Standard basis

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

Table of contents

Definition
Proof that the standard basis is a basis
Standard basis and identity matrix
Equivalent basis
Solved exercises
1. Exercise 1

Definition

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

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

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

Example Let be the space of all vectors. Then, the standard basis of 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 basisis a basis of the space of all -dimensional vectors.

Proof

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

Example Let be the space of all vectors. Then, the standard basis of is formed by the two vectors [eq6] Clearly, there is no scalar such that orso that the two vectors are not multiples of each other, that is, they are linearly independent. Now, take any vector : [eq9] where $s_{1}$ and $s_{2}$ are two scalars. Then, [eq10] In other words, any vector 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 be the identity matrix: [eq11] Denote by its rows and by its columns. Then, the rows are the vectors of the standard basis of the space of all vectors, and the columns are the vectors of the standard basis of the space of all vectors.

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

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

Equivalent basis

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

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

Proof

Denote the set of linearly independent vectors by Assume that all the vectors of the standard basis can be written as linear combinations of : [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 having entries can be written as [eq23] In other words, any vector can be written as a linear combination of the set of linearly independent vectors belonging to . As a consequence, is a basis for . We have proved that if A1 holds, then 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 cannot be written as a linear combination of the vectors of . Suppose without loss of generality that it is $e_{1}$ (if it's not, we can re-number the vectors). Thenis a set of linearly independent vectors. If A1 holds by replacing with $B_{1}$ , then $B_{1}$ is a basis for , 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 ) is greater than the cardinality of the standard basis (). If A1 does not hold, then form a new set of linearly independent vectors If A1 holds by replacing with $B_{2}$ , then we have a contradiction, otherwise keep going and form . When we get tothen 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 ). Thus we have demonstrated that A1 must hold with (otherwise we have a contradiction). Therefore, is a basis for .

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Write the standard basis of the set of all vectors.

Solution

The basis is [eq28]

How to cite

Please cite as:

Taboga, Marco (2021). "Standard basis", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/standard-basis.