Question

(True or false.) Let V be an n-dimensional vector space, and let S be a k-dimensional subspace of V, where k < n. If {⃗v1,...,⃗vk} is a basis for S and {⃗vk+1,...,⃗vn} is a set of n − k linearly independent vectors not in S, then {⃗v1,...,⃗vk,⃗vk+1,...,⃗vn} is a basis for V.

Answer 1

its false. but i don't understand why

Answer 2

It is indeed false.

Example:
\(V=\mathbb{R}^3\)

\(v_1=(1,0,0)\)
\(S=\text{span}\ \{v_1\}\)

\(v_2=(1,1,1)\)
\(v_3=(0,1,1)\)

Notice that \({v_2,v_3}\) is linearly independent and \(v_2,v_3 \not\in S\). Yet \({v_1,v_2,v_3}\) is not linearly independent since \(v_1 = v_2 − v_3\). Therefore \({v_1,v_2,v_3}\) is not a basis for \(\mathbb{R}^3\).

Answer 3

Oh right! Thank you

Answer 4

another question:
Also, why doesn't the combined span of a subspace, say V in Rn and its orthogonal complement Vperp, do not span Rn. It says so in my lecture notes but doesn't prove why.

Answer 5

I don't know what you mean by "combined span" but if \(V \subset \mathbb{R}^n\) is a vector subspace if \(\mathbb{R}^n\) then \(\displaystyle V \oplus V^\perp = \mathbb{R}^n\)

Answer 6

from my lecture notes:
"Suppose W a subspace of R . Show by an example that W ∪ W⊥ (The expression W ∪ W⊥ means taking W and W⊥ together) need not be the whole vector space R3

Solution: Simply consider the plane P containing the origin. We have seen that P⊥ is a
line perpendicular to P containing 0. We see that P ∪ P is not all of R3.

Answer 7

From my lecture notes:
Suppose W a subspace of R3 Show that \[W \cup W ^{⊥}\] need not be the whole vector space R .

Solution: Simply consider the plane P containing the origin. We have seen that P is a
line perpendicular to P containing 0. We see that \[P \cup P ^{\perp}\] is not all of R3 .

Answer 8

Yes, that makes sense. Do you understand why?

Answer 9

no.

Answer 10

P u P will be 3 LI vectors, right?

Answer 11

and i thought 3 LI vectors can span R3

Answer 12

You are not taking the span of the vectors in the two spaces. You are simply considering the union.

Answer 13

Think of just gluing together the two spaces.

Answer 14

Oh! ....

Answer 15

Got it! thank you very much

Answer 16

\(P \cup P^\perp\) is just the set of all vectors in \(P\) or \(P^\perp\), the union. I'm sure you know what that means.

Answer 17

The union is not necessarily even a vector space or vector subspace.

Answer 18

so, If there was a question,
Let U={v1,v2} be a plane in R3 and U(perp)={v3} , the set, {v1,v2,v3} spans R3.
This would be true, right?

Answer 19

Yes if you consider the span then it is indeed \(\mathbb{R}^3\). However, let's look at a picture of the union.