MATH 323 Lecture 12

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Substitute lecture

Chapter 3.4: Basis and Dimension

Definitions

We say that vectors ${\vec {v}}_{1},\ldots ,{\vec {v}}_{k}$ are linearly independendent if $a_{1}\,{\vec {v}}_{1}+\dots +a_{k}\,{\vec {v}}_{k}={\vec {0}}$

We say that ${\vec {w}}_{1},\ldots {\vec {w}}_{k}\in V$ span $V$ if any ${\vec {u}}\in V$ can be written as a linear combination of ${\vec {w}}_{1},\ldots ,{\vec {w}}_{k}$.

Let $V$ be a vector space.

The set ${\vec {e}}_{1},\ldots ,{\vec {e}}_{n}\in V$ is called a basis in $V$ if

1. ${\vec {e}}_{1},\ldots ,{\vec {e}}_{n}$ are linearly independent
2. They span $V$

The number of vectors in the basis is called the dimension of $V$

Theorem

All bases in $V$ have the same number of elements called the dimension of $V$.

Note: it is possible for a "finitely defined" Vector space to have infinite dimension.

Theorem 3.4.1

If $\{v_{1},\ldots ,v_{n}\}$ is a spanning set of $V$, $m>n$, then any vectors $w_{1},\ldots ,w_{m}$ are linearly independent.

Proof. (substitution/elimination process).

If $w_{j}$, $1\leq j\leq m$ is a 0 vector, then there is nothing to prove:

$0{\vec {w}}_{1}+\dots +\alpha {\vec {w}}_{j}+\dots +0{\vec {w}}_{m}={\vec {0}}$

If ${\vec {w}}_{1}\neq {\vec {0}}$, then ${\vec {w}}_{1}=a_{1}\,{\vec {v}}_{1}+\dots +a_{n}\,{\vec {v}}_{n}$. At least one of the $a$'s is nonzero. Without loss of generality, we assume that it is $a_{1}$ which is nonzero.

Therefore ${\vec {v}}_{1}={\frac {{\vec {w}}_{1}-a_{2}{\vec {v}}_{2}-\dots -a_{n}\,{\vec {v}}_{n}}{a_{1}}}$, so $\{{\vec {w}}_{1},{\vec {v}}_{2},\ldots ,{\vec {v}}_{n}\}$ is a spanning set for $V$.

Repeat the process with the above spanning set and ${\vec {w}}_{2},\ldots ,{\vec {w}}_{n}$. When we can no longer continue, we end up with $\{{\vec {w}}_{1},\ldots ,{\vec {w}}_{n}\}$ as a spanning set for $V$, and $\{{\vec {w}}_{n+1}=k_{1}\,{\vec {w}}_{1}+\dots +k_{n}\,{\vec {w}}_{n}\}$

Example

$\mathbb {R} ^{2}$ has bases ${\vec {e}}_{1}=\left\langle 1,0\right\rangle$, ${\vec {e}}_{2}=\left\langle 0,1\right\rangle$ and ${\vec {v}}_{1}=\left\langle 1,1\right\rangle$, ${\vec {v}}_{2}=\left\langle 1,0\right\rangle$.

Let ${\vec {w}}_{1}=\left\langle 1,2\right\rangle$, ${\vec {w}}_{2}=\left\langle 2,3\right\rangle$, and ${\vec {w}}_{3}=\left\langle 3,4\right\rangle$

Note: $\{{\vec {e}}_{1},{\vec {e}}_{2}\}$ is a spanning set of $\mathbb {R} ^{2}$.

$w_{1}=1{\vec {e}}_{1}+2{\vec {e}}_{2}\quad \rightarrow \quad {\vec {e}}_{1}={\vec {w}}_{1}-2{\vec {e}}$

…

Corollary 3.4.2

If $\{v_{1},\ldots ,v_{n}\}$ and $\{w_{1},\ldots ,w_{m}\}$ are two bases in $V$, then $m=n$.

Proof. Assume $v$'s are spanning set and $m>n$, then from Thm. 3.4.1, we know that $w_{1},\ldots ,w_{m}$ are linearly dependent. Not a basis (contradiction)

Flip $n>m$ and by same logic → contradiction. Therefore $m=n$.

Infinite Dimension

If the vector space $V$ has no basis of finitely many vectors, we say that $V$ has infinite dimension.

If $V=\{{\vec {0}}\}$, we say that $\dim V=0$

Example

Find the dimension of the space of solutions to the system

${\begin{aligned}x+y+z&=0\\x-2y+3z&=0\end{aligned}}$

Three variables, two constraints: $3-2=1$

Note that the solutions form a vector space, and the set of solutions will be the nullspace of the coefficient matrix ${\begin{bmatrix}1&1&1\\1&-2&3\end{bmatrix}}$.

The solution is of the form $\alpha \left\langle -5,2,3\right\rangle$, so the basis has only one vector. Therefore $\dim N=1$

Example

Find the dimension of the nullspace for the operation $A=\left(x{\frac {\mathrm {d} }{\mathrm {d} x}}-1\right)$ over $P_{2}$

In other words, $A$ does $p(x)\mapsto x\,{\frac {\mathrm {d} p(x)}{\mathrm {d} x}}-p(x)$. For a polynomial of degree ≤ 1, $a+bx\mapsto x(b)-(a+bx)=-a$. If $A(a+bx)=0$, then $a=0$, and $b$ is anything. Therefore, $N(A)=\{bx~\mid ~b\in \mathbb {R} \}$, the basis is $\{x\}$, and $\dim N(A)=1$.

Example

$\mathbb {R} ^{2\times 2}$

A basis in $\mathbb {R} ^{2\times 2}$ would be:

${\begin{aligned}{\vec {e}}_{1}&={\begin{bmatrix}1&0\\0&0\end{bmatrix}}{\vec {e}}_{2}&={\begin{bmatrix}0&1\\0&0\end{bmatrix}}{\vec {e}}_{3}&={\begin{bmatrix}0&0\\1&0\end{bmatrix}}{\vec {e}}_{4}&={\begin{bmatrix}0&0\\0&1\end{bmatrix}}\end{aligned}}$

Theorem 3.4.3

Let $V$ be a vector space, and $\dim V=n>0$. Then

1. any $n$ linearly independent vectors $f_{1},\ldots ,f_{n}$ span $V$
2. any $n$ vectors that span $V$ are linearly independent.

Proof.

1. assume they do not span $V$: there is a vector ${\vec {g}}$ which is not a linear combination of $f$'s, then $\{f_{1},\ldots ,f_{n},g\}$ would be linearly independent. this contradicts #Theorem 3.4.1 since there are $n+1$ vectors with only $n$ in a basis.
2. assume they are linearly dependent: then $c_{1}\,f_{1}+\dots +c_{n}\,f_{n}={\vec {0}}$, where $c_{1},\ldots ,c_{n}$ are not all 0. This means that one vector could be written as a linear combination of other vectors, so $n-1$ vectors would span $V$. This contradicts the definition of a basis, stating that $n$ vectors must be linearly independent.

Theorem 3.4.4

$\dim V=n>0$

1. No set of fewer than $n$ vectors spans $V$
2. Any $m<n$ linearly independent vectors could be extended (by more vectors) to form a basis.
3. If $v_{1},\ldots ,v_{N}$, $N>n$, span $V$, we can pare them down to $n$ vectors, which would be a basis.