Question

why would the following solution space have no basis:

2x_1+x_2+3x_3=0
x_1 + 0 +5x_3=0
0 + x_2+x_3=0

I know it reduces to
1 0 0
0 1 0
0 0 1

but why does that mean it has no basis? I would think that a 3x3 identity matrix of 3 vectors i, j ,k can span R^3

Answer 1

hey can you help me out later?

Answer 2

it is a full rank matrix so Ax=0 will be having only the trivial solution : (0, 0, 0)

Answer 3

what do you know about the nullspace of a matrix ?

Answer 4

Im not too sure what youre trying to ask... I would say that a nullspace is a homogeneous system?

Answer 5

nullspace is one of the four fundamenal spaces of a matrix, it is a set of all the solutions of Ax=0

Answer 6

Ah I see so the solutions to the Ax=0 matrix would be a nullspace. My book doesn't talk about nullspace, more like solution space / basis ( it interchanges them).

Answer 7

"solution space" and "basis" are two different things

Answer 8

basis of a vector space is just a set of linearly independent vectors whose linear combinations span the space

for example, the column vectors in 3x3 identity matric form a "basis" for the vector space R^n

Answer 9

Nvm what I just said, it asked me to find the basis of a solution space*

Answer 10

Then, if I solve for a solution space and it gives a 3x3 identity matrix, shouldn't it give me the basis of the solution space too?

Answer 11

those vectors in 3x3 identity matrix form a basis of "column space"

Answer 12

full rank just tells you that you can reach any vector in R^3 by taking linear combinations of the column vectors :

Ax = b has an unique solution for all vectors b

Answer 13

the homogeneous system Ax = 0 will have a trivial solution only so the dimension of nullspace would be 0 and we say there is "no basis"

Answer 14

I don't see why you're talking about Ax=b. I mean a \solution space will always be homogeneous no?

Answer 15

\[\begin{bmatrix}2&1&3\\1&0&5\\0&1&1\end{bmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}
= \begin{pmatrix}0\\0\\0\end{pmatrix}\]


equivalantly, you could also say that the column vectors are linearly independent, so the above system will only have a trivial solution : x1 = x2 = x3 = 0

Answer 16

Notice, the solution space of Ax = 0 will not span R^n. it only contains the zero vector.

Answer 17

OH I see, if it is linearly independent in means the only coordinate vector would be (0,0,0) so it would only give a dimension of 0 which doesn't have a basis