14-240/Tutorial-October28

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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.

Nikita