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Let us say that there is a linear combination $c_1\mathbb{x_1} + c_2\mathbb{x_2} + c_3\mathbb{x_3} + c_4\mathbb{x_4}$.

where $x_k$ is a $n \times 1$ matrix.

How do I figure out the basis of the space generated by this linear combination formally?

For example, if $\mathbb{x_1} = \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \mathbb{x_2}= \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1\end{bmatrix}, \mathbb{x_3} = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \mathbb{x_4} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$, how do I figure out the space of the basis of the linear combination of these matrices?

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    What do you mean by 'basis of the linear combination'? Are you looking for a basis for the space generated by the set $\{x_1,\ldots,x_4\}$?2012-08-29
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    @Sigur yes, I am.2012-08-29
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    First of all, any set with zero vector is linearly dependent, so can not be a basis. So, you have to search among the other 3 vectors the maximal linearly independent set.2012-08-29
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    you don't need $x_1$ and $x_4=x_2+x_3$ so a spanning set is $x_2$ and $x_3$. also $x_2$ and $x_3$ are linearly independent (easy) so the basis is $x_2, x_3$2012-08-29
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    You are very confused, linear algebra will be very difficult until you deal with your confusions. First, you refer to "the basis" when you should write "a basis". Second, the linear combination doesn't generate a space; the space you are interested in is the collection of all those linear combinations, and it is generated by the four things, $x_1,x_2,x_3,x_4$. Finally, you've written "space of the basis" where you mean "basis of the space". It may seem like I'm splitting hairs, but it's crucial when dealing with abstract concepts to use them carefully and correctly.2012-08-29
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    You need to read a chapter from a book or something. Alternatively, since we live in digital times, you can watch Gilbert Strang's lectures on MIT OCW (Search Youtube for: Gilbert Strang linear algebra).2012-08-29

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All you have to do is row reduce! Put everything into one matrix, and get rid of as many rows as you can.

Logically, in a basis, first of all there cannot be the zero vector. Secondly, no vector in a basis can be a linear combination of any other vectors. In your example, $X_4 = X_3 + X_2$, so you can leave out $X_4$. You are left with your basis: $X_2, X_3$

If you want to row reduce with a more complex question, you have the following matrix: $\left[ \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{array} \right]$

Move the zero row to the bottom for convenience.

$\left[ \begin{array}{cccc} 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right]$

Now subtract the first row + the second row from the third row.

$\left[ \begin{array}{cccc} 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$

We cannot row reduce anymore. Your answer is the rows that are not completely 0's. The first row, which maps to $X_2$, and the second row, which maps to $X_3$, is your basis. Remember, for the basis, you should use the original vectors, so the original $X_2$ and $X_3$ (coincidentally they are the same this time).

So your basis is $X_2, X_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}$

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    That is, of course, *a* basis.2012-08-29
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    @GerryMyerson What do you mean? Like other basis can be created through scalar multiplication of those two vectors right? or is there something else?2012-08-29
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    When you write "your basis is..." there's a possibility someone will take this to mean that's the only possible basis for the space in question. But in fact any two vectors $ax_2+bx_3,cx_2+dx_3$ will do, so long as $ad-bc\ne0$. I'm sure you know this, and I wasn't trying to say that what you wrote was wrong - I was just adding a little bit, lest anyone come away with the wrong impression.2012-08-29
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You mean a basis for the linear transformation. The linear combination you mentioned induces a linear operator $T:\mathbb{R}^n\to\mathbb{R}^m$ (in your case $m=1$). Then,

$$ T(\mathbf y)=\mathbf{X}\cdot\mathbf{y} $$

where $\mathbf{X}\in M_{m\times n}(\mathbb{R})$ and $\mathbf{y}\in \mathbb{R}^n$

$$ \mathbf X=\left[ {\begin{array}{cccc} x_1&x_2&\cdots&x_n \end{array} } \right] $$

The space generated by these vectors ($\{\mathbf{x}_i\}_i$) is their span. Formally put, it is the image of $T$:

$$ \operatorname{span}\{c_i\}_{i=1,2,\ldots,n}=\operatorname{im}T $$

Recall the definition of the image of $T$. It is:

$$ \operatorname{im}T = \left\{ \mathbf z| \mathbf z=T(\mathbf y); \mathbf y\in \mathbb{R}^n \right\} $$

which sometimes appears as $T(\mathbb{R}^n)$. The image of $T$ is a linear subspace of $\mathbb{R}^m$. The dimension of the image of $T$ equals the rank of $T$, that is the maximum number of linearly independent columns of $C$. The rank-nullity theorem is a quite important result when working with dimensions of subspaces ($\operatorname{dim}\operatorname{im}T+\operatorname{dim}\operatorname{ker}T=n$.)

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    If I well understood, he is looking for a basis for the space generated by those 4 vectors.2012-08-29
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    @Sigur I updated my answer. I will refine it a bit more, wait...2012-08-29
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    The image of $T$ is the *column* space of the matrix representing $T$, so if you want the image of $T(x)=Ax$ to be the span of the given vectors $x_1,x_2,x_3,x_4$ then you had better let $A$ be the matrix whose columns are those 4 vectors. Maybe that's what you're saying, but you lost me when you switched from the $x_i$ to the $c_i$.2012-08-29
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    @GerryMyerson You are right. I updated my reply to follow the same notation.2012-08-29