# Thread: Finding the span in v3

1. ## Finding the span in v3

For what values of h is v3 in Span(v1, v2) and for what values of h is {v1, v2, v3} linearly dependent? Justify each answer

v1 = [1 -3 2], v2 = [-3 9 -6], v3 = [5 -7 h]

2. ## Re: Finding the span in v3

Do you have any idea how to solve the problem?

Have you tried anything so far?

3. ## Re: Finding the span in v3

Yes, I've tried reducing it down to echelon form, but I have
{1 0 0}
{3 0 0}
{5 8 10+h}

(Sorry, I don't know how to format the matricies)

4. ## Re: Finding the span in v3

\begin{bmatrix}
1 & -3 & 5 \\
0 & 0 & 8 \\
0 & 0 & 10+h\\
\end{bmatrix}

^^ this

5. ## Re: Finding the span in v3

Originally Posted by Cammy
For what values of h is v3 in Span(v1, v2) and for what values of h is {v1, v2, v3} linearly dependent? Justify each answer
v1 = [1 -3 2], v2 = [-3 9 -6], v3 = [5 -7 h]
Because $-3v_1=v_2$ there is no value of $h$ for which $v_3$ is a multiple of $v_1$.

6. ## Re: Finding the span in v3

Originally Posted by Plato
Because $-3v_1=v_2$ there is no value of $h$ for which $v_3$ is a multiple of $v_1$.
That's how my book states it

7. ## Re: Finding the span in v3

Originally Posted by Cammy
That's how my book states it
Plato is correct. There is no $h$ that will give you a set of linearly independent vectors.

$\left |\begin{pmatrix}1 &-3 &2 \\-3 &9 &-6 \\5 &-7 &h \end{pmatrix}\right | = 0,~\forall h$

8. ## Re: Finding the span in v3

Originally Posted by Cammy
For what values of h is v3 in Span(v1, v2) and for what values of h is {v1, v2, v3} linearly dependent? Justify each answer v1 = [1 -3 2], v2 = [-3 9 -6], v3 = [5 -7 h]

Originally Posted by romsek
Plato is correct. There is no $h$ that will give you a set of linearly independent vectors.

But because $-3v_1=v_2$ it follows that $Span(v_1,v_2)=Span(v_1)$.
There is no value of $h$ for which $v_3\in Span(v_1)$
Therefore for every value of $h$ the set $\{v_1,v_2,v_3\}$ is dependent.

9. ## Re: Finding the span in v3

More fundamentally, vectors v1, v2, and v3 are "linearly independent" if the only way av1+ bv2+cv3= 0 is to have a= b= c. They are "linearly dependent" if there exist values of a, b, and c other than a= b= c= 0 that make that true.

With v1 = [1 -3 2], v2 = [-3 9 -6], v3 = [5 -7 h], that becomes a[1 -3 2]+ b[-3 9 -6]+ c[5 -7 h]= [a- 3b+ 5c -3a+ 9b- 7c 2a- 6b+ hc]= [0 0 0].

So we must have a- 3b+ 5c= 0, -3a+ 9b- 7c= 0, 2a- 6b+ hc= 0.

If we multiply the first equation by 3 and add to the second equation, we get (3- 3)a+ (9- 3)b- (7- 5)c= 6b- 2c= 0. If we multiply the first equation by 2 and subtract from the third equation, we get (2- 2)a- (6- 6)b+ (h- 10)c= (h- 10)c= 0. If h is any number other than 10, h- 10 is not 0 so we can divide by it to get c= 0. Then 6b- 2c= 6b= 0 so b= 0 and the first equation becomes a- 3b+ 5c= a= 0. If h= 10 then c can be any number and there are "non-trivial" solutions.