Theorem 29.1 - dimensions and rank of a matrix's subspace

• This carries over nicely from Finding bases for a matrix’s subspaces.

Null Space

• If we have a unique solution for $A\vec{x}=0$, it implies that the only $\vec{x}$ that satisfies the equation is the zero vector.
• Visually, that means that only the zero vector lands on the origin after the transformation, all other vectors fall somewhere else.
• If we have a more than one solution, with free variables, then rank $\ne n$, and our set (the null space) ends up with more elements.
• As we get each new free variable, we get more bases as each free var contributes to one entry of the vector.

Col and Row Space

• The theorem makes sense for these two since we use the number of pivots to find the bases. (see Finding bases for a matrix’s column space) for more.
• The columns without the pivots can be expressed as a linear combination of the columns with the pivots.