Lecture 3 - Proofs in Linear Algebra

We'll do some 'baby proofs' of some common Linear Algebra findings. But first, some examples of vector spaces:

Example

$F [x]$ is the set of all polynomials with coefficients in $F$ . In fact, we claim $F [x]$ is a vector space over $F$ under the usual addition and scaling of polynomials. For instance:

In $C [x]$ , an example is $p (x) = 1 - i x + (2 - 3 i) x^{3} \in C [x]$ . Another polynomial is $q (x) = 2 + i x - x^{2} \in C [x]$ .
We can add $p + q$ to get $(p + q) (x) = 3 - x^{2} + (2 - 3 i) x^{3} \in C [x]$ .
We can scale $q$ by $i$ to get: $(i q) (x) = 2 i - x - i x^{2} \in C [x]$ .
The zero polynomial would be $\vec{0} (x) = 0 + 0 x + 0 x^{2} + . . .$

Example

$P_{n} (F)$ is the set of all polynomials of degree $n$ or less, with coefficients from $F$ . We have the usual polynomial addition and scalar multiplication as defined in the example prior.

$p (x) = 1 + x - x^{3} \in P_{4} (R)$
$q (x) = 2 - x^{2} - x^{4} \in P_{4} (R)$
$(p + q) (x) = 3 + x - x^{2} - x^{3} - x^{4} \in P_{4} (R)$

Note that for Example 2 above, you need to have the condition of " $n$ or less" since, if it was strictly $n$ (say $n = 2$ for our case), you could take say $x^{2} \in P_{2} (R)$ and $- x^{2} \in P_{2} R$ but $x^{2} - x^{2} = 0 \notin P_{2} (R)$ .

Also note that we can create a isomorphism from $P_{4} (R)$ to $R^{5}$ by:

Matching $x_{1}$ to our constant coefficient
Matching $x_{2}$ to our $x$ coefficient
...

So, if we consider $p (x), q (x)$ in Example 2 above, we can turn them into vectors in $R^{5}$ , and then add them together:

\vec{p} = (1, 1, 0, - 1, 0), \vec{q} = (2, 0, - 1, 0, - 1)

\vec{p + q} = (3, 1, - 1, - 1, - 1) \equiv 3 + x - x^{2} - x^{3} - x^{4} = (p + q) (x)

So $P_{4} (R)$ behaves like $R^{5}$ .

Proofs in Linear Algebra

Note that $0 \vec{v} = \vec{0}$ . This isn't part of our definition, so we can't say it's a fact yet. But we need to show that this comes as a consequence of our definition. First, a note on the cancellation law, which says that:

Cancellation Law

Suppose $\vec{x}, \vec{y}, \vec{z} \in V$ where $V$ is a vector space. Then:

\vec{x} + \vec{y} = \vec{x} + \vec{z} \to \vec{y} = \vec{z}

This also isn't from our definitions, so let's prove it:

Proof
Suppose $\vec{x} + \vec{y} = \vec{x} + \vec{z}$ . Then, we know the additive inverse for $\vec{x}$ exists, which we call $\vec{- x} \in V$ . Then, notice the following:

\begin{aligned} \vec{x} + \vec{y} & = \vec{x} + \vec{z} \\ \vec{- x} + \vec{x} + \vec{y} & = \vec{- x} + \vec{x} + \vec{z} & (Add \vec{- x} to both sides) \\ \vec{0} + \vec{y} & = \vec{0} + \vec{z} & (Definition of Additive Inverse) \\ \vec{y} & = \vec{z} & (Definition of \vec{0}) \end{aligned}

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Note that, from now on, we can just say:

\vec{x} - \vec{y} = \vec{x} + (- \vec{y})

(ie: you can you the $-$ sign on it's own now. Yay!)

Now we prove $0 \vec{v} = \vec{0}$ :
Proof

\begin{aligned} 0 \vec{v} & = (0 + 0) \vec{v} \\ = 0 \vec{v} + 0 \vec{v} \end{aligned}

So then you can replace $0 \vec{v} = \vec{0} + 0 \vec{v}$ to show that:

\begin{aligned} \vec{0} + 0 \vec{v} & = 0 \vec{v} + 0 \vec{v} \\ \vec{0} & = 0 \vec{v} & (Subtract from both sides) \end{aligned}

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Let's do another proof:

Theorem

$α \vec{0} = \vec{0}$ .

Proof
Note that:

\begin{aligned} α \vec{0} & = α (\vec{0} + \vec{0}) \\ = α \vec{0} + α \vec{0} \end{aligned}

So since $α \vec{0} = \vec{0} + α \vec{0}$ , then we can subtract from both sides and get $α \vec{0} = \vec{0}$ .
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Theorem

$\vec{- v} = (- 1) \vec{v}$ . We note the $\vec{- v}$ is the additive inverse of $\vec{v}$ .

Proof
$(- 1) \vec{v}$ is indeed the additive inverse if, when we add $\vec{v}$ , we get the zero vector. Notice:

\begin{aligned} \vec{v} + (- 1) \vec{v} & = 1 \vec{v} + (- 1) \vec{v} \\ = (1 + (- 1)) \vec{v} \\ = 0 \vec{v} \\ = \vec{0} \end{aligned}

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