Basis vectors belonging to a subspace

Question

Let's suppose that we have $n$-dimensional vector space with a known complete basis $e_1,e_2, .. e_n$ and some $k$-dimensional subspace ( $k<n$ ) with basis $v_1,v_2, .. v_k$ . Intuitively it seems to be true that maximally $k$ of vectors $e_i$ can belong to this subspace, at the worst case not a single $e_i$ vector belong to it , but it is just an intuition..

How to prove the statement that this number of vectors is between $0$ and $k$? ( consequently the number of vectors $e_i$ not belonging to the subspace is between $n-k$ and $n$.

Answer 1

To prove this you have to note that if you have strictly more that $k$ of the $e_{i}$'s in your subspace spanned by $v_{1}, \ldots , v_{k}$ then the $e_{i}$'s are (still) linearly independent and span the set of all linear combinations of the $e_{i}$'s. So a minimal spanning set for this space contains the number of $e_{i}$'s (which is strictly greater than $k$). However every element of this space can be written as a linear combination of $v_{1}, \ldots , v_{k}$ (since it is a basis) and so a minimal spanning set for this space must have at most $k$ vectors, a contradiction. Thus there there are at most $k$ $e_{i}$'s in the space spanned by $v_{1}, \ldots , v_{k}$.

If I have understood your comment this should answer it. Note that for a particular vector space $V$, particular $e_{i}$'s and particular $v_{1}, \ldots , v_{k}$ then there will be only one specific number of $e_{i}$'s that belong to $v_{1}, \ldots , v_{k}$, and from above paragraph we know this number is between 0 and $k$ inclusive. To show there are cases where each of these values is possible you need to construct examples and show they satisfy your desired conditions. So let $e_{1}, \ldots , e_{n}$ be a basis for $V$. Now pick a $k$ between 0 and $n$. Now follow P Vanchinathan's construction for your $v_{1}, \ldots , v_{k}$. You can then prove that the space spanned by $v_{1}, \ldots , v_{k}$ contains exactly $k$ of $e_{1}, \ldots , e_{n}$ and so you have an example that you require. But since $k$ was arbitrary between 0 and $k$ you will have all the examples you are seeking (ie you have shown for $V$ there are subspaces (notice not a single subspace) that contain exactly 0 or 1 or 2 or ... or n of the given basis vectors).

I am now adding another bit to the answer to answer (hopefully) your latest comment. There are two things/questions you may be thinking

Question A: Can we show these conditions happen at least once in some vector space?

Question B: Given any vector space and any basis of it can we find subspaces such that the conditions hold (this is a much stronger question that question A).

To answer (A) take $R^{n}$ the standard basis $e_{1}, \ldots , e_{n}$ and for your subspaces the zero subspace, subspace spanned by $e_{1}$, the subspace spanned by $e_{1} , e_{2}$ and so on. Notice that whatever subspace a pick to try and answer this question then ANY basis for this subspace can (and has to be) written in terms of the basis I am using for $R^{n}$, since that is how we are describing the elements of $R^{n}$. Notice this answer does not (at first sight) tell me if the result is true if we change the basis for $R^{n}$ that we are using (though you can see it does with a little thought). However it absolutely does not tell us anything about what happens in any other vector space (like $\mathbb{R}_{3}[x]$ say, the set of all real polys with degree less than or equal to 3).

To answer (B): This is what the answer to your first follow-up question does (which notice was not the same as your original question). It is inevitable that the subspace basis vectors have some relationship to the basis vectors of the whole space since for two subspaces of the same dimension they will in general contain different numbers of basis vectors for the whole space. (think of $\mathbb{R}^{3}$ with standard basis and then the two 2d subpaces of the x-y plane and any other plane through the origin other than the z-x or z-y planes)

Hope this has helped

Answer 2

Given $V$ finite-dimensional vector space (over a certain field $K$), let $\{e_1,...,e_n\}$ be a basis for $V$. Consider now $v_1,...,v_k\in V$ linearly independent vectors with $k\leq n$. Define $W:=\mathrm{span}\{v_j\mid j=1,...,k\}$. The claim is $\#\{e_i\mid e_i\in W, 0\leq i\leq n\}\leq k$. If not, then you can find at least $k+1$ idenpendent vectors, say, without loss, the first $k+1$, which belong to $W$: $e_1,...,e_{k+1}\in W$. Then, for each $i=1,...,k+1$, you can write $e_i=\sum_{j=1}^k\alpha_j^i v_j$, since $W=\mathrm{span}\{v_j\mid j=1,...,k\}$. Consider now the matrix $A=[\alpha_j^i]$: it must hold $\mathrm{rk}(A)\leq k$, so that the system $AX=0$ must have a non-trivial solution, which gives the contradiction: the vectors $e_1,...,e_{k+1}$ would be linearly dependent, since we found a linear combination of them giving the null vector.

Answer 3

If $U$ is a subspace of $V$ (all vector spaces are supposed to be finite dimensional), then $\dim U\le \dim V$.

Suppose $e_{i_1},\dots,e_{i_r}$ are $r$ distinct vectors among $e_1,e_2,\dots,e_n$ and that they belong to the span of $\{v_1,\dots,v_k\}$. Since any subset of a linearly independent set is linearly independent, we know that

1. $\operatorname{span}\{e_{i_1},\dots,e_{i_r}\}$ is a subspace of $\operatorname{span}\{v_1,\dots,v_k\}$

2. $\dim\operatorname{span}\{e_{i_1},\dots,e_{i_r}\}=r$

3. $\dim\operatorname{span}\{v_1,\dots,v_k\}=k$

Therefore $r\le k$.

The “worst case” can indeed happen: $$ e_1=\begin{bmatrix}1\\0\end{bmatrix} \quad e_2=\begin{bmatrix}0\\1\end{bmatrix} \quad v_1=\begin{bmatrix}1\\1\end{bmatrix} $$ in $\mathbb{R}^2$. None among $e_1$ and $e_2$ belongs to the span of $\{v_1\}$.