Linear independent and combination

In summary: So we have the equations p+ q+ sr^2+ 3t+ 2tr= 0, p+ 4q+ s+ 4t+ 12rt= 0, and q+ sr^2+ 2tr= 0 which we must solve for p, q, s, and t ("r" is just a parameter). Since there is no "p" in the last equation we can subtract the first equation from the second to eliminate p from those and have 3q+ s(1- r^2)+ t+ 10rt= 0 and q+ sr^2+ 2tr= 0.
  • #1
vcb003104
19
0

Homework Statement


let r be an element of R
... 1.... 1 ......r^2.....3 + 2r
u =( 1 )...v = ( 4 )...w = (1 )...b = ( 5 + 12r)
...0.....1......r^2 ...... 2r


(sorry don't know how to type matrices)

1. For which values of r is the set {u, v, w} linearly independent?

2. For which values of r is the vector b a linear combination of u. v, w?

3. For which of these values of r can b be written as a linear combination of u, v and w in more than one way

Homework Equations



(the matrices)

The Attempt at a Solution



So for 1, I reduced the matrix ( u v w ) to become something like this:

_1_1_r^2
(_1_4_1)
0_1_r^2

__1_1_r^2
→ (0_1_r^2)
__0_0_1-4r^2


so for it to be linearly independent 1 - [itex] 4r^2 [/itex] =/= 0

so r =/= [itex]\pm 1/2[/itex]

for part b)

We want something like:

c1(u) + c2(v) + c3(w) = b

I reduced everything and got:


c1 = 3
c2 = 2r - [itex] r^2 [/itex] (2/(1 - 2r))
c3 = [itex]\frac{2(1 + 2r)}{(1 + 2r)(1 - 2r)}[/itex]

Is it alright to say that for it to be a linear comb. r can't = [itex]\pm1/2[/itex]? (Is it correct that I didn't cancel out the 1 + 2r ?)

and I don't really get part C
 
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  • #2
vcb003104 said:

Homework Statement


let r be an element of R
... 1.... 1 ......r^2.....3 + 2r
u =( 1 )...v = ( 4 )...w = (1 )...b = ( 5 + 12r)
...0.....1......r^2 ...... 2r
Using the "tex" and "/tex" tags (in "[" and "]")
"u= \ begin{pmatrix}1 \\ 1 \\ 0 \ end{pmatrix} v= \ begin{pmatrix}1 \\ 4 \\ 1 \ end{pmatrix} w= \ begin{pmatrix}r^2 \\ 1 \\ r^2\ end{pmatrix} b= \ begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\ end{pmatrix}" (without the spaces) will give

[tex]u= \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix} v= \begin{pmatrix}1 \\ 4 \\ 1 \end{pmatrix} w= \begin{pmatrix}r^2 \\ 1 \\ r^2\end{pmatrix} b= \begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}[/tex]


(sorry don't know how to type matrices)

1. For which values of r is the set {u, v, w} linearly independent?

2. For which values of r is the vector b a linear combination of u. v, w?

3. For which of these values of r can b be written as a linear combination of u, v and w in more than one way

Homework Equations



(the matrices)

The Attempt at a Solution



So for 1, I reduced the matrix ( u v w ) to become something like this:

_1_1_r^2
(_1_4_1)
0_1_r^2

__1_1_r^2
→ (0_1_r^2)
__0_0_1-4r^2


so for it to be linearly independent 1 - [itex] 4r^2 [/itex] =/= 0

so r =/= [itex]\pm 1/2[/itex]
The basic definition of "linear independent" says that four vectors u, v, w, and b, will be linearly independent if and only if the only way we can have pu+ qv+ sw+ tb= 0 is if p= q= s= t= 0. Here, for any numbers, p, q, s, and t
[tex]pu+ qv+ sw+ tb= \begin{pmatrix}p \\ p \\ 0 \end{pmatrix}+ \begin{pmatrix}q \\ 4q \\ q \end{pmatrix}+ \begin{pmatrix}sr^2 \\ s \\ sr^2\end{pmatrix}+ \begin{pmatrix}t(3+ 2r) \\ t(5+ 12r) \\ t(2r)\end{pmatrix}= \begin{pmatrix}p+ q+ r^2+ 3t+ 2tr \\ p+ 4q+ s+ 5t+12rt \\ q+ sr^2+ 2tr\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}[/tex]

so we have the equations [itex]p+ q+ sr^2+ 3t+ 2tr= 0[/itex], [itex]p+ 4q+ s+ 4t+ 12rt= 0[/itex], and [itex]q+ sr^2+ 2tr= 0[/itex] which we must solve for p, q, s, and t ("r" is just a parameter). Since there is no "p" in the last equation we can subtract the first equation from the second to eliminate p from those and have [itex]3q+ s(1- r^2)+ t+ 10rt= 0[/itex] and [itex]q+ sr^2+ 2tr= 0[/itex]. Multiply that last equation by 3 and subtract from the previous equation to eliminate q and we have [itex]s(1- 2r^2)+ t(1+ 8r)= 0[/itex]. If either [itex]1- 2r^2= 0[/itex] or [itex]1+ 8r=0[/itex] that will be true without s and t being equal to 0 so those vectors will NOT be "linearly independent". For any other value of r, they will be which also answers (b).

for part b)

We want something like:

c1(u) + c2(v) + c3(w) = b

I reduced everything and got:


c1 = 3
c2 = 2r - [itex] r^2 [/itex] (2/(1 - 2r))
c3 = [itex]\frac{2(1 + 2r)}{(1 + 2r)(1 - 2r)}[/itex]

Is it alright to say that for it to be a linear comb. r can't = [itex]\pm1/2[/itex]? (Is it correct that I didn't cancel out the 1 + 2r ?)

and I don't really get part C
"Writing b as a linear combination of u, v, and w" means writing
[tex]\begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}= x\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}+ y\begin{bmatrix}1 \\ 4 \\ 1 \end{bmatrix}+ z\begin{bmatrix}r^2 \\ 1 \\ r^2\end{bmatrix}= \begin{bmatrix}x+ y+ zr^2 \\ x+ 4y+ z \\ y+ zr^2\end{bmatrix}[/tex]
for some numbers x, y, and z.

That is the same as the equations [itex]3+ 2r= x+ y+ zr^2[/itex], [itex]5+ 12r= x+ 4y+ z[/itex], and [itex]2r= y+ zr^2[/itex].

The question is "for what values of r do those equations have more than one solution for x, y and z?"
 
  • #3
HallsofIvy said:
Using the "tex" and "/tex" tags (in "[" and "]")
"u= \ begin{pmatrix}1 \\ 1 \\ 0 \ end{pmatrix} v= \ begin{pmatrix}1 \\ 4 \\ 1 \ end{pmatrix} w= \ begin{pmatrix}r^2 \\ 1 \\ r^2\ end{pmatrix} b= \ begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\ end{pmatrix}" (without the spaces) will give

[tex]u= \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix} v= \begin{pmatrix}1 \\ 4 \\ 1 \end{pmatrix} w= \begin{pmatrix}r^2 \\ 1 \\ r^2\end{pmatrix} b= \begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}[/tex]
The basic definition of "linear independent" says that four vectors u, v, w, and b, will be linearly independent if and only if the only way we can have pu+ qv+ sw+ tb= 0 is if p= q= s= t= 0. Here, for any numbers, p, q, s, and t
[tex]pu+ qv+ sw+ tb= \begin{pmatrix}p \\ p \\ 0 \end{pmatrix}+ \begin{pmatrix}q \\ 4q \\ q \end{pmatrix}+ \begin{pmatrix}sr^2 \\ s \\ sr^2\end{pmatrix}+ \begin{pmatrix}t(3+ 2r) \\ t(5+ 12r) \\ t(2r)\end{pmatrix}= \begin{pmatrix}p+ q+ r^2+ 3t+ 2tr \\ p+ 4q+ s+ 5t+12rt \\ q+ sr^2+ 2tr\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}[/tex]

so we have the equations [itex]p+ q+ sr^2+ 3t+ 2tr= 0[/itex], [itex]p+ 4q+ s+ 4t+ 12rt= 0[/itex], and [itex]q+ sr^2+ 2tr= 0[/itex] which we must solve for p, q, s, and t ("r" is just a parameter). Since there is no "p" in the last equation we can subtract the first equation from the second to eliminate p from those and have [itex]3q+ s(1- r^2)+ t+ 10rt= 0[/itex] and [itex]q+ sr^2+ 2tr= 0[/itex]. Multiply that last equation by 3 and subtract from the previous equation to eliminate q and we have [itex]s(1- 2r^2)+ t(1+ 8r)= 0[/itex]. If either [itex]1- 2r^2= 0[/itex] or [itex]1+ 8r=0[/itex] that will be true without s and t being equal to 0 so those vectors will NOT be "linearly independent". For any other value of r, they will be which also answers (b). "Writing b as a linear combination of u, v, and w" means writing
[tex]\begin{pmatrix}3+ 2r \\ 5+ 12r \\ 2r\end{pmatrix}= x\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}+ y\begin{bmatrix}1 \\ 4 \\ 1 \end{bmatrix}+ z\begin{bmatrix}r^2 \\ 1 \\ r^2\end{bmatrix}= \begin{bmatrix}x+ y+ zr^2 \\ x+ 4y+ z \\ y+ zr^2\end{bmatrix}[/tex]
for some numbers x, y, and z.

That is the same as the equations [itex]3+ 2r= x+ y+ zr^2[/itex], [itex]5+ 12r= x+ 4y+ z[/itex], and [itex]2r= y+ zr^2[/itex].

The question is "for what values of r do those equations have more than one solution for x, y and z?"

Hi there, but how do I find the ones that have more than one way of writing it?

Do I solve the matrices for [itex]3+ 2r= x+ y+ zr^2[/itex], [itex]5+ 12r= x+ 4y+ z[/itex], and [itex]2r= y+ zr^2[/itex] and make it to have a free parameter so that there are infinite solutions?
 
Last edited:

Related to Linear independent and combination

1. What does it mean for vectors to be linearly independent?

When a set of vectors is linearly independent, it means that no vector in the set can be written as a linear combination of the other vectors. In other words, each vector in the set is unique and contributes something new to the span of the set.

2. How do you determine if a set of vectors is linearly independent?

To determine if a set of vectors is linearly independent, you can use the linear independence test. This involves setting up a system of equations using the coefficients of the vectors and solving for their variables. If the only solution is the trivial solution (all variables equal to 0), then the vectors are linearly independent. If there are other non-trivial solutions, then the vectors are linearly dependent.

3. What is a linear combination of vectors?

A linear combination of vectors is a sum of scalar multiples of those vectors. For example, if we have two vectors v1 and v2 and we multiply v1 by a scalar c1 and v2 by a scalar c2, then their linear combination would be c1v1 + c2v2. This can be extended to any number of vectors.

4. How do you know if a vector can be written as a linear combination of other vectors?

If a vector v can be written as a linear combination of other vectors, it means that v is in the span of those vectors. To determine if this is the case, you can use the spanning set test. This involves setting up a system of equations using the vectors and solving for the variables. If there is at least one solution, then v is in the span of the vectors and can be written as a linear combination of them.

5. Can a set of linearly dependent vectors be linearly independent?

No, a set of linearly dependent vectors cannot be linearly independent. If a set of vectors is linearly dependent, then it means that at least one vector in the set can be written as a linear combination of the others. This violates the definition of linear independence, which requires that no vector can be written as a linear combination of the others.

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