keywords:linear algebra, basis, color, chemistry.

Basis Explanation

by Joseph Ruan, proud member of the Math Squad

Supplementary Explanations of a Basis

It is important to first check out the original basis page for the more rigorous definition of a Basis. If you still don't understand, then come back to this page. This page assumes that one already fully understands the terms "span", "linear independence" and "subspace".

What is a Basis?

From the rigorous definition of a Basis, we know that a group of vectors $ v_1, v_2... v_n $ are defined as a basis of a Subspace V if they fulfill two requirements:

  • The vectors span V. In other words, every vector in V can be written as a linear combination of the basis vectors.
  • The vectors are linearly independent. In other words none of the basis vectors can be written as a linear combination of the other basis vectors.

Note: Putting it loosely, the "subspace V" is fancy math-speak for a specific collection of vectors.

This previous definition is shamelessly copied from the rigorous definition of a Basis.

However, what does this even mean? Let's start with a conceptual method of understanding this. As a starting point, it is possible to think of basis vectors as building blocks and their corresponding vector space V is every possible product. But let's move onto some analogies and examples.

Conceptual explanations & analogies

Conceptually, we can analogize the basis to other similar ideas, such as molecules, colors, and food.

Chemistry analogy

Let's analogize everything we know in the abstract magical world of math into the more tangible world of Chemistry.

Arbitrarily, let's call our subspace V as every molecule made of only Carbon and Hydrogen, or in chemical terms every vector in V is a hydrocarbon. And let our "vectors" be molecules and "linearly independent vectors" would just mean that each molecule cannot be made up of other molecules. In the end, all we've done is turn vectors into molecules.

Now, consider this: to make every possible hydrocarbon, you only need two molecules, Hydrogen and Carbon! For example, given an arbitrary hydrocarbon, such as Octane (CH8), you can make Octane with Carbon + 8* Hydrogen! This is common sense, since by definition every hydrocarbon is made of Hydrogen and Carbon. Moreover since every hydrocarbon is just a combination of Hydrogen and Carbon, our "vectors", Hydrogen and Carbon span the "subspace" of Hydrocarbons. Finally, since Hydrogen and Carbon are clearly chemically different, they are also "linearly independent".

So what does this all this analogizing show? That Carbon and Hydrogen are the basis vectors for the subspace of hydrocarbons! In other words, you can imagine basis vectors as this specific type of building block: they are capable of generating every possible vector in the subspace V and are the minimum number of "building blocks" necessary to do so.

This second part is important to note. You only need Carbon and Hydrogen. To make every Hydrocarbon, you could have Carbon, Hydrogen and Methane (CH4). However, basis vectors also need to be linearly independent; since Methane can be made with Hydrogen and Carbon, it is redundant and needs to be removed. A sign that a set of vectors is a basis is that any vector you add to it from vector space V would be redundant.

Hopefully from this, you can understand that Basis Vectors

  • can produce every possibility of the given "goal/group" of vectors
  • include only the minimum number of vectors

Colors Analogy

If you like art, you'd know that there are three primary colors: red, green and blue.

By definition, primary colors can make up every possible color out there. For example, an arbitrary color, like purple, is made of red and blue, or white, which is made up of all three. Moreover, this set of "vectors" is "linearly independent", because none of the colors can be made by adding the other two. You can imagine this by trying to make red out of green and blue. Logically, since they still manage to generate every color and don't have any redundant colors among their own set, then they serve as a basis for all colors.

So, these primary colors are a basis for all colors. In other words, by varying the amounts of each, you can make every possible color.

For those of you who've taken a course in classical physics, you will know that there is another set of primary colors often used to make all colors: yellow, magenta and cyan. These three colors also can make every possible color, (arbitrarily, if we take green, we can make this by subtracting yellow from cyan) and they are also "linearly independent" (it's impossible to make Cyan out of Magenta and yellow). So therefore, these three are ANOTHER set of basis vectors. In fact, there are many possible sets of basis vectors that can still correspond to original set of vectors. Or in these color terms, there is more than one triad of colors that can paint the entire spectrum of colors without having any redundancies.

Moreover, basis vectors have other attributes. First is that the set of them is within the original set. What do I mean? Remember Red, blue and green? They're also colors, and are therefore part of the original set. Second, a set of basis vectors can form every other set of basis vectors. For example, the triad of Cyan, Magenta and yellow can be made by mixing the triad of Red, blue and green (Cyan = blue +green, Magenta = Red + blue, yellow = Green + red with varying amounts of each). Third, they are the maximum number of linearly independent vectors. In this case, Red and green are linearly independent and same with blue and green; however red, green and blue is the maximum number of linearly independent vectors in V I can manage before adding redundant ones.

So from this analogy I hope you more clearly understand that basis vectors are:

  • not unique
  • can form all other sets of basis vectors
  • are part of the original set that they produce
  • contain the maximum number of linearly independent vectors.


One last, abstract way of thinking about basis vectors is with cooking. Let's say that each meal is only made of meat, bread, vegetables, sauce and cheese. By definition, the list we just gave is a list of "vectors" which span the set of burgers. Moreover, it's very hard to make any element on the list out of the other four (I have no clue how to make meat out of vegetables, bread and cheese. If someone does, please tell me so I can horrify vegetarians).

This analogy will be short, but there are two things to note: first, with this list of basis vectors, there is only ONE way to make a burger with one slice of bread, two slices of meat, no sauce, no vegetables and one cheese square. With this list of ingredients, there is only one recipe to make this exact burger. The same can be said for any distinct burger: there is only one specific combination to make it.

Moreover, you can't use these ingredients to make anything else. For example, let's say, added onto the list of ingredients was a rock. However, as we learned earlier, all basis vectors need to also be part of V. You can even think of it as reducing V to a minimum number of elements so that the rest can be filled in with linear combinations. Rocks aren't food and therefore cannot be added as a basis vector. In other words, your basis vector's span must be exactly V. You cannot make things outside of V. In this case, you can't make something with meat, bread, vegetables, sauce and cheese that is NOT food; you can only make food with them.

So from this analogy I hope you finally understand that basis vectors:

  • have only a UNIQUE way of making every vector in set V
  • the space constructed by the basis vectors must be EXACTLY V

Physical explanations & examples

Now that we're through conceptual examples, let's get some more physical examples i.e. actually math-related ones. We will go through several properties:

  • basic examples of what are sets of basis vectors and what aren't
  • the determinant of a matrix of basis vectors
  • Dimensions
  • the Basis Matrix and its properties

Let's say you are given two vectors,

$ \begin{pmatrix}1 \\0 \end{pmatrix} $ and $ \begin{pmatrix}0 \\1 \end{pmatrix} $.

First thing to note is that these two vectors are the columns of the Identity matrix. Moreover, it should be relatively easy to see that every two dimensional vector can be written as a sum of these two vectors and that they are linearly independent. So from a backwards point of view (normally we look at the set first and then construct the basis), these two vectors serve as a basis for all two dimensional vectors.

However, given three vectors,

$ \begin{pmatrix}1 \\0 \end{pmatrix} $, $ \begin{pmatrix}0 \\1 \end{pmatrix} $, and $ \begin{pmatrix}2 \\2 \end{pmatrix} $,

we know for sure $ \begin{pmatrix}1 \\1 \end{pmatrix} $

is twice the sum of the first two vectors, and is therefore redundant. Therefore, this set of three vectors is NOT a basis for dimension 2. More interestingly, note, for now that the dimension of the set of two dimensional vectors is two, and the number of basis vectors needed is two.

Also, however, given two vectors:

$ \begin{pmatrix}0 \\1 \end{pmatrix} $ and $ \begin{pmatrix}0 \\2 \end{pmatrix} $.

These do NOT serve as a basis for vectors of dimension two because it is impossible to make

$ \begin{pmatrix}1 \\0 \end{pmatrix} $

out of them. Moreover, the two are NOT linearly independent.

Finally, given these two vectors,

$ \begin{pmatrix}1 \\1 \end{pmatrix} $ and $ \begin{pmatrix}0 \\1 \end{pmatrix} $,

we know that these vectors are first, linearly independent. But do they span the entire space of two dimensional vectors?

Yes. Note that since

$ \begin{pmatrix}1 \\0 \end{pmatrix} $ and $ \begin{pmatrix}0 \\1 \end{pmatrix} $

span the space of two dimensional vectors. This means that every vector,

$ \begin{pmatrix}a \\b \end{pmatrix} $

can be written as

$ a_1 \begin{pmatrix}1 \\0 \end{pmatrix} + a_2 \begin{pmatrix}0 \\1 \end{pmatrix} $ ,

in this case

$ a_1 $ =a and $ a_2 $=b.

However notice that

$ \begin{pmatrix}1 \\1 \end{pmatrix} = \begin{pmatrix}1 \\0 \end{pmatrix} + \begin{pmatrix}0 \\1 \end{pmatrix} $. \

This means that given a vector

$ \begin{pmatrix}a \\b \end{pmatrix} = a_1 \begin{pmatrix}1 \\0 \end{pmatrix} + a_2 \begin{pmatrix}0 \\1 \end{pmatrix} = a_1* \begin{pmatrix}1 \\0 \end{pmatrix} + a_1* \begin{pmatrix}0 \\1 \end{pmatrix} + a_2 \begin{pmatrix}0 \\1 \end{pmatrix} $.

This simplifies to

$ k_1* \begin{pmatrix}1 \\0 \end{pmatrix} + k_2* \begin{pmatrix}0 \\1 \end{pmatrix} $.

In other words, it spans and is linearly independent. Therefore it is a basis.

Diagonal Matrices

Let's move on to a more complicated situation. Let's say that we're given the space of diagonal 2x2 matrices. For example,

$ \begin{pmatrix}2 & 0 \\0 & 1 \end{pmatrix} $

is such a matrix. What could possibly serve as a basis?

To save space, I will omit how the solution is found over here, but you can find the solution at Basis_Problems. One possible set of Basis vectors are

$ \begin{pmatrix}1 & 0 \\0 & 0 \end{pmatrix} $, $ \begin{pmatrix}0 & 0 \\0 & 1 \end{pmatrix} $.

Note that every diagonal matrix can be made by adding those two. Moreover, the two are linearly independent. Therefore, they serve as a basis. It is important to note that as long as the vectors span space V and are linearly independent, they are automatically a basis for that space.

An example of a a set of vectors that do NOT serve as a basis of all 2x2 diagonal matrices is :

$ \begin{pmatrix}1 & 0 \\1 & 1 \end{pmatrix} $, $ \begin{pmatrix}0 & 1 \\0 & 0 \end{pmatrix} $, $ \begin{pmatrix}2 & 0 \\2 & 2 \end{pmatrix} $, $ \begin{pmatrix}0 & 0 \\1 & 0\end{pmatrix} $

There're many problems. First, they aren't even linearly independent. Second, they don't span the space. Third they span outside of the space. Fourth is that none of the vectors are even in the space. Any of these attributes easily indicate that a set of vectors may not be a basis for a certain space.


Basis vectors are, quite simply, the minimum number vectors which can construct another space.

For examples problems involving bases and explanations of how to find a basis given a space, see Basis_Problems.

For some change of Bases, see Change_of_Bases

Questions and comments

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