Course Notes for Linear Algebra

There are three vectors here, but they are all vectors in \(\RR^2\text{.}\) In two-dimensional space, there are only two independent directions of movement, so three vectors can never be linearly independent. I conclude that this is a linearly dependent set.

The second and third vectors are multiplied (by \(3\) and \((-5)\text{,}\) respectively) of the first vector. That means that all three vectors point along the same line, so these are not different directions and cannot be linearly independent.

The second vector is clearly independent of the other two since it only has a \(z\) coordinate and the others have 0 in the \(z\) coordinate. It points in the \(z\) direction, while the others are restricted to the \(x-y\) plane. For the first and third vector, the dot product shows that they are perpendicular, so they point in fundamentally different directions. I conclude that this is a linearly independent set.

Working by inspection again, I can see that the sum of the first two vectors is the third vector. Any relation like this, where some linear combinations of some of the vectors produce another vector, means that the set is linearly dependent.

This is in \(\RR^2\text{,}\) so the maximum number of linearly independent vectors is 2. These are not all in the same direction, so this span is all of \(\RR^2\text{.}\) I can take any two of these as a basis since no vector is a multiple of any other vector. Since the span is \(\RR^2\text{,}\) we could also take the standard basis \(\{ e_1, e_2\}\text{.}\)

None of the vectors have a non-zero \(x\) component, so this set cannot span all of \(\RR^3\text{.}\) However, I do have different directions in the \(yz\) plane, so this span does span all of the \(yz\) plane. Therefore, since none of the vectors is a multiple of the others, any two vectors form a basis. I can also take the subset of the standard basis that spans the \(yz\) plane: \(\{e_2, e_3\}\text{.}\)

All of the vector point in axis directions in \(\RR^5\text{,}\) and all the axis directions are different. I conclude that this is a linearly independent set, so I can take all four vectors as a basis. Since I get the first, second, third and fifth axis directions, I could also take a portion of the standard basis that spans those four directions: \(\{e_1, e_2, e_3, e_5\}\text{.}\)

All four vectors point in the second axis direction, so they are all multiples of the standard basis vector \(e_2\text{.}\) Therefore, the span is one-dimensional and any individual vector, or for that matter \(e_2\text{,}\) could be taken as a basis.

There are a number of possible approaches here. First, I can talk about fundamentally different directions. If all three vectors have a 0 \(x\) component, then they are all found in the \(yz\text{.}\) In that plane (like any plane), there are only two fundamentally different directions. Therefore, a set of three vectors cannot have fundamentally different directions, hence cannot be linearly independent.

Alternatively, I could delete the first coordinate of each vector and think of the resulting vectors in \(\RR^2\text{.}\) Then I would have three vectors in \(\RR^2\text{,}\) so they are linearly independent, and there exists a way to write one of the vectors as a linear combination of the others: \(u = av + bw\text{.}\) Since all I did was deleted a zero component from each vector, this relationship will still work with the original vectors in \(\RR^3\text{,}\) proving that they are linearly independent.

Again, there are several possible approaches. First, I can think about fundamentally different directions. Since all the vectors in the set are perpendicular to all the other vectors, there is no way that they can share any kind of common direction. This means that they all need a fundamentally new, perpendicular-to-all-others direction. Therefore, they must be a basis.

Any linear subspace has a basis of non-zero vectors. For any linear subspace, take any basis whatever. Then, for every vector \(v\) in that basis, replace \(v\) with \(\frac{1}{|v|} v\text{.}\) This new vector is a scalar multiple of the old, so it points in the same direction. Since the directions have not changed, these new replacement vectors are still a basis: they still all point in fundamentally different directions. But, by dividing by the original lengths, the lengths of these new vectors are all 1. Therefore, I have constructed a basis of vectors of length 1. Since I made no assumption on the linear subspace, I conclude that all linear subspaces have a basis of vectors of length 1.