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Let $u,v,w,t \in V$

V - Vector space. Prove or disprove:

$Sp\{u+v-3w,2v-w,t+w,v+w\} = Sp\{u,v,w,t\}$


What i did was to take a linear combination of the right side and compare it, element wise, to the left side:

Let $a_1,a_2,a_3,a_4 \in R$

$$u+v-3w = a_1u$$ $$2v-w = a_2v$$ $$t+w = a_3w$$ $$v+w = a_4t$$

4 equations with 4 variables, i can make this an homogeneous system, so there must be at least 1 solution, so the final answer must be yes, the spans are equall(because there is at least one solution)?

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3 Answers 3

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To show that two spans are equal, it is usually easiest to show that the elements in the basis of the first span can be written as a linear combination of elements in the basis of the second span, and then do the same with the elements of the basis of the second being written as elements of the basis of the first. That may seem like a lot but in your case its not so bad.

To show the elements in the first can be written as a linear combination of the second is pretty trivial since they are all already written as linear combinations of the elements in the second. The tricky part is doing it the other way.

We can do it like this:

$$u = (u + v - 3w) -4/3(2v - w) + 5/3(v+w)$$ $$v = 1/3((2v-w) + (v+w))$$ $$w = -1/3(2v - w) + 2/3(v+w)$$ $$t = (t+w) + 1/3(2v - w) - 2/3(v+w)$$

Now, you might ask how I came up with those, and the trick is to use the Gaussian Elimination technique for inverting matrices. The reason this works is that one basis is the same as the identity basis, so to find the linear combination that gets to the identity is equivalent to taking the inverse.

It looks like you had an idea that was in the right direction, but it isn't always enough to create a system like you've done because sometimes the vectors in the span are not all linearly independent, and sometimes the subspaces they define are not the same, especially in higher dimensions.

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  • $\begingroup$ Can you explain more about the "trick", i mean, the way you came up with those equations that you wrote, using Gaussian Elimination and the connection to inverse? $\endgroup$
    – Alon
    Dec 20, 2019 at 23:37
  • $\begingroup$ Since you have as your second basis u,v,w,t, those can be [1,0,0,0], ..., [0,0,0,1] respectively, stacking them gives the identity Since your other basis is a lin. comb. of these you can write them as [1,1,-3,0],..., [0,1,1,0], and create a matrix A with them as rows. To get [1,0,0,0], (aka u), as a lin. comb. of the four vectors, you can use [1,-4/3,0,5/3] as weights in the lin. comb. which becomes a column in matrix notation. Repeating for v,w,t will give three more columns which make the matrix B. we get by construction AB = I, so B =A^-1. Gauss. Elim. is just how you find the inverse $\endgroup$ Dec 20, 2019 at 23:54
  • $\begingroup$ This technique of inverting the matrix will only work when the number of vectors in both the spanning sets are equal. $\endgroup$
    – Anurag A
    Dec 21, 2019 at 0:52
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What you have done is not correct. To check if the two spans are equal, you have to see if the vectors, $\mathbf{L}_1=u+v-3w, \, \mathbf{L}_2=2v-w, \, \mathbf{L}_3=t+w, \, \mathbf{L}_4=v+w$ can be written as a linear combination of the vectors $\mathbf{R}_1=u, \, \mathbf{R}_2=v, \, \mathbf{R}_3=w, \, \mathbf{R}_4=t$ and vice versa. Observe that $$\mathbf{L}_1=\mathbf{R}_1+\mathbf{R}_2-3\mathbf{R}_3, \,\, \mathbf{L}_2=2\mathbf{R}_2-\mathbf{R}_3, \text{ and so on}$$ Now you have to check if you can do the same by expressing $\mathbf{R}_i$'s in terms of $\mathbf{L}_j$'s. For example, $$\mathbf{R}_2=\frac{1}{3}(\mathbf{L}_2+\mathbf{L}_3).$$


What I showed above was just to motivate you and it may have appeared as trial and error. However we can go about this very systematically as follows: suppose we want to express $\mathbf{R}_1=u$ in terms of $\mathbf{L}_j$'s, so we want to find solutions to the following system: \begin{align*} u & = c_1(u+v-3w)+c_2(2v-w)+c_3(t+w)+c_4(v+w)\\ u & = u(c_1)+v(c_1+2c_2+c_4)+w(-3c_1-c_2+c_3+c_4)+t(c_3)\\ \end{align*} Assuming that $u,v,w,t$ are linearly independent we compare the coefficients to get \begin{align*} c_1 & =1\\ c_1+2c_2+c_4 & =0\\ -3c_1-c_2+c_3+c_4 & =0\\ c_3 & =0 \end{align*} This gives us $c_1=1, c_2=-4/3, c_3=0, c_4=5/3$. Now you can do the same for each of the other vectors $\mathbf{R}_i$s.

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  • $\begingroup$ Thank you, so i need to show that each vector from one space can be written as a combination of the vectors from the other span and vice versa. But that looks a little bit like try and error, there is a technique maybe (for more difficult cases for example)? $\endgroup$
    – Alon
    Dec 20, 2019 at 23:38
  • $\begingroup$ @Alon I have added more to my solution to show how we can find the coefficients systematically and there is no need of trial and error. $\endgroup$
    – Anurag A
    Dec 21, 2019 at 20:01
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Let $A =$ Span {$u+v-3w,2v-w,t+w,v+w$} and $B =$ Span {$u,v,w,t$}.

Here we can use row operations to prove the sets are equal. First adopt a coordinate system: let \begin{align} u &= (1, 0, 0, 0) \\ v &= (0, 1, 0, 0) \\ w &= (0, 0, 1, 0) \\ t &= (0, 0, 0, 1). \\ \end{align} Rewrite $A$ and $B$ as the row space of some matrices: \begin{align} A &= \text{row space } \begin{bmatrix} 1 & 1 & -3 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ \end{bmatrix} \\ B &= \text{row space } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}. \\ \end{align} Row operations do not affect the row space because each row operation is equivalent to multiplication by an invertible matrix. Therefore it is natural to reduce both matrices to RREF, and then compare the spans of the rows.

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