Basis and Dimension

Bases for Vector Spaces

Example: Show that $\left\{\left(1,2\right),\left(3,4\right)\right\}\subset {\mathrm{ℝ}}^2$ is a basis for ${\mathrm{ℝ}}^2$.

Let $ℬ=\left\{{\mathrm{𝐛}}_1,{\mathrm{𝐛}}_2\right\}$, where ${\mathrm{𝐛}}_1=\left(1,2\right)\in {\mathrm{ℝ}}^2$ and ${\mathrm{𝐛}}_2=\left(3,4\right)\in {\mathrm{ℝ}}^2$. We must show that $ℬ$ is a linearly independent set and a spanning set for ${\mathrm{ℝ}}^2$.

1. Linear independence. Suppose that ${𝑐}_1,{𝑐}_2\in \mathrm{ℝ}$ are scalars such that ${𝑐}_1{\mathrm{𝐛}}_1+{𝑐}_2{\mathrm{𝐛}}_2=\mathrm{𝟎}$. Then we have:

   $$\begin{array}{cc}{𝑐}_1\left(1,2\right)+{𝑐}_2\left(3,4\right)& =\left(0,0\right)\\ \left({𝑐}_1+3{𝑐}_2,2{𝑐}_1+4{𝑐}_2\right)& =\left(0,0\right).\end{array}$$

   This gives us the system of equations:

   $$\begin{array}{cc}{𝑐}_1+3{𝑐}_2& =0\\ 2{𝑐}_1+4{𝑐}_2& =0.\end{array}$$

   Using elementary algebra, we can solve for ${𝑐}_1$ and ${𝑐}_2$ to find that ${𝑐}_1=0$ and ${𝑐}_2=0$ is the only solution. By Theorem 16, $ℬ$ is a linearly independent set.

2. Spanning set. Let $\mathrm{𝐯}=\left(𝑥,𝑦\right)$ be an arbitrary vector in ${\mathrm{ℝ}}^2$. We will show that there exist scalars ${𝑐}_1,{𝑐}_2\in \mathrm{ℝ}$ such that $\mathrm{𝐯}={𝑐}_1{\mathrm{𝐛}}_1+{𝑐}_2{\mathrm{𝐛}}_2$, that is:

   $$\begin{array}{cc}\left(𝑥,𝑦\right)& ={𝑐}_1\left(1,2\right)+{𝑐}_2\left(3,4\right)\\ \left(𝑥,𝑦\right)& =\left({𝑐}_1+3{𝑐}_2,2{𝑐}_1+4{𝑐}_2\right).\end{array}$$

   This gives us the system of equations:

   $$\begin{array}{cc}{𝑐}_1+3{𝑐}_2& =𝑥\\ 2{𝑐}_1+4{𝑐}_2& =𝑦.\end{array}$$

   Using elementary algebra, we find that ${𝑐}_1=3𝑦/2-2𝑥$ and ${𝑐}_2=𝑥-𝑦/2$. Thus, by choosing ${𝑐}_1$ and ${𝑐}_2$ as such, we have $\mathrm{𝐯}={𝑐}_1{\mathrm{𝐛}}_1+{𝑐}_2{\mathrm{𝐛}}_2$. Since $\mathrm{𝐯}$ was an arbitrary vector in ${\mathrm{ℝ}}^2$, we have shown that every vector in ${\mathrm{ℝ}}^2$ can be expressed as a linear combination of the vectors in $ℬ$, so $ℬ$ is a spanning set for ${\mathrm{ℝ}}^2$.

Since $ℬ$ is a linearly independent set and a spanning set for ${\mathrm{ℝ}}^2$, $ℬ$ is a basis for ${\mathrm{ℝ}}^2$. ∎

Nonexample.

The set $\left\{\left(1,2\right),\left(2,4\right)\right\}\subset {\mathrm{ℝ}}^2$ is not a basis for ${\mathrm{ℝ}}^2$ since it is not linearly independent.

Nonexample.

The set $\left\{\left(1,2\right)\right\}\subset {\mathrm{ℝ}}^2$ is not a basis for ${\mathrm{ℝ}}^2$ since it is not a spanning set for ${\mathrm{ℝ}}^2$.

Combining the two criteria from Definition 19, we can deduce the following:

This theorem states that a basis $ℬ$ for a vector space $𝑉$ is a set of vectors from $𝑉$ such that every vector in $𝑉$ can be uniquely expressed as a linear combination of the vectors in $ℬ$.

This is a statement of uniqueness and existence:

• Uniqueness. Every vector in $𝑉$ can be expressed as a linear combination of the vectors in $ℬ$ in at most one way.
• Existence. Every vector in $𝑉$ can be expressed as a linear combination of the vectors in $ℬ$ in at least one way.

Proof: Unique Representation Theorem.

Suppose that $ℬ=\left\{{\mathrm{𝐛}}_1,{\mathrm{𝐛}}_2,\dots ,{\mathrm{𝐛}}_{𝑛}\right\}$ is a basis for $𝑉$ over $𝐹$. By the definition of a basis, $ℬ$ is a linearly independent set and a spanning set for $𝑉$. Particularly, $ℬ$ spans $𝑉$, so for any $\mathrm{𝐯}\in 𝑉$, there exists at least one solution ${𝑐}_1,{𝑐}_2,\dots ,{𝑐}_{𝑛}\in 𝐹$ for the equation:

$$\mathrm{𝐯}={𝑐}_1{\mathrm{𝐛}}_1+{𝑐}_2{\mathrm{𝐛}}_2+\cdots +{𝑐}_{𝑛}{\mathrm{𝐛}}_{𝑛}.$$

Suppose ${𝑑}_1,{𝑑}_2,\dots ,{𝑑}_{𝑛}$ is another solution to this equation. That is, suppose that:

$$\mathrm{𝐯}={𝑑}_1{\mathrm{𝐛}}_1+{𝑑}_2{\mathrm{𝐛}}_2+\cdots +{𝑑}_{𝑛}{\mathrm{𝐛}}_{𝑛}.$$

By the definition of linear independence, we must have ${𝑐}_1={𝑑}_1,{𝑐}_2={𝑑}_2,\dots ,{𝑐}_{𝑛}={𝑑}_{𝑛}$. Thus, for any $\mathrm{𝐯}\in 𝑉$, there is exactly one solution ${𝑐}_1,{𝑐}_2,\dots ,{𝑐}_{𝑛}\in 𝐹$ for the equation

$$\mathrm{𝐯}={𝑐}_1{\mathrm{𝐛}}_1+{𝑐}_2{\mathrm{𝐛}}_2+\cdots +{𝑐}_{𝑛}{\mathrm{𝐛}}_{𝑛}.$$

The converse is trivial. ∎

Proof: Theorem 21.

Let $𝑉$ be a vector space, and let $𝑆$ be a finite spanning set for $𝑉$. By the Spanning Set Theorem,

Proof: Theorem 22.

Let $𝑉$ be a vector space, and let $𝑆$ be a finite linearly independent set of vectors from $𝑉$. By the Characterization of Linearly Independent Sets,

Dimension of Vector Spaces

Proof: Unique Size Theorem.

TODO

As a result of this theorem, we can define the dimension of a finite-dimensional vector space as the number of vectors in any basis for the vector space:

Combining Theorem 23 and Definition 20 together, we see that although a finite-dimensional vector space $𝑉$ can have many different bases, they all have the same number of vectors, namely $\dim \left(𝑉\right)$ number of vectors.

That is, if we know the dimension $𝑛$ of a vector space and we have a set of exactly $𝑛$ vectors from that vector space, then we only need to check one of the two criteria from Definition 19 to determine if that set is a basis for the vector space.

If the set of $𝑛$ vectors is linearly independent, it is a basis, so we get spanning for free. If the set of $𝑛$ vectors is a spanning set, it is a basis, so we automatically get linear independence for free.

Proof: Basis Theorem.

Suppose $𝑉$ is a finite-dimensional vector space with dimension $𝑛$.

1. Suppose $ℬ=\left\{{\mathrm{𝐛}}_1,{\mathrm{𝐛}}_2,\dots ,{\mathrm{𝐛}}_{𝑛}\right\}\subseteq 𝑉$ is a linearly independent set of $𝑛$ vectors. For the sake of contradiction, assume that $ℬ$ is not a basis for $𝑉$. Then $ℬ$ is not a spanning set for $𝑉$ ($ℬ$ is already linearly independent, so the other condition of a basis cannot be true.) That is, there exists some vector $\mathrm{𝐯}\in 𝑉$ such that $\mathrm{𝐯}\notin \mathrm{span}\left(ℬ\right)$. By Corollary 8, then, the set $\left\{{\mathrm{𝐛}}_1,{\mathrm{𝐛}}_2,\dots ,{\mathrm{𝐛}}_{𝑛},\mathrm{𝐯}\right\}$ is linearly independent.

Proof: Proposition 6.

Let $𝐻$ be a subspace of a finite-dimensional vector space $𝑉$. Let $\dim \left(𝐻\right)=𝑚$ and $\dim \left(𝑉\right)=𝑛$. Let ${ℬ}_{𝐻}$ be a basis for $𝐻$, and let ${ℬ}_{𝑉}$ be a basis for $𝑉$.

By Theorem 22, there exists an improper superset of ${ℬ}_{𝐻}$ that is a basis for $𝑉$. Since ${ℬ}_{𝑉}$ is a basis for $𝑉$, it must have the same number of vectors as any other basis for $𝑉$, including the improper superset of ${ℬ}_{𝐻}$. Thus, the number of vectors in the improper superset of ${ℬ}_{𝐻}$ is equal to the number of vectors in ${ℬ}_{𝑉}$, which is $𝑛$. Since the improper superset of ${ℬ}_{𝐻}$ contains all the vectors in ${ℬ}_{𝐻}$, the number of vectors in the improper superset of ${ℬ}_{𝐻}$ is greater than or equal to the number of vectors in ${ℬ}_{𝐻}$, which is $𝑚$. Therefore, we have:

$$𝑛\ge 𝑚\iff 𝑚\le 𝑛\iff \dim \left(𝐻\right)\le \dim \left(𝑉\right).$$

Furthermore, if $𝑚=𝑛$, then the improper superset of ${ℬ}_{𝐻}$ that is a basis for $𝑉$ has the same number of vectors as ${ℬ}_{𝐻}$, so ${ℬ}_{𝐻}$ must be the same set as the improper superset of ${ℬ}_{𝐻}$, which is a basis for $𝑉$. Thus, ${ℬ}_{𝐻}$ is a basis for $𝑉$, so $𝐻=𝑉$. ∎

Proof: Dimension Formula.

TODO

If $𝑋+𝑌$ is a direct sum, then $𝑋\cap 𝑌=\left\{\mathrm{𝟎}\right\}$ which has dimension $0$, so:

Basis Coordinates and the Standard Basis

In other words, an ordered basis is a basis which defines precisely which basis vector is “first,” which is “second,” and so on.

Recall that by the Unique Representation Theorem, every vector in a vector space $𝑉$ can be represented as a unique linear combination of vectors from a basis $ℬ$. With $ℬ$ ordered, the weights of that linear combination can be properly defined:

A basis for $𝑉$ over a field $𝐹$ has $\dim \left(𝑉\right)$ number of vectors, so there would be $\dim \left(𝑉\right)$ number of coordinates for each vector in $𝑉$ with respect to that basis. Thus, we can represent each vector in $𝑉$ as an $𝑛$-tuple of coordinates in $𝐹$, where $𝑛=\dim \left(𝑉\right)$. An $𝑛$-tuple of coordinates in $𝐹$ is an element of ${𝐹}^{𝑛}$, so we can represent each vector in $𝑉$ as a vector from ${𝐹}^{𝑛}$. Specifically:

This naturally leads to the fact that the vector space ${𝐹}^{𝑛}$ should have dimension $𝑛$, since it takes $𝑛$ coordinates to represent each vector in ${𝐹}^{𝑛}$:

Since ${𝐹}^{𝑛}$ is a vector space, it has a basis. In particular, the basis that self-describes the coordinates of vectors in ${𝐹}^{𝑛}$ is called the standard basis for ${𝐹}^{𝑛}$:

Example.

The vector $\left(1,2,3\right)\in {\mathrm{ℝ}}^3$ has coordinates $\left(1,2,3\right)$ under the standard basis $ℰ$, since $1{\mathrm{𝐞}}_1+2{\mathrm{𝐞}}_2+{\mathrm{𝐞}}_3=1\left(1,0,0\right)+2\left(0,1,0\right)+3\left(0,0,1\right)=\left(1,2,3\right)$.

Exercises

Exercise 35.

Describe the set of all vector spaces with exactly one basis.

Exercise 36.

Show that if $𝑆$ is a set of $𝑛$ vectors from an $\left(𝑛+1\right)$-dimensional vector space $𝑉$, then $𝑆$ cannot span $𝑉$.

Exercise 37: Extension of Theorem 13.

Suppose $𝑉$ is a finite-dimensional vector space with dimension $𝑛$, and $𝐻$ is a subspace of $𝑉$. Show that there exists a set of at most $𝑛$ vectors from $𝑉$ that spans $𝐻$.

Exercise 38.

Show that if $\left\{{\mathrm{𝐯}}_1,{\mathrm{𝐯}}_2,{\mathrm{𝐯}}_3,{\mathrm{𝐯}}_4\right\}$ is a basis for a vector space $𝑉$, then $\left\{{\mathrm{𝐯}}_1+{\mathrm{𝐯}}_2,{\mathrm{𝐯}}_2+{\mathrm{𝐯}}_3,{\mathrm{𝐯}}_3+{\mathrm{𝐯}}_4,{\mathrm{𝐯}}_4\right\}$ is also a basis for $𝑉$.

Exercise 39.

Let ${\mathrm{\mathbb{P}}}_2\left(\mathrm{ℝ}\right)$ be the vector space of all polynomials with real coefficients of degree at most 2. Define two bases for ${\mathrm{\mathbb{P}}}_2\left(\mathrm{ℝ}\right)$: $ℬ=\left\{1,𝑥,{𝑥}^2\right\}$ and $𝒞=\left\{1,1+𝑥,1+𝑥+{𝑥}^2\right\}$. Find ${\left[3+2𝑥+{𝑥}^2\right]}_{ℬ}$ and ${\left[3+2𝑥+{𝑥}^2\right]}_{𝒞}$.

Exercise 40.

Suppose $𝑉$ is a vector space and $ℬ$ is a basis for $𝑉$. Show that for any $\mathrm{𝐯},\mathrm{𝐰}\in 𝑉$,

$${\left[\mathrm{𝐯}+\mathrm{𝐰}\right]}_{ℬ}={\left[\mathrm{𝐯}\right]}_{ℬ}+{\left[\mathrm{𝐰}\right]}_{ℬ}.$$

Exercise 41.

Suppose $𝑉$ is a vector space and ${\mathrm{𝐯}}_1,{\mathrm{𝐯}}_2,{\mathrm{𝐯}}_3\in 𝑉$ are linearly independent. Show that $\dim \left(\mathrm{span}\left\{{\mathrm{𝐯}}_1+{\mathrm{𝐯}}_2,{\mathrm{𝐯}}_1-{\mathrm{𝐯}}_3,{\mathrm{𝐯}}_2+{\mathrm{𝐯}}_3\right\}\right)=2$.

Exercise 42.

Suppose $\left\{{\mathrm{𝐯}}_1,{\mathrm{𝐯}}_2,\dots ,{\mathrm{𝐯}}_{𝑝}\right\}$ is a linearly independent set in $𝑉$, and $\mathrm{𝐰}\in 𝑉$. Show that:

$$\dim \left(\mathrm{span}\left\{{\mathrm{𝐯}}_1+\mathrm{𝐰},{\mathrm{𝐯}}_2+\mathrm{𝐰},\dots ,{\mathrm{𝐯}}_{𝑝}+\mathrm{𝐰}\right\}\right)\ge 𝑝-1.$$

Exercise 43.

For any three subspaces $𝑋$, $𝑌$, and $𝑍$ of a finite-dimensional vector space $𝑉$, show that:

$$\begin{array}{cc}\dim \left(𝑋+𝑌+𝑍\right)& =\dim \left(𝑋\right)+\dim \left(𝑌\right)+\dim \left(𝑍\right)\\ & -\dim \left(𝑋\cap 𝑌\right)-\dim \left(𝑋\cap 𝑍\right)-\dim \left(𝑌\cap 𝑍\right)\\ & +\dim \left(𝑋\cap 𝑌\cap 𝑍\right).\end{array}$$

Exercise 44.

For any three subspaces $𝑋$, $𝑌$, and $𝑍$ of a finite-dimensional vector space $𝑉$, show that:

$$\begin{array}{cc}\dim \left(𝑋+𝑌+𝑍\right)& =\dim \left(𝑋\right)+\dim \left(𝑌\right)+\dim \left(𝑍\right)\\ & -\frac{1}{3}\left(\dim \left(𝑋\cap 𝑌\right)+\dim \left(𝑋\cap 𝑍\right)+\dim \left(𝑌\cap 𝑍\right)\right)\\ & -\frac{1}{3}\left(\dim \left(\left(𝑋+𝑌\right)\cap 𝑍\right)+\dim \left(\left(𝑋+𝑍\right)\cap 𝑌\right)+\dim \left(\left(𝑌+𝑍\right)\cap 𝑋\right)\right).\end{array}$$

Exercise 45.

Suppose $𝑉$ is finite-dimensional, and $𝑋$ and $𝑌$ are subspaces of $𝑉$ such that $𝑉=𝑋+𝑌$. Show that there exists a basis $ℬ$ for $𝑉$ such that $ℬ\subseteq 𝑋\cup 𝑌$.

Exercise 46.

Suppose $𝑉$ is a finite-dimensional vector space, and $𝑈$ and $𝑊$ are subspaces of $𝑉$ such that $\dim \left(𝑈\right)+\dim \left(𝑊\right)>\dim \left(𝑉\right)$. Show that $𝑈\cap 𝑊$ contains a nonzero vector.

Exercise 47.

Let $𝑋$ and $𝑌$ be subspaces of a vector space $𝑉$ such that $𝑉=𝑋\oplus 𝑌$. Let $\left\{{\mathrm{𝐱}}_1,{\mathrm{𝐱}}_2,\dots ,{\mathrm{𝐱}}_{𝑚}\right\}$ be a basis for $𝑋$ and $\left\{{\mathrm{𝐲}}_1,{\mathrm{𝐲}}_2,\dots ,{\mathrm{𝐲}}_{𝑛}\right\}$ be a basis for $𝑌$. Show that $\left\{{\mathrm{𝐱}}_1,{\mathrm{𝐱}}_2,\dots ,{\mathrm{𝐱}}_{𝑚},{\mathrm{𝐲}}_1,{\mathrm{𝐲}}_2,\dots ,{\mathrm{𝐲}}_{𝑛}\right\}$ is a basis for $𝑉$. (This is an extension of Theorem 15.)

Exercise 48.

Suppose $𝑉$ is a finite-dimensional vector space, and $𝑋$ is a subspace of $𝑉$. Show that there exists a subspace $𝑌$ of $𝑉$ such that $𝑉=𝑋\oplus 𝑌$.

Exercise 49.

Recall from Exercise that $\mathrm{ℝ}$ is a vector space over $\mathrm{\mathbb{Q}}$. Show that in particular, $\mathrm{ℝ}$ is an infinite-dimensional vector space over $\mathrm{\mathbb{Q}}$.

Exercise 50.

Show that the vector space ${𝐶}_{\left[0,1\right]}\left(\mathrm{ℝ}\right)$, the set of all real-valued functions that are continuous on the closed interval $\left[0,1\right]$, is infinite-dimensional.