Which collection is a vector space whose elements are not arrows?
- All polynomials of degree at most 3, with the usual addition and scaling
- The set {1, 2, 3} with ordinary arithmetic
- All 2 by 2 invertible matrices
- All arrows in the plane
Which collection is a vector space whose elements are not arrows?
- The set {1, 2, 3} with ordinary arithmetic
- All polynomials of degree at most 3
- All 2 by 2 invertible matrices
Which of these subsets of the plane is a subspace?
- The x-axis, points of the form (x, 0)
- The line y equals x plus 1
- The first quadrant, both coordinates non negative
Sam says the solutions of A times x equals b, with b nonzero, also form a subspace. Is Sam right?
Circle one: True False
Why do the solutions of a homogeneous system always form a subspace?
- Every homogeneous system has exactly one solution
- Zero is the only solution of any system
- Linearity keeps sums and scalings of solutions as solutions
Why do all 2 by 2 symmetric matrices form a vector space, while the invertible 2 by 2 matrices do not?
- Sums and scalings of symmetric matrices stay symmetric, but sums of invertible matrices can be singular
- Symmetric matrices are square and invertible ones are not
- Invertible matrices are too large to form a space
- Symmetric matrices are arrows
Which of these subsets of the plane is a subspace?
- The x-axis, all points of the form (x, 0)
- The line y equal to x plus 1
- The first quadrant, where both coordinates are non negative
- The unit circle
Why do all 2 by 2 symmetric matrices form a vector space, while the invertible ones do not?
- Symmetric matrices are arrows and invertible ones are not
- Invertible matrices are too large to form a space
- Sums and scalings of symmetric matrices stay symmetric, but sums of invertible ones can be singular
Zoe claims the first quadrant is a subspace because it holds zero and is closed under addition. What is her mistake?
- It fails scaling: minus 1 times (1, 1) leaves it
- It is not closed under addition
- It misses the zero vector
Sam says the polynomials of degree exactly 3 form a vector space. Is Sam right?
Circle one: True False