# Thread: True or false span questions

1. ## True or false span questions

Hi,

1.
I think that it would be false but I can't really tell.

2.
n x (n+1) matrix is merely suggesting that there is one more row than column. I don't see any reason here why Ax = 0 would not be solvable. Therefore the question is true?

- otownsend

2. ## Re: True or false span questions

(1) certainly is true. To say that u is in the span of v1 and v2 means u can be written as a linear combination of v1 and v2, u= av1+ bv2 for some scalars a and b.. Any vector, v, in the span of v1 and v2 can be written v= cv1+ dv2= (c- a)v1+ (d- b)v2+ av1+ bv2= (c- a)v1+ (d- b)v2+ u.

(2). Actually A has one more column than row. That has to be true so that Ax, for x in R^(n+ 1), to exist. But A times the 0 vector is the 0 vector for any A so such an equation has the "trivial solution", x= 0. This is true.

3. ## Re: True or false span questions

(1) but wouldn't it be "redundant" to include the vector u within the span, as in span{v1,v2,u} since it doesn't not add any dimensionality and therefore doesn't communicate anything new. Correct? I guess more what I'm asking here is just because the vectors within the span are linearly dependent doesn't make the span incorrect per say, right?
(2) could you elaborate on what you mean by "This has to be true so that Ax, for x in R^(n+1), to exist"?

4. ## Re: True or false span questions

Originally Posted by otownsend
(1) but wouldn't it be "redundant" to include the vector u within the span, as in span{v1,v2,u} since it doesn't not add any dimensionality and therefore doesn't communicate anything new. Correct?
The is no requirement that the vectors in a spanning set be independent.
So what harm is done by redundancy? This is a spanning set, not a bases.

Originally Posted by otownsend
(2) could you elaborate on what you mean by "This has to be true so that Ax, for x in R^(n+1), to exist"?
Example: $AX = \left( {\begin{array}{*{20}{c}} a&b&c&d \\ e&f&g&h \\ i&j&k&l \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \\ z \\ w \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \\ 0 \end{array}} \right)$

Doesn't that have a solution? The trivial zero vector?

5. ## Re: True or false span questions

Originally Posted by Plato
The is no requirement that the vectors in a spanning set be independent.
So what harm is done by redundancy? This is a spanning set, not a bases.

Example: $AX = \left( {\begin{array}{*{20}{c}} a&b&c&d \\ e&f&g&h \\ i&j&k&l \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \\ z \\ w \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \end{array}} \right)$

Doesn't that have a solution? The trivial zero vector?
P.S. correction of a cut-paste.