A is a 3 x 3 matrix, then adj(2A) = 2adj(A).
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If A is a 5 × 3 matrix, then null(A) forms a subspace of R⁵.
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If A is a diagonal matrix, then all of the minors of A are also diagonal.
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If A is a matrix with linearly independent columns, then Ax = 0 has nontrivial solutions.
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If A is a matrix with linearly independent columns, then Ax = b has a solution for all b.
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If A is a matrix with more columns than rows, then the columns of A are linearly independent.
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If A is a matrix with more rows than columns, then the columns of A are linearly independent.
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If A is a square matrix that has all positive entries, then so does adj(A).
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If A is a square matrix, then row(A) = col(A).
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If A is an invertible n × n matrix, then the number of solutions to Ax = b depends on the vector b in Rⁿ.
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If A is an n x n matrix with all entries equal to 1, then det(A) = n.
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If A is an n x n matrix with all positive entries, then det(A) > 0.
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If Ax = b is a consistent linear system, then b is in row(A).
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If E is an elementary matrix, then det(E) = 1.
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If S = span{u₁, u₂, u₃}, then dim(S) = 3.
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If S₁ and S₂ are subsets of Rⁿ but not subspaces, then the intersection of S₁ and S₂ cannot be a subspace of Rⁿ.
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If S₁ and S₂ are subsets of Rⁿ but not subspaces, then the union of S₁ and S₂ cannot be a subspace of Rⁿ.
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If a linear system has four equations and seven variables, then it must have infinitely many solutions.
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If a linear system has seven equations and four variables, then it must be inconsistent.
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If a linear system has the same number of equations and variables, then it must have a unique solution.
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If a set of vectors U is linearly independent in a subspace S, then vectors can be removed from U to create a basis for S.
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If a set of vectors U spans a subspace S, then vectors can be added to U to create a basis for S.
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If a set of vectors in Rⁿ is linearly dependent, then the set must span Rⁿ.
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If a set of vectors includes 0, then is cannot span Rⁿ.
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If an n × n matrix A is singular, then the columns of A must be linearly independent.
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If the cofactors of an n x n matrix are all nonzero, then det(A) ≠ 0.
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If the columns of an n × n matrix A span Rⁿ, then A is singular.
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If u is a nonzero eigenvector of A, then u and Au point in the same direction.
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If u₄ is a linear combination of {u₁, u₂, u₃}, then span{u₁, u₂, u₃, u₄} ≠ span{u₁, u₂, u₃}.
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If u₄ is a linear combination of {u₁, u₂, u₃}, then {u₁, u₂, u₃, u₄} is linearly independent.
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If u₄ is not a linear combination of {u₁, u₂, u₃}, then span{u₁, u₂, u₃, u₄} = span{u₁, u₂, u₃}.
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If u₄ is not a linear combination of {u₁, u₂, u₃}, then {u₁, u₂, u₃, u₄} is linearly dependent.
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If u₄ is not a linear combination of {u₁, u₂, u₃}, then {u₁, u₂, u₃, u₄} is linearly independent.
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If {u1, u2, u3} does not span R³, then neither does {u₁, u₂, u₃, u₄}.
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If {u₁, u₂, u₃, u₄} is linearly dependent, then so is {u₁, u₂, u₃}.
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If {u₁, u₂, u₃, u₄} spans R³, then so does {u₁, u₂, u₃}.
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If {u₁, u₂, u₃} is linearly independent, then so is {u₁, u₂, u₃, u₄}.
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If λ₁ and λ₂ are eigenvalues of a matrix, then so is λ₁ + λ₂.
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Interchanging the rows of a matrix have no effect on its determinant.
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Let T : R² → R⁷ be a linear transformation. Then range(T) is a subspace of R².
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Let T : R⁵ → R⁸ be a linear transformation. Then ker(T) is a subspace of R⁸.
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Let m > n. Then U = {u₁, u₂,...,um} in Rⁿ can form a basis for Rⁿ if the correct m - n vectors are removed from U.
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Let m > n. Then U = {u₁, u₂,...,um} in Rⁿ can form a basis for Rⁿ if the correct n - m vectors are added to U.
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Rⁿ has exactly one subspace of dimension m for each of m = 0,1,2,...,n.
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Suppose A is a matrix with n rows and m columns. If m < n, then the columns of A span Rⁿ.
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Suppose A is a matrix with n rows and m columns. If n < m, then the columns of A span Rⁿ.
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Suppose that A is a 9 x 5 matrix, and that T(x) = Ax is a linear transformation. Then T can be onto.
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The set of integers forms a subspace of R.
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The set {0} forms a basis for the zero subspace.
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The union of two subspaces of Rⁿ forms another subspace of Rⁿ.
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Three nonzero vectors that lie in a plane in R³ might form a basis for R³.
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A must be a square matrix to be invertible.
T
Different sequences of row operations can lead to different echelon forms for the same matrix.
T
If 0 is an eigenvalue of A, then nullity(A) > 0.
T
If 0 is the only eigenvalue of A, then A must be the zero matrix.
T
If A and B are invertible n × n matrices, then the inverse of AB is B⁻¹A⁻¹.
T
If A is a 4 x 13 matrix, then the nullity of A could be equal to 5.
T
If A is a 4 x 4 matrix and det(A) = 4, then nullity(A) = 0.
T
If A is a 4 × 7 matrix, then null(A) forms a subspace of R⁷.
T
If A is a diagonal matrix, then the eigenvalues of A lie along the diagonal.
T
If A is a matrix with columns that span Rⁿ, then Ax = b has a solution for all b in Rⁿ.
T
If A is a matrix, then the dimension of the row space of A is equal to the dimension of the column space of A.
T
If A is a square matrix with integer entries, then so is adj(A).
T
If A is a square matrix, then (adj(A))^T = adj(A^T).
T
If A is an n x n matrix with det(A) = 1, then A⁻¹ = adj(A).
T
If A is an n × n matrix and b ≠ 0 is in Rⁿ, then the solutions to Ax = b do not form a subspace.
T
If A is invertible, then (A⁻¹)⁻¹=A.
T
If S₁ is a subspace of dimension 3 in R⁴, then there cannot exist a subspace S₂ of R⁴ such that S₁ ⊂ S₂ ⊂ R⁴ but S₁ ≠ S₂ ≠ R⁴.
T
If a set of vectors U is linearly independent in a subspace S, then vectors can be added to U to create a basis for S.
T
If a set of vectors U spans a subspace S, then vectors can be removed from U to create a basis for S.
T
If an n x n matrix A has n distinct eigenvectors, then A is diagonalizable.
T
If det(A) ≠ 0, then the columns of A are linearly independent.
T
If each eigenspace of A has dimension equal to the multiplicity of the associated eigenvalue, then A is diagonalizable.
T
If m < n, then a set of m vectors cannot span Rⁿ.
T
If m > n, then a set of m vectors in Rⁿ is linearly dependent.
T
If u₄ is a linear combination of {u₁, u₂, u₃}, then span{u₁, u₂, u₃, u₄} = span{u₁, u₂, u₃}.
T
If u₄ is a linear combination of {u₁, u₂, u₃}, then {u₁, u₂, u₃, u₄} is linearly dependent.
T
If u₄ is not a linear combination of {u₁, u₂, u₃}, then span{u₁, u₂, u₃, u₄} ≠ span{u₁, u₂, u₃}.
T
If {u₁, u₂, u₃, u₄} does not span R³, then neither does {u₁, u₂, u₃}.
T
If {u₁, u₂, u₃, u₄} is linearly independent, then so is {u₁, u₂, u₃}.
T
If {u₁, u₂, u₃} is a basis for R³, then span{u₁, u₂} is a plane.
T
If {u₁, u₂, u₃} is linearly dependent, then so is {u₁, u₂, u₃, u₄}.
T
Let A, B, and S be n x n matrices, and S be invertible. If B = S⁻¹AS, then det(A) = det(B).
T
Let S₁ and S₂ be subspaces of Rⁿ, and define S to be the set of all vectors of the form s₁ + s₂, where S₁ is in s₂ is in S₂. Then S is a subspace of Rⁿ.
T
Let S₁ and S₂ be subspaces of Rⁿ, and define S to be the set of all vectors of the form s₁ - s₂, where s₁ is in S₁ and s₂ is in S₂. Then S is a subspace of Rⁿ.
T
Let T : R³ → R⁹ be a linear transformation. Then range(T) is a subspace of R⁹.
T
Let T : R⁶ → R³ be a linear transformation. Then ker(T) is a subspace of R⁶.
T
Suppose that A is a 4 x 4 matrix, and that B is a matrix obtained by multiplying the third column of A by 2. Then det(B) = 2det(A).
T
Suppose that A is a 9 x 5 matrix, and that T(x) = Ax is a linear transformation. Then T can be one-to-one.
T
The dimension of an eigenvalue is always less than or equal to the multiplicity of the associated eigenvalue.
T
The intersection of two subspaces of Rⁿ forms another subspace of Rⁿ.
T
The rank of a matrix A can not exceed the number of rows of A.
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