Properties of Dimension

Once the dimension of a vector space is known, then the determination of whether or not a set of vectors is linearly independent, or if it spans the vector space, can often be much easier. In this section we will state a workhorse theorem and then apply it to the column space and row space of a matrix. It will also help us describe a super-basis for $\complex{m}$.

1. Why does have the title it does?
2. Why is so surprising ?
3. A= \begin{bmatrix} 1 & -1 & 2 & 8 & 5 \\ 1 & 1 & 1 & 4 & -1 \\ 0 & 2 & -3 & -8 & -6 \\ 2 & 0 & 1 & 8 & 4 \end{bmatrix}

We begin with a useful theorem that we will need later, and in the proof of the main theorem in this subsection. This theorem says that we can extend linearly independent sets, one vector at a time, by adding vectors from outside the span of the linearly independent set, all the while preserving the linear independence of the set.

In the story Goldilocks and the Three Bears, the young girl Goldilocks visits the empty house of the three bears while out walking in the woods. One bowl of porridge is too hot, the other too cold, the third is just right. One chair is too hard, one too soft, the third is just right. So it is with sets of vectors some are too big (linearly dependent), some are too small (they do not span), and some are just right (bases). Here is Goldilocks' Theorem.

There is a tension in the construction of a basis. Make a set too big and you will end up with relations of linear dependence among the vectors. Make a set too small and you will not have enough raw material to span the entire vector space. Make a set just the right size (the dimension) and you only need to have linear independence or spanning, and you get the other property for free. These roughly-stated ideas are made precise by .

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

We now prove one of the most surprising theorems about matrices. Notice the paucity of hypotheses compared to the precision of the conclusion.

This says that the row space and the column space of a matrix have the same dimension, which should be very surprising. It does not say that column space and the row space are identical. Indeed, if the matrix is not square, then the sizes (number of slots) of the vectors in each space are different, so the sets are not even comparable.

It is not hard to construct by yourself examples of matrices that illustrate , since it applies equally well to any matrix. Grab a matrix, row-reduce it, count the nonzero rows or the number of pivot columns. That is the rank. Transpose the matrix, row-reduce that, count the nonzero rows or the pivot columns. That is the rank of the transpose. The theorem says the two will be equal. Every time. Here is an example anyway.

That the rank of a matrix equals the rank of its transpose is a fundamental and surprising result. However, applying we can easily determine the dimension of all four fundamental subspaces associated with a matrix.

There are many different ways to state and prove this result, and indeed, the equality of the dimensions of the column space and row space is just a slight expansion of . However, we have restricted our techniques to applying and then determining dimensions with bases provided by and . This provides an appealing symmetry to the results and the proof.

The theorems in this section about the dimensions of various subspaces associated with matrices can be tested easily in Sage. For a large arbitrary matrix, we first verify , followed by the four conclusions of .