# Proving Linear Independence in Real Vector Spaces

• bugatti79
In summary, the conversation discusses how to show that the span of a set of vectors in a real vector space can also be expressed as the span of another set of vectors, and vice versa. The conversation went through the process of equating the two spans and showing that each vector in one span can be expressed as a linear combination of the vectors in the other span. The final step was to express the original set of vectors as linear combinations of the second set.
bugatti79

## Homework Statement

Let V be a real vector space and {b_1,b_2,b_3,b_4} a linearly independent set of vectors in V

## The Attempt at a Solution

Show that the span $(b_1,b_2,b_3,b_4)=span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$

If I equate the LHS and RHS as

$\alpha_1b_1=+\alpha_1b_1-\alpha_3b_3$ implies $\alpha_3=0$

$\alpha_2b_2=\alpha_2b_2-\alpha_1b_1$ implies $\alpha_1=0$

$\alpha_3b_3=\alpha_3b_3$ but $\alpha_3=0$

$\alpha_4b_4=\alpha_4b_4-\alpha_2b_2$ implies $\alpha_2=0$

This correct? What about $\alpha_4$?

bugatti79 said:

## Homework Statement

Let V be a real vector space and {b_1,b_2,b_3,b_4} a linearly independent set of vectors in V

## The Attempt at a Solution

Show that the span $(b_1,b_2,b_3,b_4)=span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$

If I equate the LHS and RHS as

$\alpha_1b_1=+\alpha_1b_1-\alpha_3b_3$ implies $\alpha_3=0$

$\alpha_2b_2=\alpha_2b_2-\alpha_1b_1$ implies $\alpha_1=0$

$\alpha_3b_3=\alpha_3b_3$ but $\alpha_3=0$

$\alpha_4b_4=\alpha_4b_4-\alpha_2b_2$ implies $\alpha_2=0$

This correct? What about $\alpha_4$?
No, this isn't correct. Let v be an arbitrary vector in $span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$, which means that v is some linear combination of these vectors. Show that this same vector is a linear combination of $(b_1,b_2,b_3,b_4)$.

Now go the other way. Let u be an arbitrary vector in $span(b_1, b_2, b_3, b_4)$. Show that u is also a linear combination of the vectors in the second set.

Mark44 said:
No, this isn't correct. Let v be an arbitrary vector in $span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$, which means that v is some linear combination of these vectors. Show that this same vector is a linear combination of $(b_1,b_2,b_3,b_4)$.

Now go the other way. Let u be an arbitrary vector in $span(b_1, b_2, b_3, b_4)$. Show that u is also a linear combination of the vectors in the second set.

$$(b_1,b_2,b_3,b_4)=\alpha_1\left ( b_1 - b_3 \right )+\alpha_2(b_2-b_1)+\alpha_3(b_3)+\alpha_4(b_4-b_2)$$

Is this the correct start?

thanks

Mark44 said:
No, this isn't correct. Let v be an arbitrary vector in $span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$, which means that v is some linear combination of these vectors. Show that this same vector is a linear combination of $(b_1,b_2,b_3,b_4)$.

Now go the other way. Let u be an arbitrary vector in $span(b_1, b_2, b_3, b_4)$. Show that u is also a linear combination of the vectors in the second set.

bugatti79 said:
$$(b_1,b_2,b_3,b_4)=\alpha_1\left ( b_1 - b_3 \right )+\alpha_2(b_2-b_1)+\alpha_3(b_3)+\alpha_4(b_4-b_2)$$

Is this the correct start?
No. Let v = <v1, v2, v3, v4> be an arbitrary vector in $span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$.

Mark44 said:
No. Let v = <v1, v2, v3, v4> be an arbitrary vector in $span(b_1-b_3,b_2-b_1,b_3,b_4-b_2)$.

Not sure what to do...?

$$(v_1,v_2,v_3,v_4)=\alpha_1\left ( b_1 - b_3 \right )+\alpha_2(b_2-b_1)+\alpha_3(b_3)+\alpha_4(b_4-b_2)$$

bugatti79 said:
Not sure what to do...?

$$(v_1,v_2,v_3,v_4)=\alpha_1\left ( b_1 - b_3 \right )+\alpha_2(b_2-b_1)+\alpha_3(b_3)+\alpha_4(b_4-b_2)$$
Show that v is a linear combination of $(b_1,b_2,b_3,b_4)$.

Now go the other way. Let u be an arbitrary vector in $span(b_1, b_2, b_3, b_4)$. Show that u is also a linear combination of the vectors in the second set.

Mark44 said:
Show that v is a linear combination of $(b_1,b_2,b_3,b_4)$.

but how do I show this? I realize that a vector v is a linear combination of the vectors b_1, b_2, b_3 and b_4 if it can be expressed in the form

$$v=\alpha_1b_1 +\alpha_2b_2+\alpha_3b_3+\alpha_4b_4 +...+\alpha_nb_n$$

Dont know how to proceed.. thanks

Work with the right side of the equation in post #6.

Mark44 said:
Work with the right side of the equation in post #6.

$$V=\left \{ v_1,v_2,v_3,v_4 \right \}=b_1(\alpha_1-\alpha_2)+b_2(\alpha_2-\alpha_4)+b_3(\alpha_3-\alpha_1)+b_4(\alpha_4)$$

$$U=\left \{ u_1,u_2,u_3,u_4 \right \}=\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4$$

I don't know how one would be a linear combination of the other? Thanks

bugatti79 said:
$$V=\left \{ v_1,v_2,v_3,v_4 \right \}=b_1(\alpha_1-\alpha_2)+b_2(\alpha_2-\alpha_4)+b_3(\alpha_3-\alpha_1)+b_4(\alpha_4)$$
$$=(\alpha_1-\alpha_2) b_1 +(\alpha_2-\alpha_4)b_2 + (\alpha_3-\alpha_1)b_3+ (\alpha_4)b_4$$

Doesn't this show that v (not V) is a linear combination of b1, b2, b3, and b4?
bugatti79 said:
$$U=\left \{ u_1,u_2,u_3,u_4 \right \}=\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4b_4$$

I don't know how one would be a linear combination of the other? Thanks

Mark44 said:
$$=(\alpha_1-\alpha_2) b_1 +(\alpha_2-\alpha_4)b_2 + (\alpha_3-\alpha_1)b_3+ (\alpha_4)b_4$$

Doesn't this show that v (not V) is a linear combination of b1, b2, b3, and b4?

Sorry I don get it..Arent we to show that

if $$v=(v_1,v_2...v_n)$$ and $$u=(u_1,u_2...u_n)$$ then span v = span u if and only if v is a linear combination of those in u and vice versa?

Mark44 said:
$$=(\alpha_1-\alpha_2) b_1 +(\alpha_2-\alpha_4)b_2 + (\alpha_3-\alpha_1)b_3+ (\alpha_4)b_4$$

Doesn't this show that v (not V) is a linear combination of b1, b2, b3, and b4?

OK, I think I see what is required. I now need to express the vectors b_1, b_2, b_3 and b_4 as linear combinations of $$(b_1-b_3), (b_2-b_1), (b_3)$$ and $$(b_4-b_2)$$

Hmmmm...how is that done?!

bugatti79 said:
OK, I think I see what is required. I now need to express the vectors b_1, b_2, b_3 and b_4 as linear combinations of $$(b_1-b_3), (b_2-b_1), (b_3)$$ and $$(b_4-b_2)$$

Hmmmm...how is that done?!

From post #2
Mark44 said:
Now go the other way. Let u be an arbitrary vector in $span(b_1, b_2, b_3, b_4)$. Show that u is also a linear combination of the vectors in the second set.

Then u = ?

$$u=\alpha_1(b_1-b_3)+\alpha_2(b_2-b_1)+\alpha_3b_3+\alpha_4(b_4-b_2)$$

I don't know why its coming up like this...I didnt use the strike tags at all...
Mod Note: Fixed LaTeX.

Last edited by a moderator:
OK, so now you have proved what you needed to per your post #1.

Thank you Mark,

I am slowly learning :-)

Actually, should the vector u be represented by beta scalars and v by the alpha scalars?

I don't think that matters. What matters is that for a given vector will have a different set of coordinates for the two bases.

Ok, the last section

Is the span $$\left \{ b_1+b_2,b_2+b_3,b_3 \right \}=\left \{ b_1,b_2,b_3,b_4 \right \}$$

Let $$v=\left \{ v_1,v_2,v_3,v_4 \right \}$$ span $$\left \{ b_1+b_2,b_2+b_3,b_3 \right \}$$

rearranging

$$v=\alpha_1b_1+(\alpha_1+\alpha_2)b_2+(\alpha_2+\alpha_3)b_3$$

Let $$u=\left \{ u_1,u_2,u_3,u_4 \right \}$$ span $$\left \{ b_1,b_2,b_3,b_4 \right \}$$

$$u=\alpha_1b_1+\alpha2b_2+\alpha_3b_3+alpha_4b_4$$

Hence v does not span u because v does not contain $$b_4$$
How do I state this correctly?

I don't know why you are still asking. You have already proved the two parts of this. The second part was in post #14.

No I don't think so. Its another question, ie its a different span we are asked to check...?

OK, I didn't realize this was a different question.

Are b1, b2, b3, and b4 linearly independent?

If so, span{b1, b2, b3, and b4} couldn't possibly be the same as the span of the three vectors you listed on the left side. More specifically, the group of three vectors doesn't include b4.

Yes,

THey are linearly independant. Ok, thanks for the clarification!
bugatti

bugatti79 said:
Ok, the last section

Is the span $$\left \{ b_1+b_2,b_2+b_3,b_3 \right \}=\left \{ b_1,b_2,b_3,b_4 \right \}$$
Better: Is Span{b1 + b2, b2 + b3, b3} = Span{b1, b2, b3, b4}?
bugatti79 said:
Let $$v=\left \{ v_1,v_2,v_3,v_4 \right \}$$ span $$\left \{ b_1+b_2,b_2+b_3,b_3 \right \}$$
Let v = <v1, v2, v3, v4>. What you have written makes no sense. A vector (v) doesn't span a set of other vectors. A set of vectors spans a subspace of some vector space.

If V is the vector space (or subspace) that is spanned by {b1 + b2, b2 + b3, b3}, then v can be written as some specific linear combination of these vectors.

Clearly there's going to be a problem, because your set of vectors doesn't include b4.
bugatti79 said:
rearranging

$$v=\alpha_1b_1+(\alpha_1+\alpha_2)b_2+(\alpha_2+\alpha_3)b_3$$

Let $$u=\left \{ u_1,u_2,u_3,u_4 \right \}$$ span $$\left \{ b_1,b_2,b_3,b_4 \right \}$$

$$u=\alpha_1b_1+\alpha2b_2+\alpha_3b_3+alpha_4b_4$$

Hence v does not span u because v does not contain $$b_4$$
How do I state this correctly?

ok, thanks for clarification of the nonsense I was writing :-)

## What is linear independence in real vector spaces?

Linear independence in real vector spaces refers to a set of vectors that cannot be expressed as a linear combination of each other. In simpler terms, it means that no vector in the set is redundant or can be "created" by adding or multiplying other vectors in the set.

## Why is proving linear independence important?

Proving linear independence is important because it helps us understand the relationships between vectors and determine if they can form a basis for the vector space. This is crucial in many areas of mathematics and physics, such as solving systems of equations and determining the dimensions of a vector space.

## How do you prove linear independence?

To prove linear independence in real vector spaces, we need to show that the only solution to the equation c1v1 + c2v2 + ... + cnvn = 0 is when all the coefficients (c1, c2, ..., cn) are equal to zero. This can be done by setting up a system of equations and solving for the coefficients, or by using other techniques such as the determinant test or the Gram-Schmidt process.

## What are some common methods for proving linear independence?

Besides the determinant test and the Gram-Schmidt process, other common methods for proving linear independence include using the definition of linear independence, using the principle of mathematical induction, and using the rank-nullity theorem.

## Can a set of vectors be both linearly independent and linearly dependent?

No, a set of vectors cannot be both linearly independent and linearly dependent. These two concepts are mutually exclusive. A set of vectors can either be linearly independent or linearly dependent, but not both at the same time.

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