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

Avoid the Einstellung Effect

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 $R^{3}$.