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I have a vector space $L$, which is a subspace of $\mathbb{R}^4$, spanned by these vectors: $$(4, 1, 1, 2), (2, 3, 1, 0), (-10, 35, 5, -20), (2, 13, 3, -4).$$

I need to find two vectors from $\mathbb{R}^4$ which don't belong to $L$.

I attempted to solve this by firstly reducing the $L$ matrix.

I found that $L$ has a dimension of $2$.

Then, I took the $2$ linearly independent vectors from the reduced matrix: $$(1, 0, -10, -2), (0, 1, 15, 5),$$ and tried finding $3^\text{rd}$ and $4^\text{th}$ vectors with which the $2$ vectors would still be linearly independent.

I found that $2$ such vectors are $$(0, 0, 1, 0) \text{ and } (0, 0, 0, 1).$$

Did I get this right?

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If you want to explain it further please do, so I can accept an answer. –  Tool May 5 '12 at 23:52
    
A bit nitpicky, but saying "the 2 vectors that don't belong to L" (which Henry edited out) is imprecise, as there are several independent vectors that don't belong to $L$. In fact, $(1,0,0,0)$ and $(0,1,0,0)$ also don't belong to $L$. –  TMM May 5 '12 at 23:53

1 Answer 1

up vote 4 down vote accepted

It seems you made a mistake in reducing your matrix. You want to perform row-operations on the matrix (which corresponds to adding multiples of one vector to another) as any linear combination of the vectors that span $L$ is also in $L$. So if you want to perform row-operations, you put them in $A$ as rows (and not as columns):

$$A = \left(\begin{matrix}4 & 1 & 1 & 2 \\ 2 & 3 & 1 & 0 \\ -10 & 35 & 5 & -20 \\ 2 & 13 & 3 & -4\end{matrix}\right).$$

Performing elementary row-operations on $A$, you will again find that $L$ is spanned by two basis vectors only (since the dimension of the rowspace is the same as the dimension of the columnspace). One reduced version of $A$ is given by:

$$A' = \left(\begin{matrix}5 & 0 & 1 & 3 \\ 0 & 5 & 1 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{matrix}\right).$$

Since we have only performed elementary row-operations on $A$ to obtain $A'$, they span the same subspace $L$. This means that $L$ is also spanned by the two vectors:

$$(5, 0, 1, 3) \text{ and } (0, 5, 1, -2).$$

Several vectors in $\mathbb{R}^4$ are not a linear combination of these two vectors, such as the vectors $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$ and $(0,0,0,1)$. Any two of these will do.

So your solution strategy is completely correct, but you made a mistake in reducing the matrix.

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