14-240/Tutorial-October28

DBN: Classes: 2014-15: 14-240 / Navigation Welcome to Math 240! (additions to this web site no longer count towards good deed points) # Week of... Notes and Links 1 Sep 8 About This Class, What is this class about? (PDF, HTML), Monday, Wednesday 2 Sep 15 HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf 3 Sep 22 HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf 4 Sep 29 HW3, Wednesday, Tutorial, HW3_solutions.pdf 5 Oct 6 HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf 6 Oct 13 No Monday class (Thanksgiving), Wednesday, Tutorial 7 Oct 20 HW5, Term Test at tutorials on Tuesday, Wednesday 8 Oct 27 HW6, Monday, Why LinAlg?, Wednesday, Tutorial 9 Nov 3 Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial 10 Nov 10 HW8, Monday, Tutorial 11 Nov 17 Monday-Tuesday is UofT November break 12 Nov 24 HW9 13 Dec 1 Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial F Dec 8 The Final Exam Register of Good Deeds Add your name / see who's in!

Boris

Be Efficient

By this point in the course, we become good at solving systems of linear equations. However, we should not use this same old problem-solving strategy over and over if a more efficient one exists. Consider the following problems:

Q1: Determine if [math]\displaystyle{ S = \{(1, 4, -6), (1, 5, 8), (2, 1, 1), (0, 1, 0)\} }[/math] is linearly independent in [math]\displaystyle{ R^3 }[/math].

We can solve this linear equation to find the answer:

[math]\displaystyle{ c_1(1, 4, -6) + c_2(1, 5, 8) + c_3 (2, 1, 1) + c_4(0, 1, 0) = (0, 0, 0) }[/math] where [math]\displaystyle{ c_i \in R }[/math].

Yet there is a less time-consuming approach that relies on two observations:

(1) The dimension of [math]\displaystyle{ R^3 }[/math] is [math]\displaystyle{ 3 }[/math] so the size of a basis is also [math]\displaystyle{ 3 }[/math].

(2) No linearly independent set can have more vectors than a generating set (by the Replacement Theorem).

Since a basis is a generating set and the size of [math]\displaystyle{ S }[/math] is [math]\displaystyle{ 4 }[/math], then the Replacement Theorem tells us that [math]\displaystyle{ S }[/math] cannot be linearly independent. Hence, the problem can be solved without solving any linear equations.

Q2: Determine if the polynomials [math]\displaystyle{ x^3 - 2x ^2 + 1, 4x^2-x+3, 3x-2 }[/math] generate [math]\displaystyle{ P_3(R) }[/math].

Once again, we can solve a linear equation but we do not have to. Observe:

(1) The dimension [math]\displaystyle{ P_3(R) }[/math] is [math]\displaystyle{ 4 }[/math] so the size of a basis is also [math]\displaystyle{ 4 }[/math].

(2) No generating set can have fewer vectors than a basis (by a Corollary to the Replacement Theorem).

Since there are only [math]\displaystyle{ 3 }[/math] polynomials, then the Corollary tells us that it cannot generate [math]\displaystyle{ P_3(R) }[/math]. Once again, we used a more efficient strategy.

Extending a Linearly Independent Set to a Basis

Boris's tip (for concrete sets and vector spaces only):

If a problem requires us to extend a linearly independent set to a basis, then the easiest approach is to add vectors from the standard ordered basis. Here is an example:

Let [math]\displaystyle{ S = \{(-3, -6, 0), (0, 7, 0)\} }[/math] be a linearly independent subset of [math]\displaystyle{ R^3 }[/math]. To extend [math]\displaystyle{ S }[/math] to a basis, add vectors from [math]\displaystyle{ \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\} }[/math]. The only question is which vector(s) should we add?

We see that both vectors in [math]\displaystyle{ S }[/math] have a [math]\displaystyle{ 0 }[/math] as the third component so a safe choice is to add [math]\displaystyle{ (0, 0, 1) }[/math]. Since [math]\displaystyle{ R^3 }[/math] has a dimension of [math]\displaystyle{ 3 }[/math], then [math]\displaystyle{ \{(-3, -6, 0), (0, 7, 0), (0, 0, 1)\} }[/math] is a basis of [math]\displaystyle{ R^3 }[/math].