Consider the set of vectors

Part 1.

Express the vector \(\left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]\) as a linear combination of the vectors in \(S\text{,}\) i.e. find scalars such that

Consider the set of vectors

Part 2.

Find a different way to express the vector \(\left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]\) as a linear combination of the vectors in \(S\text{.}\)

Consider the set of vectors

Part 3.

Consider another vector \(\left[\begin{array}{c} 8 \\ 6 \\ 7 \\ 5 \end{array} \right]\text{.}\) Without computing the RREF of another matrix, how many ways can this vector be written as a linear combination of the vectors in \(S\text{?}\)

1. Zero.

2. One.

3. Infintiely-many.

4. Computing a new matrix RREF is necessary.

Let’s review some of the terminology we’ve been dealing with...

Part 1.

If every vector in a space can be constructed as one or more linear combination of vectors in a set \(S\text{,}\) we can say...

1. the set \(S\) spans the space.

2. the set \(S\) fails to span the space.

3. the set \(S\) is linearly independent.

4. the set \(S\) is linearly dependent.

Let’s review some of the terminology we’ve been dealing with...

Part 2.

If the zero vector \(\vec 0\) can be constructed as a unique linear combination of vectors in a set \(S\) (the combination multiplying every vector by the scalar value \(0\)), we can say...

1. the set \(S\) spans the space.

2. the set \(S\) fails to span the space.

3. the set \(S\) is linearly independent.

4. the set \(S\) is linearly dependent.

Let’s review some of the terminology we’ve been dealing with...

Part 3.

If every vector of a space can either be constructed as a unique linear combination of vectors in a set \(S\text{,}\) or not at all, we can say...

1. the set \(S\) spans the space.

2. the set \(S\) fails to span the space.

3. the set \(S\) is linearly independent.

4. the set \(S\) is linearly dependent.

A basis of a vector space is a set of vectors \(S\) for which

1. Every vector of the space can be expressed as a linear combination of the vectors in \(S\text{.}\)

2. For each vector \(\vec{v}\) in the space, there is only one way to write it as a linear combination of the vectors in \(S\text{.}\)

These two properties may be expressed more succintly as the statement "Every vector in \(V\) can be expressed uniquely as a linear combination of the vectors in \(S\)".

In terms of a vector equation, a set \(S=\left\{\vec{v}_1,\ldots,\vec{v}_n\right\}\) is a basis of a space if the vector equation

Put another way, a basis may be thought of as a minimal set of “building blocks” that can be used to construct any other vector of the space.

The standard basis of \(\IR^n\) is the set \(\{\vec{e}_1, \ldots, \vec{e}_n\}\) where

In particular, the standard basis for \(\mathbb R^3\) is \(\{\vec e_1,\vec e_2,\vec e_3\}=\{\hat\imath,\hat\jmath,\hat k\}\text{.}\)

Take the RREF of an appropriate matrix to determine if each of the following sets is a basis for \(\IR^4\text{.}\)

Part 1.

1. A basis because it both spans and is linearly independent.

2. Spans, but not a basis as it is linearly dependent.

3. Linearly independent, but not a basis as it fails to span.

4. Fails to span and linearly independent, so not a basis.

Take the RREF of an appropriate matrix to determine if each of the following sets is a basis for \(\IR^4\text{.}\)

Part 2.

1. A basis because it both spans and is linearly independent.

2. Spans, but not a basis as it is linearly dependent.

3. Linearly independent, but not a basis as it fails to span.

4. Fails to span and linearly independent, so not a basis.

Take the RREF of an appropriate matrix to determine if each of the following sets is a basis for \(\IR^4\text{.}\)

Part 3.

1. A basis because it both spans and is linearly independent.

2. Spans, but not a basis as it is linearly dependent.

3. Linearly independent, but not a basis as it fails to span.

4. Fails to span and linearly independent, so not a basis.

Take the RREF of an appropriate matrix to determine if each of the following sets is a basis for \(\IR^4\text{.}\)

Part 4.

1. A basis because it both spans and is linearly independent.

2. Spans, but not a basis as it is linearly dependent.

3. Linearly independent, but not a basis as it fails to span.

4. Fails to span and linearly independent, so not a basis.

Take the RREF of an appropriate matrix to determine if each of the following sets is a basis for \(\IR^4\text{.}\)

Part 5.

1. A basis because it both spans and is linearly independent.

2. Spans, but not a basis as it is linearly dependent.

3. Linearly independent, but not a basis as it fails to span.

4. Fails to span and linearly independent, so not a basis.

If \(\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}\) is a basis for \(\IR^4\text{,}\) that means \(\RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]\) has a pivot in every row (because it spans), and has a pivot in every column (because it’s linearly independent).

What is \(\RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]\text{?}\)

The set \(\{\vec v_1,\dots,\vec v_m\}\) is a basis for \(\IR^n\) if and only if \(m=n\) and \(\RREF[\vec v_1\,\dots\,\vec v_n]= \left[\begin{array}{cccc} 1&0&\dots&0\\ 0&1&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&1 \end{array}\right] \text{.}\)

That is, a basis for \(\IR^n\) must have exactly \(n\) vectors and its square matrix must row-reduce to the so-called identity matrix containing all zeros except for a downward diagonal of ones. (We will learn where the identity matrix gets its name in a later module.)