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author | Holden Rohrer <hr@hrhr.dev> | 2021-09-21 17:12:46 -0400 |
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committer | Holden Rohrer <hr@hrhr.dev> | 2021-09-21 17:12:46 -0400 |
commit | 32f4af5f369fa9f0b2988ecad7797f4bec3661c3 (patch) | |
tree | 7ce1c56011914681d6e2ffb5737dcdf1078d3930 /li/05_linear_span | |
parent | b8433c9909bc5d29df16fd3011251a0a214d2b1a (diff) |
notes and homework
Diffstat (limited to 'li/05_linear_span')
-rw-r--r-- | li/05_linear_span | 17 |
1 files changed, 17 insertions, 0 deletions
diff --git a/li/05_linear_span b/li/05_linear_span new file mode 100644 index 0000000..ed5fdf2 --- /dev/null +++ b/li/05_linear_span @@ -0,0 +1,17 @@ + Linear Independence and Linear Span +Solution set for Ax=0, Ax=B, and the dimension of null space/rank of a +matrix. + +Def. linear independence over vector space (V, +, *) +Let v = {v1, ..., vk} be vectors in V. We say v is linearly independent +iff whenever a1v1 + a2v2 + ... akvk = 0, a1 = a2 = ... = ak = 0. + +Where v_i in v in R^n, we can find linear independence by writing it as + +[ v1 | v2 | ... | vk ][a1 a2 ... ak]^T = 0 + +Prove: basis of a vector space has the same dimension regardless of the +vectors. + +Def basis: a linearly independent set of vectors which spans a vector +space. |