Question

Can R^4 be spanned by 3 linearly independent vectors?
Please, explain me

Answer 1

No, it can't. You'd need 4 vectors.

Answer 2

must be? what if the 4th one is a combination of the given 2?

Answer 3

If the 4th vector is a linear combination of the other 3, then it is already part of the span made by the other 3.

Answer 4

please, wait few minutes, I scan the problem which drove me crazy and its answer from the book.

Answer 5

Problem 25

Answer 6

at the second page, it is section4.5

Answer 7

Basically create a matrix and find out if the null space is 0 or something like that.

Answer 8

create a matrix from the vectrors.

Answer 9

Reduce it to an upper echelon matrix.

Answer 10

If it has all 4 pivots, then they're linearly independent and thus span the whole thing.

Answer 11

I did and the set of linear independent vectors is just 3 vectors as shown in answer paper. However, in my mind, just 3 vectors cannot span a 4 dimensions space. but they say yes, you see?

Answer 12

I see 4 vectors on problem 25

Answer 13

to previous problems, all are ok, since they give out 4 vectors and original vector space is R^3, so the dependent vector can be get rid of and the leftover span the original space,too. this case is weird

Answer 14

There are 4 vectors man.

Answer 15

yup, and they ask us whether the linearly independent vectors ( the leftover after you get rid of the dependent vector if there is) spans the original vector space or not? take reduce echelon form , i got 3 linear independent vectors as what the answer book says.

Answer 16

and those 3 cannot span R^4. I think exactly you think. but they say yes.??? look at the answer page, they say yes "one possible answer for those 3 vectors"

Answer 17

So you're saying that the 4th vector is a linear combination of the other 3?

Answer 18

hey, friend, I ask you because I don't understand. heheheh... you ask me back like this , how can I answer

Answer 19

\[
\begin{bmatrix}
1&2&1&2\\
1&-1&1&-1\\
-1&3&2&2\\
1&1&1&1
\end{bmatrix}
\]

Answer 20

the reduce echelon form of that matrix is