MATH-M 303 Lecture Notes - Lecture 16: Gaussian Elimination, Linear Combination, If And Only If
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M303 section 4. 2 notes- null spaces, column spaces, and linear maps. Matrices and linear maps give many examples of subspaces. =[(cid:882)(cid:882): so (cid:2203) (cid:1873)(cid:1864, theorem 2- if is (cid:1865) (cid:1866) matrix, then (cid:1873)(cid:1864) is a subspace of (cid:3041) Condition to be in (cid:1873)(cid:1864) : (cid:2206)=(cid:2777) Let (cid:2206),(cid:2207) (cid:1873)(cid:1864) , so (cid:2206)=(cid:2777)=(cid:2207: (cid:4666)(cid:2206)+(cid:2207)(cid:4667)=(cid:2206)+(cid:2207)=(cid:2777)+(cid:2777)=(cid:2777, so (cid:2206)+(cid:2207) (cid:1873)(cid:1864) Let (cid:1855) : (cid:4666)(cid:1855)(cid:2206)(cid:4667)=(cid:1855)(cid:4666)(cid:2206)(cid:4667)=(cid:1855)(cid:2777)=(cid:2777, so (cid:1855)(cid:2206) (cid:1873)(cid:1864) (cid:1873)(cid:1864) is a subspace of (cid:3041: ex. Let be set of vectors (cid:4666)(cid:1853),(cid:1854),(cid:1855),(cid:1856)(cid:4667) (cid:2872) such that (cid:1853) (cid:884)(cid:1854)+(cid:887)(cid:1855)=(cid:1856) and (cid:1855) (cid:1853)=(cid:1854). Set in question is null space of =[(cid:883) (cid:884) (cid:887) (cid:883) Thus is subspace of (cid:2872: (cid:1873)(cid:1864) defined implicitly- easy to determine if vector is in null space, but more work needed to find. Solve (cid:2206)=(cid:2777); turns implicit definition into explicit description of (cid:1873)(cid:1864) Find a spanning set for (cid:1873)(cid:1864) for : Parametric vector form: since free variables can be anything, we can have any linear combination of these.
