Math 2500: Linear Algebra

Subsubsection 3.6.3.1 Other Vector Spaces

Once you understand these terms as they apply to the Euclidean spaces \(\mathbb{R}^n\) and their subspaces (especially those associated to matrices), you should pause to admire your achievement.

But next, realize that those terms apply generally to all sorts of vector spaces! Can you make examples and non-examples in some of these other situtations?

• The set \(\mathcal{P}_3\) of polynomials in \(x\) of degree no more than 3?
• The set \(M_{m,n}\) of \(m\times n\) matrices.
• The set \(\mathcal{C}(\mathbb{R})\) of continuous functions.
• The set of all functions which are solutions to the differential equation \(y'' = y\text{.}\)

Subsubsection 3.6.3.2 Two Methods of Sorting out Linear Independence

We have enough information to collect two ways to answer the question: “Is this set linearly independent?” Well, at least when working with vectors in some Euclidean space \(\mathbb{R}^n\text{.}\)

Given a set of vectors \(\{v_1, v_2, \ldots, v_k\}\) from \(\mathbb{R}^n\text{:}\)

• Form the \(n\times k\) matrix \(A = \left( \begin{smallmatrix} v_1 & v_2 & \dots & v_k\end{smallmatrix}\right)\text{.}\)
• Put \(A\) into reduced row echelon form \(R = \mathrm{rref}(A)\text{.}\) (Really, you only need to go to echelon form, here.)
• Read out pivot columns and free columns of \(R\text{.}\) Those columns of \(A\) which are free columns are linear combinations of previous columns to the left! So, if any column of \(R\) (and \(A\)) is a free columns, the set of vectors is linearly dependent. If all of the columns of \(R\) (and \(A\)) are pivot columns, then the set is linearly independent.

This method is particularly good for identifying which subset of our original set of vectors would form a basis of the vector subspace \(\mathrm{span}(\{v_1, v_2, \ldots, v_k\})\text{.}\)

(We will see the reason for the name soon.) Given a set of vectors \(\{v_1, v_2, \ldots, v_k\}\) from \(\mathbb{R}^n\text{:}\)

• Form the \(k\times n\) matrix \(A = \left( \begin{smallmatrix} v_1^T \\ v_2^T \\ \vdots \\ v_k^T \end{smallmatrix}\right)\text{.}\)
• Put \(A\) into reduced row echelon form \(R = \mathrm{rref}(A)\text{.}\)
• The rows of \(R\) will contain a basis for the vector subspace \(\mathrm{span}(\{v_1, v_2, \ldots, v_k\})\) (written as rows instead of columns, of course). If \(R\) has any zero rows, the original set was linearly dependent, otherwise it was linearly independent.

This method is good at picking out a simple basis of the vector subspace \(\mathrm{span}(\{v_1, v_2, \ldots, v_k\})\text{,}\) but the resulting vectors \emph{probably won't come from your original set}.

Subsubsection 3.6.4.1 Commands for span, dimension, and basis

This first command will construct a vector subspace of \(3\)-space which is spanned by the two vectors we pass in as arguments.

Note that SageMath already gives us some information:

• dimension
• a basis

SageMath has chosen its preferred basis, as usual.

Subsubsection 3.6.4.2 The Algorithms for Checking Linear Independence

SageMath does not have built-in commands with names for checking linear independence or linear dependence. Instead, you have to just use the algorithms.

(Where have we seen that before?) From this, it is clear that \(\left\{ v_1, v_2 \right\}\) is linearly dependent because both rows have pivots.

We work much the same way here.

This tells us that our first two columns are pivot columns, so we should keep those as part of our basis for \(\mathrm{span}\left(v_1, v_2\right)\text{.}\)

Subsubsection 3.6.4.3 Spaces Associated to a Matrix

SageMath knows about the column space and row space associated to a matrix. For this next example, we will work over the ring AA of “algebraic numbers”, so we can include \(\sqrt{2}\text{.}\)

Or we could do this in other ways.