Linear Combination - clarifying statement.

[charlies1902]

In my textbook, The vector v is a linear combination of the vectors v1, v2, and v3 if there are scalars c1, c2, and c3, such that v=c1*v1+c2*v2+c3*v3.

So c has to be a "scalar."


To find these c values you can set up the augmented matrix (v1, v2, v3, v) and find the RREF. I'm a little confused. If you have a free variable in in the RREF matrix, that means c1, c2, and c3 aren't "scalars." does that mean v is not a linear combination of v1, v2, v3?
It would seem so from that question.

However in the next section in my textbook, it shows that you can write v as a linear combination of v1, v2, v3 even if you have a free variable in the augmented matrix. Isn't that contradictory?

 

[Dick]

In my textbook, The vector v is a linear combination of the vectors v1, v2, and v3 if there are scalars c1, c2, and c3, such that v=c1*v1+c2*v2+c3*v3.

So c has to be a "scalar."


To find these c values you can set up the augmented matrix (v1, v2, v3, v) and find the RREF. I'm a little confused. If you have a free variable in in the RREF matrix, that means c1, c2, and c3 aren't "scalars." does that mean v is not a linear combination of v1, v2, v3?
It would seem so from that question.

However in the next section in my textbook, it shows that you can write v as a linear combination of v1, v2, v3 even if you have a free variable in the augmented matrix. Isn't that contradictory?

Having a free variable in a scalar doesn't mean it not a scalar. It just means it's not a definite scalar. It could be any number of particular scalars.

 

[Ray Vickson]

In my textbook, The vector v is a linear combination of the vectors v1, v2, and v3 if there are scalars c1, c2, and c3, such that v=c1*v1+c2*v2+c3*v3.

So c has to be a "scalar."


To find these c values you can set up the augmented matrix (v1, v2, v3, v) and find the RREF. I'm a little confused. If you have a free variable in in the RREF matrix, that means c1, c2, and c3 aren't "scalars." does that mean v is not a linear combination of v1, v2, v3?
It would seem so from that question.

However in the next section in my textbook, it shows that you can write v as a linear combination of v1, v2, v3 even if you have a free variable in the augmented matrix. Isn't that contradictory?

If v1, v2 and v3 are linearly independent, and if v is a linear combination of them, the coefficients c1, c2 and c3 will be unique, which means that there will not be any "free variables". However, if v1, v2 and v3 are linearly dependent (and v is a linear combination of them) the scalars will not be unique, so there will be "free variables". There is no contradiction here.

RGV

 

[charlies1902]

Okay thanks. I got another question.

Let S1 be the set of all linear combinations of the
vectors v1, v2, . . . , vk in R^n, and S2 be the set of
all linear combinations of the vectors v1, v2, . . . ,vk, cvk, where c is a nonzero scalar. Show that
S1 = S2.


The book's solution is attached.

I'm not quite sure what it's saying at the end when it says "v∈S1. Therefore S1=S2"

 

[Ray Vickson]

Okay thanks. I got another question.

Let S1 be the set of all linear combinations of the
vectors v1, v2, . . . , vk in R^n, and S2 be the set of
all linear combinations of the vectors v1, v2, . . . ,vk, cvk, where c is a nonzero scalar. Show that
S1 = S2.


The book's solution is attached.

I'm not quite sure what it's saying at the end when it says "v∈S1. Therefore S1=S2"

If v1, v2, ... vn is one set of vectors and u1, u2, ..., un is another, but with u1 = v1, u2 = v2, ..., u_{n-1} = v_{n-1} and u_n = c*v_n, then using either set {u} or {v} we get the same "span" (where the span of a set of vectors is the set of all linear combinations of them). That is all it is saying.

Another way of saying it is that you can re-scale vn without affecting the span. Of course, it then follows that you can re-scale every one of the vectors without affecting the span. This matters sometimes, because it implies that instead of using a particular set of vectors to span a linear space, you can "normalize" them, for example, and still get the same spanned space.

RGV

 

[charlies1902]

If v1, v2, ... vn is one set of vectors and u1, u2, ..., un is another, but with u1 = v1, u2 = v2, ..., u_{n-1} = v_{n-1} and u_n = c*v_n, then using either set {u} or {v} we get the same "span" (where the span of a set of vectors is the set of all linear combinations of them). That is all it is saying.

Another way of saying it is that you can re-scale vn without affecting the span. Of course, it then follows that you can re-scale every one of the vectors without affecting the span. This matters sometimes, because it implies that instead of using a particular set of vectors to span a linear space, you can "normalize" them, for example, and still get the same spanned space.

RGV

We haven't covered span yet in linear algebra, but I remember it from differential equations.

Can I word it this way?
The constants associated with Vk are ck/c*c for the former, and c*ck for the latter. To get from ck/c*c to c*ck, it requires a constant multiple, thus S1=S2?