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Standard basis

The standard basis is the simplest basis of the space of all K-dimensional vectors. It is made up of vectors that have one entry equal to 1 and the remaining $K-1$ entries equal to 0.

Table of Contents

Definition

In what follows we deal with the space of all K-dimensional vectors, which we denote by $S$. We do not specify whether the vectors are row vectors or column vectors, or whether their entries are real or complex numbers.

Definition Let $S$ be the space of all K-dimensional vectors. Denote by $e_{k}$ a vector whose k-th entry is equal to 1 and whose remaining $K-1$ entries are equal to 0. Then, the set of K vectors[eq1]is called the standard basis of $S$.

The standard basis is also often called canonical or natural basis.

Example Let $S$ be the space of all $3	imes 1$ vectors. Then, the standard basis of $S$ is formed by the three vectors[eq2]

Proof that the standard basis is a basis

We have defined the standard basis, but we have not proved that it is indeed a basis.

Proposition The standard basis[eq3]is a basis of the space $S$ of all K-dimensional vectors.

Proof

Remember that a basis of $S$ is a set of linearly independent vectors spanning $S$. Take any vector $e_{k}$. It cannot be written as a linear combination of the other vectors of E because the k-th entry of all the other vectors is 0, while the k-th entry of $e_{k}$ is 1. Since no vector of E can be written as a linear combination of the others, then they are linearly independent. Take any vector $sin S$ and denote its entries by [eq4]. Then $s$ can be written as[eq5]that is, as a linear combination of the canonical basis (see the next example for a concrete example). Thus, we have proved that the canonical basis is a set of linearly independent vectors that span $S$. Therefore, the canonical basis is indeed a basis for $S$.

Example Let $S$ be the space of all $2	imes 1$ vectors. Then, the standard basis of $S$ is formed by the two vectors[eq6]Clearly, there is no scalar a such that [eq7]or[eq8]so that the two vectors are not multiples of each other, that is, they are linearly independent. Now, take any vector $sin S$: [eq9]where $s_{1}$ and $s_{2}$ are two scalars. Then,[eq10]In other words, any vector $s$ can be written as a linear combination of $s_{1}$ and $s_{2}$.

Standard basis and identity matrix

There is a simple relation between standard bases and identity matrices.

Proposition Let I be the $K	imes K$ identity matrix:[eq11]Denote by [eq12] its rows and by [eq13] its columns. Then, the rows [eq14] are the K vectors of the standard basis of the space of all $1	imes K$ vectors, and the columns [eq15] are the K vectors of the standard basis of the space of all $K	imes 1 $ vectors.

The proposition does not need to be proved because it is self-evident.

Example Let I be the $3	imes 3$ identity matrix[eq16]Then,[eq17]which is the standard basis of the space of $3	imes 1$ vectors.

Equivalent basis

Which bases are equivalent to the standard basis, in the sense that they span the same space $S$ (of all K-dimensional vectors) that is spanned by the standard basis? The next proposition answers this question.

Proposition Any set of K linearly independent vectors is a basis for the space $S$ of all K-dimensional vectors.

Proof

Denote the set of K linearly independent vectors by [eq18]Assume that all the vectors [eq19] of the standard basis can be written as linear combinations of [eq20]:[eq21]where $eta _{j}^{k}$ are the (scalar) coefficients of the combination. We are going to call this assumption A1. If A1 holds, then any vector $sin S$ having entries [eq22] can be written as[eq23]In other words, any vector $sin S$ can be written as a linear combination of the set of linearly independent vectors belonging to $B$. As a consequence, $B$ is a basis for $S$. We have proved that if A1 holds, then $B$ is a basis. We now need to prove that A1 holds. The proof is by contradiction. Suppose A1 does not hold. Then, one of the vectors of the standard basis of $S$ cannot be written as a linear combination of the vectors of $B$. Suppose without loss of generality that it is $e_{1}$ (if it's not, we can re-number the vectors). Then[eq24]is a set of linearly independent vectors. If A1 holds by replacing $B$ with $B_{1}$, then $B_{1}$ is a basis for $S$, which leads to a contradiction because 1) by the Dimension Theorem, all bases must have the same cardinality, but 2) the cardinality of $B_{1}$ (equal to $K+1$) is greater than the cardinality of the standard basis (K). If A1 does not hold, then form a new set of linearly independent vectors [eq25]If A1 holds by replacing $B$ with $B_{2}$, then we have a contradiction, otherwise keep going and form [eq26]. When we get to[eq27]then we know that A1 certainly holds because the natural basis belongs to $B_{K}$. But this too leads to a contradiction (cardinality of $B_{K}$ greater than K). Thus we have demonstrated that A1 must hold with $B$ (otherwise we have a contradiction). Therefore, $B$ is a basis for $S$.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Write the standard basis of the set of all $1	imes 4$ vectors.

Solution

The basis is[eq28]

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