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 $S=\{(1,4,-6),(1,5,8),(2,1,1),(0,1,0)\}$ is linearly independent in $R^{3}$.

We can solve this linear equation to find the answer:

$c_{1}(1,4,-6)+c_{2}(1,5,8)+c_{3}(2,1,1)+c_{4}(0,1,0)=(0,0,0)$ where $c_{i}\in R$.

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

(1) The dimension of $R^{3}$ is $3$ so the size of a basis is also $3$.

(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 $S$ is $4$, then the Replacement Theorem tells us that $S$ cannot be linearly independent. Hence, the problem can be solved without solving any linear equations.

Q2: Determine if the polynomials $x^{3}-2x^{2}+1,4x^{2}-x+3,3x-2$ generate $P_{3}(R)$.

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

(1) The dimension $P_{3}(R)$ is $4$ so the size of a basis is also $4$.

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

Since there are only $3$ polynomials, then the Corollary tells us that it cannot generate $P_{3}(R)$. 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 $S=\{(-3,-6,0),(0,7,0)\}$ be a linearly independent subset of $R^{3}$. To extend $S$ to a basis, add vectors from $\{(1,0,0),(0,1,0),(0,0,1)\}$. The only question is which vector(s) should we add?

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