Suppose we are given the set . We would like to determine if this set is linearly independent. We know that a set is linearly independent if

(1)

implies

(2)

Also note that if any of the constants is not equal to zero, say , one can write

(3)

Of course, this means that when a set is linearly dependent, then at least one member can be written as a linear combination of other members. In this example, we see

(4)

And therefore, the set is linearly dependent.

Suppose we are given the set . We would like to determine if is linearly independent. Write

(5)

This implies

(6)

(7)

Of course is a solution. If this is the only case, then the set is linearly independent; otherwise dependent. As you get from the first and second equations

(8)

From the last two equations

(9)

And finally

(10)

Therefore, set is linearly independent.

Suppose we are given in the vector space of all polynomials with degree less than and would like to know if the set is linearly dependent. Start with writing

(11)

Note that this is true for every We get

(12)

which implies . Also note that since we only have two vectors in we can quickly check that no vector is a constant multiple of another and therefore, the set is linearly independent.

Suppose we are given in the vector space of all polynomials with degree less than and would like to know if the set is linearly dependent. Start with writing

(13)

This implies

(14)

We can also write this system as followed

(15)

Determinant of the coefficient matrix is zero which means, the option of having a unique solution is out. Since there is always the solution we see immediately that there are infinitely many solutions and therefore the set is linearly dependent. This also implies one of the polynomials in can be written as a linear combination of other members in Can you find at least one such case?

Suppose we are given the set . We know the set is a basis for the vector space of all polynomials of degree 2 or less. Therefore, any other basis for must obey the minimum condition of having 3 elements (even before we check to see if the candidate set spans the space and is linearly independent.). However, set has members and so it cannot be a basis for

Suppose we are given the set . We want to check if set is a basis for the vector space of all 2 by 2 matrices. Two conditions to be checked.

1. Does span the space of by matrices?

2. Is linearly independent.

In this case, we know dimension of the vector space of 2 by 2 matrices is 4 and so any one out of the two conditions is enough to show is a basis. We show both here (but 1 is enough as explained.)

Take an arbitrary 2 by 2 matrix

(16)

and write

(17)

This implies

(18)

The associated coefficient matrix is

(19)

Determinant of this matrix is equal to and so the associate system has a unique solution. Therefore spans the vector space of 2 by 2 matrices.

Next, when we see that is the only solution and so is linearly independent

Suppose we are given . Observe that given any vector in , we can write

(20)

This system has a unique solution since

(21)

Therefore, spans . It is easy to see that the two members of are not constant multiple of each other. And so is a basis for Like the previous example, in the case that dimension of the vector space is known, only one of the conditions will do the job.

Suppose we are given the set . We want to find a basis for this space and in result the dimension of the subspace of We start by writing a typical member of this vector space

(22)

Therefore the set spans We check if this set is linearly independent. Note that the two members of this set are not constant multiple of one another. Therefore we know this set is linearly independent. You can try other ways to conclude this. We see that the set is a basis for and since it has two members, has dimension .