Basis and Dimension

Basis

Analogous to the definition of a basis for

, we can define a basis for a general vector space as follows: Definition: A set of vectors

is a basis for

is a linearly independent set, and

Remark: The above definition still works well if the set of vectors is an infinite set. But in this course, we will mainly focus on vector spaces having a finite basis. There are several important theorems related to basis: Theorem: Suppose a vector space

can be spanned by

vectors. If any set of

vectors in

is linearly independent, then

. Proof: Suppose

and suppose that

is a linearly independent set in

. Then

for some real numbers

. Since

, not all

are zero. Assume

. Then we have

and it is not hard to see that

can replace

in the spanning set i.e.

. Then

for some real numbers

. Since

is linearly independent, not all

are zero. Assume

. Then by similar argument,

can replace

in the spanning set i.e.

. Assume

, we can repeat the above process until all the vectors

are replaced by vectors

. Hence

. In other words,

is a linear combination of

. But this is impossible because

is a linearly independent set in

. Therefore,

Dimension

If a vector space

is spanned by a finite set of vectors, then

is said to be finite-dimensional. Starting from that finite set, if a vector is a linear combination of the others, we can throw it away and the remaining vectors will still span

. Repeat this process until you cannot throw away any more vector in the set. It means that no vector in the set is a linear combination of others i.e. the set is now linearly independent. Since it still spans

, the set of remaining vectors forms a basis for

. Basis Theorem: Suppose

and

are both bases for

. Then

. Proof: By the above theorem,

. Interchange the roles of two bases and apply the theorem again, we have

. Hence,

. Thanks to the Basis Theorem, the number of vectors in any basis for

are the same. We define this number to be the dimension of

, denoted by

. By convention, the dimension of the zero vector space is zero. The dimension of a vector space can be interpreted as

the smallest number of vectors that can span the vector space, or
the largest number of linearly independent vectors in the vector space.

Dimension is also a number that measures the "size" of a vector space - the greater the dimension, the larger the vector space. Suppose

is a subspace of

. We expect that

. The following theorem confirms this: Theorem: Let

be a vector space such that

and let

and

be subspaces of

. Then

is finite-dimensional and .
Any basis of is part of a basis for .
If and , then .

(It will be proved in class.)

Some Examples

Example 1:

is a basis for

. Example 2: Let

. We already proved that

is a subspace of

. Find a basis for

. If you express the equation

as an augmented matrix

. It is a matrix already in echelon form. And the variables

and

are regarded as free variables. Let

. Then the solutions are

Therefore,

and the two vectors are linearly independent (because one is not a scalar multiple of another). In other words,

is a basis for

. Example 3:

is a basis for

. Example 4: For any

. We define the trace of

, denoted by

, to be the sum of all diagonal entries of

. It can be shown that

is a subspace of

. Find a basis for

when

and hence

. Idea: There are 6 off-diagonal entries that can be chosen freely. As for the diagonal entries, two of them can be chosen freely and the remaining one is determined because the diagonal sum is zero. Therefore, it is expected that

. As for the basis, we will find it out in class.

Exercise

Let .

Show that is a subspace of .
Prove that is a basis for .

Check my answer

Basis and Dimension

Basis

Dimension

Some Examples

Exercise

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