## Podcast Beta

## Questions and Answers

If the columns of A are dependent, then Ax=0 has a nonzero solution.

True

If the rows of A are dependent, then Ax=0 has a nonzero solution.

False

Every set of six vectors in R4 is linearly dependent.

True

Suppose Ax=b has two different solutions for some b, then the columns of A are dependent.

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For a square matrix A, the map T(x) = Ax is invertible if and only if det(A) â‰ 0.

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Let T:R8->R8 be a linear map. If T is injective, then T is onto.

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Let T:R8->R4 be a linear map. If T is injective, then T is onto.

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Let T:R4->R8 be a linear map. If T is injective, then T is onto.

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If v1, v2, v3 in Rn are independent, then v2 and v3 are also independent.

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For a square matrix A, A^(-1) exists if and only if Ax=0 has no nonzero solution x.

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If f: R3->R is a linear map such that f(1,0,0) = 2 and f(0,1,0) = 1, then f(1,-2,0) = 0.

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If f: R3->R is a linear map such that f(1,0,0) = 2 and f(0,1,0) = 1, then f(0,0,-2) = 0.

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For square matrices, if B+XA = CA and det(A) â‰ 0, then X = (C-B)A^(-1).

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For square matrices, det(ABA^(-1)C) = det(B)det(C) where A^(-1) exists.

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If Ax=0 has a unique solution, then Ay=b cannot have two solutions.

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A linear map from Rn to Rn defined by a matrix C is a bijection if and only if det(C) â‰ 0.

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Det((P^(-1)AP)^3110) = (det(A))^3110.

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Suppose (v1,...,vk) = (u1,...,uk)P. Then, for every invertible k x k matrix P, v1,...,vk form a basis for a vector space V if and only if u1,...,uk form a basis for V.

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The kernel of an arbitrary linear map is a linear subspace.

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Every linearly independent set of a vector space V can be expanded (by adding more vectors) to a basis for V.

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If W = span{v1,...,vm} and the vectors in W are independent, then dim(W) >= m.

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There exists a vector space that has two bases with different numbers of vectors.

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The columns of an invertible n x n matrix A always form a basis for Rn.

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For an m x n matrix M of rank r, its column span has dimension n-r.

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For an m x n matrix M of rank r, its null space has dimension n-r.

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## Study Notes

### Linear Algebra Key Concepts

- Column dependence in matrix A implies Ax=0 has nonzero solutions.
- Row dependence in matrix A does not guarantee nonzero solutions for Ax=0.
- In R4, any set of six vectors is guaranteed to be linearly dependent.
- Existence of two distinct solutions for Ax=b indicates column dependence of matrix A.
- A linear transformation T(x) = Ax is invertible if the determinant of A is non-zero.
- Injective linear maps from R8 to R8 are also onto due to equal dimensions.
- Injective linear maps from R8 to R4 can be onto; dimension differences allow for this.
- Injective linear maps from R4 to R8 do not guarantee onto status, as dimension exceeds.
- Independence among vectors v1, v2, and v3 in Rn extends to pairs v2 and v3 being independent as well.
- Existence of an inverse for square matrix A requires Ax=0 to only have the trivial solution.
- A defined linear functional f retains linearity, with f(1,-2,0) yielding zero based on scaling of f(1,0,0) and f(0,1,0).
- The statement concerning X in matrices involving B and C is false when determining X=(C-B)A^(-1).
- For square matrices, the determinant relationship holds: det(ABA^(-1)C) = det(B)det(C) given A is invertible.
- A unique solution for Ax=0 leads to the conclusion that Ay=b can also only have a single solution.
- A linear transformation defined by matrix C is a bijection if the determinant of C is non-zero.
- The determinant of combined transformed matrices follows the relationship det((P^(-1)AP)^k) = (det(A))^k, preserving determinant properties.
- Basis formations of vectors via invertible matrices remain valid; (v1,...,vk) and (u1,...,uk) are both bases if one set forms a basis.
- The kernel of a linear map is always a linear subspace.
- Any linearly independent set in a vector space can be augmented to create a full basis for that space.
- For a linear span of independent vectors in subspace W, its dimension is at least the number of vectors m.
- It is impossible for a vector space to have two bases with differing cardinalities.
- Invertible n x n matrix columns form a basis for Rn due to spanning properties.
- For an m x n matrix M of rank r, the statement about column span dimension is incorrect; it instead has dimension r.
- The null space of an m x n matrix M of rank r has dimension n-r, capturing the essence of free variables and solutions.

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## Description

Prepare for your final exam in Linear Algebra with these practice flashcards. Each card tests your understanding of key concepts such as linear dependence and solutions to Ax=0. Use these effective study tools to reinforce your knowledge.