# 14-240/Tutorial-October28

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

We can solve this linear equation to find the answer:

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

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

(1) The dimension of ${\displaystyle R^{3}}$ is ${\displaystyle 3}$ so the size of a basis is also ${\displaystyle 3}$.
(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 ${\displaystyle S}$ is ${\displaystyle 4}$, then the Replacement Theorem tells us that ${\displaystyle S}$ cannot be linearly independent. Hence, the problem can be solved without solving any linear equations.

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

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

(1) The dimension ${\displaystyle P_{3}(R)}$ is ${\displaystyle 4}$ so the size of a basis is also ${\displaystyle 4}$.
(2) No generating set can have fewer vectors than a basis (by a Corollary to the Replacement Theorem).

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

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