generating sets, linear independency?

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Frankenstein143

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The question is whether these sets are generating sets and linearly (in)dependent
I tried to solve a) and b) but I don't know whether I`m right or not, can someone check it?
c) I could not solve, can someone help me?IMG_E1F403F4AEB9-1.png
 
You didn’t check their linear dependency in a), and you didn’t check if the vectors are generators in b).
The way you checked linear dependency in b) is ok but doesn’t generalize well, for a), you should see for what scalar values of c1, c2, c3 does c1v1 + c2v2 + c3v3 equal 0.
as for c), I’m not sure what the (x) means, could you explain?
 
Hey, for c) (x) is a random vector (x1,x2,x3) BUT we cannot chose (a,0,0) because of the Definition
 
You should really ask your professor about this, this is the first time I see this kind of notation. If its a single vector, there is no point in testing the linear dependence, and it can’t generate R3 because dim(R3) = 3. If it’s a set of all vectors which satisfy the condition, I don’t understand why didn’t just put R3\{....}

EDIT
I think I figured it out, they used the (x) to indicate that it’s an ordered set (the same way you denote an ordered triplet (x,y,z) ). Now, I would choose some vectors from that set and see how they behave. Note that there are a lot of vectors in this set, all of R3 excluding (1,0,0), (2,0,0), generally, without (a,0,0), for all real numbers a. Try to find a couple of linearly dependent vectors, and try to express any vector from R3 ( (a,b,c) ) as a linear combination of vectors in H.
That is if I’m right about what the (x) notation represents, you should still ask your prof
 
Last edited:
Frankenstein143 said:
Oh no. I`m not sure whether (x) is a single Vector or not
Click to expand...
I think that it is clear that \(\mathcal{H}\) is the set of all three dimensional column vectors such a vector like
\(\left( {\begin{array}{*{20}{c}} a \\ 0 \\ 0 \end{array}} \right)\) where \(a\in\mathbb{R}\) cannot be in \(\mathcal{H}\)
Is it clear that \(\left\{ {\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 0 \end{array}} \right),\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 0 \end{array}} \right),\left( {\begin{array}{*{20}{c}} 0 \\ 0 \\ 1 \end{array}} \right)} \right\}\subseteq\mathcal{H}~?\) BUT \({\left( {\begin{array}{*{20}{c}}
1 \\
0 \\
0
\end{array}} \right)}\notin \mathcal{H}~?\)
 
In the text, they ask about the family of H. I don't know how you say it in English but it means a family of all vectors in R3 except (a,0,0).
I think it means the whole 3-dimensional space without (a,0,0). But I´m not sure whether H is linearly dependent or not. Because there exist linearly dependent vectors and linearly independent vectors too. For example Vectors (0,1,0) and 0(0,2,0) are linearly dependet Vectors. does it mean that H is linearly dependent?
To be honest, I do not understand why the Vector (a,0,0) was taken out, ist there any sense?

Does it mean, that H cannot generate R3 because it is linearly dependent?
 
What you did (that is not do) in part a was quite rude. You did row operations without stating what you were trying to do! Luckily you did not make any mistakes so I was easily able to figure out what you were doing.
 
Jomo said:
What you did (that is not do) in part a was quite rude. You did row operations without stating what you were trying to do! Luckily you did not make any mistakes so I was easily able to figure out what you were doing.
Click to expand...

It would be helpful if you can write (what I did) it correctly. I just wanted to show that the set F can generate any Vector in R3?
In order to show that Set F is linearly independent, I just put a=b=c=0.
 
row2 -->row2-row1 (just what you did, but right it so others know what you tried to do)
row 3--> row 3 - row1
 
In order to show that Set F is linearly independent, I just put a=b=c=0. No.
The trivial solution, a=b=c=0, will always satisfy av1 + bv2 + cv3 = 0.
The question is whether or not the trivial solution is the only solution to av1 + bv2 + cv3 = 0. If it is not the only solution then we say that v1, v2, and v3 are linearly dependent.
 
But this is the only solution?!
If I want a Zero Vecto I have to put a=b=c=0 and this is only true if c1=c2=c3=0:
c1=1/2(0+0)=0, c2=1/2(0+0)+0, c3=-1/2(0-0)=0
 
Frankenstein143 said:
But this is the only solution?!
If I want a Zero Vecto I have to put a=b=c=0 and this is only true if c1=c2=c3=0:
c1=1/2(0+0)=0, c2=1/2(0+0)+0, c3=-1/2(0-0)=0
Click to expand...
Based on what you wrote, you have c1v1+c2v2+c3v3=0.
To solve that v1, v2 and v3 is a linearly independent set you need to show that c1=c2=c3=0 is the only solution to c1v1+c2v2+c3v3=0. Is it the only solution? As you showed in your last post, it is the only solution.
 
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