Linear Algebra - Lecture 14 - Midterm exam solutions

Overview

The lecture works through 11 linear algebra midterm problems, from vector and matrix arithmetic to systems of equations, linear independence, spans, and subspaces. The central theorem is that a homogeneous system with fewer equations than variables has a nonzero solution; it supports the result that any N vectors in a space generated by M vectors are dependent when M < N.

Key takeaways

A system with fewer homogeneous equations than variables has a nonzero solution; this follows by eliminating a variable and reducing to fewer equations in the remaining variables.
If M vectors generate a space, any collection of N vectors in that space must be linearly dependent when M < N, because their combination coefficients satisfy M homogeneous equations in N unknowns.
Row reduction exposes free variables and gives parametric descriptions of solution sets, such as the affine line obtained from the three-equation system.
In R², two linearly independent vectors automatically form a basis, while a set containing only scalar multiples of one nonzero vector spans a one-dimensional subspace.
A vector outside a subspace extends any linearly independent set within that subspace: a dependence involving the outside vector would force it to belong to the subspace.
The intersection of any collection of subspaces is a subspace because zero belongs to every member and the intersection preserves closure under addition and scalar multiplication.

Chapters

0:00 Vector Combinations and Matrix Arithmetic

Computes 3(-2, 4, 0) + 5(0, 1, 6) + 2(-5, 0, 3) = (4, 17, 24) by scaling and adding coordinates.
Adds matrices entry by entry and checks matrix-product dimensions before multiplying a 2 × 3 matrix by a 3 × 2 matrix.
The product of a 2 × 3 matrix and a 3 × 2 matrix has dimensions 2 × 2.

4:12 An Affine Line from a Linear Equation

Rearranges 91x − 7y = 42 to obtain y = 13x − 6.
Writes the solution set as (x, y) = (0, −6) + x(1, 13), an affine line in R².
Distinguishes the translated solution set from its one-dimensional direction subspace, spanned by (1, 13).

6:29 Solving a Homogeneous System in Three Variables

Solves x − 3y + z = 0 and 2x + 3y − 13z = 0 by substitution, finding 12y = 5x and z = x/4.
Expresses the solution space as all scalar multiples of (12, 5, 3).
Checks that (12, 5, 3) satisfies both equations: 12 − 15 + 3 = 0 and 24 + 15 − 39 = 0.
Uses row operations as an alternative method and reaches the same one-dimensional solution space.

16:31 Testing Independence and Finding Basis Coordinates in R²

Tests vectors f₁ = (2, 1) and f₂ = (3, 2) by solving 2x + 3y = 0 and x + 2y = 0; only x = y = 0 works.
Concludes that the two independent vectors form a basis for the two-dimensional space R².
Finds e₁ = (1, 0) = 2f₁ − f₂ by solving the corresponding two-equation system.

21:31 Row-Reducing a Matrix and Describing Its Solution Line

Reduces the augmented matrix with rows (1, 1, 2, 1), (2, −1, 3, −3), and (4, 1, 7, −1) to reduced row echelon form.
Obtains rows (1, 0, 5/3, −2/3), (0, 1, 1/3, 5/3), and (0, 0, 0, 0), leaving z free.
The resulting equations give x = −2/3 − (5/3)z and y = 5/3 − (1/3)z, an affine line of solutions.

26:59 Proving a Transformed Pair Remains Independent

Assumes v₁ and v₂ are linearly independent and considers 4v₁ + 7v₂ alongside v₁ + 2v₂.
A zero combination of the new vectors implies 4x + y = 0 and 7x + 2y = 0 by independence of v₁ and v₂.
Solving those equations gives x = y = 0, proving the transformed pair is linearly independent.

29:53 Determining Spans and Dimensions in R²

For S₁ = {(1, 1), (1, 2), (2, 1)}, the first two vectors are independent and span R², so S₁ spans a two-dimensional space.
For S₂ = {(1, 1), (2, 2), (0, 0)}, the latter vectors add no new direction because (2, 2) = 2(1, 1).
The span of S₂ is the one-dimensional line generated by (1, 1), with {(1, 1)} as a basis.

32:14 Why Fewer Homogeneous Equations Than Variables Have Nonzero Solutions

Shows the two-equation, three-variable case by separating whether the coefficients of x₁ in both equations are zero.
When a₁₁ ≠ 0, solves the first equation for x₁ and substitutes into the second, leaving one homogeneous equation in x₂ and x₃.
A homogeneous equation in two variables has a nonzero solution, which extends to a nonzero solution (x₁, x₂, x₃).
Identifies this as the special case of the general theorem for m homogeneous equations in n variables when m < n.

38:27 Fewer Generators Force Dependence Among More Vectors

If a subspace is generated by M vectors, writes each of N vectors in that space as a linear combination of the M generators.
A proposed dependence among the N vectors reduces to M homogeneous equations in N coefficients.
When M < N, the fewer-equations theorem supplies a nonzero coefficient vector, proving the N vectors are linearly dependent.

44:44 Intersections of Subspaces and Adding a Vector Outside a Subspace

Proves an intersection of subspaces contains zero and is closed under vector addition and scalar multiplication, so it is a subspace.
For independent w₁ and w₂ in W and v outside W, considers xw₁ + yw₂ + zv = 0.
If z = 0, independence of w₁ and w₂ forces x = y = 0; if z ≠ 0, the equation implies v is a linear combination of vectors in W.
That second case contradicts v ∉ W, so {w₁, w₂, v} is linearly independent.

Transcript excerpt

0:20 Okay. So, in this lecture I'm just going to go through the um solutions to the problems on the midterm exam. Uh-huh. Start with number one. Um We want to compute the linear combination of these three vectors. So, if we multiply this vector by three, we have -6 12 0. Multiply this vector by five,

1:10 we get 0 5 30. Multiply this vector by two, we get -10 0 6. And when you add it up, -6 -10 is Oops. See. -6 Oh, minus minus 10. So, that's -6 + 10 is 4. 12 + 5 is 15. And 30 -6 is 24. And 12 and 5 I should say is 17. I'm not 15.

1:59 So, 4 17 24 is that. We have the matrices A and B. 2 0 4 7 8 minus 15 plus 1 3 4 5 1 0. When you add them, you get 3 3 8. Oops. Um 3 3 8 and 12.

2:53 To make this clearer. This is 5 1 0. So, 12 minus 7 15. And when we multiply B times C B is 1 3 4 5 1 0. C is 1 0 minus 2 0 2 9. So, when you multiply, you get 1 minus 8 is minus 7. So, this is 2 by 3, this is 3 by 2. So, the product will be a 2 by 2 matrix. 0

3:45 6 36 is 42. 5 plus 0 plus 0 is 5. 2 plus 0 is 2. So, it's just this 2 by 2 matrix. Okay. Problem 2, >> [clears throat] >> A Oh. Say this was A and this was B. To A, we have the equation 91x - 7y = 42,

4:35 which is the same as 7y = 91x - 42. Or, you can divide by 7. Let me write it, actually, slightly

This is the start of the transcript. Save this video free for the whole transcript, searchable and time-coded.

Related notes

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Nathanson math lectures.