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I have two sets of vectors, like these:

$v_1 = (1, 6, 4)$
$v_2 = (2, 4, -1)$
$v_3 = (-1, 2, 5)$ in set $V$

$w_1 = (1, -2, 5)$
$w_2 = (0, 8, 9)$ in set $W$

I need to show that $V$ and $W$ span the same subspace. A theorem states that I have to verify that each vector in one set is equal to the linear equation of all the vectors in the others.

I figure there should be a faster way to do this than doing Gaussian operations on five different matrices (matrices with columns made up of the vectors in the equation). I tried to see if using determinants could get me the answer, so I translated this equation into an augmented matrix so I could use Kramer's Rule to find the unknowns:

$k_1(1, 6, 4) + k_2(2, 4, -1) + k_3(-1, 2, 5) = (1, -2, 5)$

And I found out that the matrix of the vectors on the left (technically the denominator of the Kramer equation) is zero, which means there isn't a solution of $(k_1, k_2, k_3)$ in this equation, which shouldn't make any sense because there has to be for the span theorem to be true, which the question tells me it is.

I guess I'm confused about what the determinant coming out to zero means in this context. Does it mean there's no solution, or does it mean the only solution is the trivial solution (0). I'm not sure if I can even use determinants here because the matrix isn't homogeneous, so I don't know if the rules about using determinants to verify independent vectors still applies. If it does apply, and if the solution is trivial, that would mean that the vectors are independent, and that the equation still respects the span theorem, so I can use determinants to verify the theorem. But I'm not sure if that's true or not.

Any help is appreciated

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2 Answers

up vote 0 down vote accepted

Determinant equaling $zero$ just means that there are vectors which cannot be written as a linear combination of your form, but it does not tell you whether your vector is one such. To do that, you might, alas, have to do gaussian elimination.

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Clearly the rank of $\{w_1,w_2\}$ is two. In order to prove $V=W$ show that the rank of $\{v_1,v_2,v_3,w_1,w_2\}$ is $2$.

Edit (for that Cramer-thing): consider $Ax=b$, $A$ a quadratic $n\times n$ matrix, and let $A_i$ denote the matrix where the $i^{\text{th}}$ column of $A$ is replaced by $b$. Then $(x_1,\dots,x_n)^T$ is a solution if $\det(A)x_i=\det(A_i)$ is valid for all $i$. So if in our case $\det(A)=0$ you'll have to show that all $\det(A_i)=0$. I presume you don't want to do this, do you?

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