# work check please... check if a set spans a vector space and linear dependance

• Mar 11th 2011, 03:04 PM
battleman13
work check please... check if a set spans a vector space and linear dependance
Determine whether the set S = {v1,v2,v3} = {(1,-2,2) , (3,-5,3) , (-1,1,1) } spans R^3.

If the set does not span R^3, give a geometrical description of the subspace that it does span.

Here is my work but I think its wrong:

Let a vector u = (u1,u2,u3) in R^3.

Are there scalars c1,c2,c3 such that U = c1u1,c2u2,c3u3 is that system with augmented matrix containing each vector as a column consistent?

Let A = $\begin{array}{ccc}1&3&-1\\-2&-5&1\\2&3&1\end{array}$

Let C = $\begin{array}{ccc}c1\\c2\\c3\end{array}$

Let U = $\begin{array}{ccc}u1\\u2\\u3\end{array}$

Is there a solution C in R^3 to Ac = u for every u in R3 ?

by finding the det A using cofactor expansion I found the determinate to be -24 which is not 0. Therefore Ac = u has a solution in c in R^3 for every u in R^3

That would mean this set spans the subspace?

My notes from class are :

The following three statements are equivalent:

1) A is invertible

2) Ac = u has a solution c in R^n for every u in R^n

3) det A is not equal to 0

My second question:

Determine whether or not the set S = {v1,v2,v3} = {(1,-2,1) , (2,-3,5) , (1,-3,-2)} is linearly dependent or independent. If the set is dependent, then express one of the vectors in the set as a linear combination of the other vectors in the set.

According to my notes the set S = { V1,V2,....Vk } is linearly dependent if the equation c1V1 + c2V2 + .... + ciVk = 0 vector has a solution where not all of ci's are equal to zero.

In laymans terms there needs to be scalars c1,c2,...ck such that when multiplied with v1,v2,...vk you get the zero vector AND at least one of these scalars cannot be 0. The set S can only be independent if the trivial solution is the only solution to
c1V1 + c2V2 + ... + ck-1Vk-1 + ckVk = zero vector

I took my three vector and the zero vector and made the augmented matrix containing v1,v2,v3 and the zero vector as its columns. I then put that system in reduced row echelon form and got the following matrix:

$\begin{array}{cccc}1&0&3&0\\0&1&-1&0\\0&0&0&0\end{array}$

by looking at the matrix you can see that:

c1 + 3(c3) = 0
c2 - c3 = 0

this is our system of linear equations

I then let c3 = t and using back substituion I found c2 = t and c1 = -3t

then letting t = 1 I found c3 = 1, c2 =1, c1 = -3

you can conclude that -3V1 + V2 + V3 = 0

V1 = 1/3(V2) + 1/3(V3)
V2 = 3V1 - V3
V3 = 3V1 - V2

I can conclude that S is linearly dependent.

How does my work look? Thanks in advance
• Mar 11th 2011, 03:23 PM
tonio
Quote:

Originally Posted by battleman13
Determine whether the set S = {v1,v2,v3} = {(1,-2,2) , (3,-5,3) , (-1,1,1) } spans R^3.

If the set does not span R^3, give a geometrical description of the subspace that it does span.

Here is my work but I think its wrong:

Let a vector u = (u1,u2,u3) in R^3.

Are there scalars c1,c2,c3 such that U = c1u1,c2u2,c3u3 is that system with augmented matrix containing each vector as a column consistent?

Let A = $\begin{array}{ccc}1&3&-1\\-2&-5&1\\2&3&1\end{array}$

Let C = $\begin{array}{ccc}c1\\c2\\c3\end{array}$

Let U = $\begin{array}{ccc}u1\\u2\\u3\end{array}$

Is there a solution C in R^3 to Ac = u for every u in R3 ?

by finding the det A using cofactor expansion I found the determinate to be -24 which is not 0. Therefore Ac = u has a solution in c in R^3 for every u in R^3

I got that the determinant is zero by calculating it via the diagonals. Check this.

Tonio

That would mean this set spans the subspace?

My notes from class are :

The following three statements are equivalent:

1) A is invertible

2) Ac = u has a solution c in R^n for every u in R^n

3) det A is not equal to 0

My second question:

Determine whether or not the set S = {v1,v2,v3} = {(1,-2,1) , (2,-3,5) , (1,-3,-2)} is linearly dependent or independent. If the set is dependent, then express one of the vectors in the set as a linear combination of the other vectors in the set.

According to my notes the set S = { V1,V2,....Vk } is linearly dependent if the equation c1V1 + c2V2 + .... + ciVk = 0 vector has a solution where not all of ci's are equal to zero.

In laymans terms there needs to be scalars c1,c2,...ck such that when multiplied with v1,v2,...vk you get the zero vector AND at least one of these scalars cannot be 0. The set S can only be independent if the trivial solution is the only solution to
c1V1 + c2V2 + ... + ck-1Vk-1 + ckVk = zero vector

I took my three vector and the zero vector and made the augmented matrix containing v1,v2,v3 and the zero vector as its columns. I then put that system in reduced row echelon form and got the following matrix:

$\begin{array}{cccc}1&0&3&0\\0&1&-1&0\\0&0&0&0\end{array}$

by looking at the matrix you can see that:

c1 + 3(c3) = 0
c2 - c3 = 0

this is our system of linear equations

I then let c3 = t and using back substituion I found c2 = t and c1 = -3t

then letting t = 1 I found c3 = 1, c2 =1, c1 = -3

you can conclude that -3V1 + V2 + V3 = 0

V1 = 1/3(V2) + 1/3(V3)
V2 = 3V1 - V3
V3 = 3V1 - V2

I can conclude that S is linearly dependent.

How does my work look? Thanks in advance

.
• Mar 11th 2011, 03:39 PM
battleman13
You are correct... the determinant is 0.

I checked it via calculator and then realized I took the wrong entries when I did my first column cofactor expansion. I recorrected it and came out with 0 for an answer.

Thanks!

Now to show the geographical description of the subspace it does span...
• Mar 11th 2011, 09:09 PM
Tinyboss
Find two that are linearly independent, and take their cross product. The subspace spanned is the plane normal to that vector.
• Mar 12th 2011, 04:05 AM
HallsofIvy
Tinyboss's method will work but I don't like it as it is restricted to vectors in $R^3$.

Here's a method more in keeping with battleman13's first idea:

If these three vectors did span $R^3$ then we must have some numbers a, b, c, such that $a(1,-2,2)+ b(3,-5,3) b(-1,1,1)= (u_1, u_2, u_3)$ for any three numbers $u_1, u_2, u_3$. And one way of determining a, b, and c would be to row reduce the augmented matrix
$\begin{bmatrix}1 & 3 & -1 & u_1 \\ -2 & -5 & 1 & u_2 \\ 2 & 3 & 1 & u_3\end{bmatrix}$
where the first three columns are the given vectors.

Start by adding twice the first row to the second and add subrtract twice the first row from the third:
$\begin{bmatrix}1 & 3 & -1 & u_1 \\ 0 & 1 & -1 & u_2+ 2u_1 \\ 0 & -3 & 3 & u_3- 2u_1\end{bmatrix}$

Finally, subract 3 times the second row from the third:
$\begin{bmatrix}1 & 3 & -1 & u_1 \\ 0 & 1 & -1 & u_2+ 2u_1 \\ 0 & 0 & 0 & u_3+ 3u_2+ 4u_1\end{bmatrix}$

The fact that the first three columns of the final row are 0 tells us these vectors do NOT span $R^3$. In order that those equations be solvable, we must also have $u_3+ 3u_2+ 4u_1= 0$. We can solve that for $u_3$: $u_3= -4u_1- 3u_2$. Any vector in that subspace is of the form $(u_1, u_2, u_3)= (u_1, u_2, -4u_1- 3u_2)= u_1(1, 0, -4)+ u_2(0, 1, -3)$ so that the subspace is spanned by those two vectors.

Geometrically, we can replace $u_1$, $u_2$, and $u_3$ with x, y, and z and say that the subspace is the plane z+ 3y+ 4x= 0.