Geometric interpretation of: $(v_1,...,v_n)$ l.i. in a vector space $V$, $w\in V$, then $(v_1+w, ..., v_n+w)$ l.d. implies $w\in$span$(v_1,...,v_n)$.

Question

The following problem is from Chapter 2 of the 2nd edition of Axler's Linear Algebra Done Right

3. Suppose $(v_1,...,v_n)$ is linearly independent in a vector space $V$ and $w\in V$. Prove that if $(v_1+w, ..., v_n+w)$ is linearly dependent, then $w\in \text{span}(v_1, ...,v_n)$.

I can prove this with the definitions of the concepts (and there is another question about how to prove it).

My question is: what is the geometric interpretation for this result?

I'm looking for an example of linearly independent $v_1$ and $v_2$ and a $w$ in span$(v_1,v_2)$ such that $(v_1+w,v_2+w)$ is linearly dependent. The only case I can think of is if $w=-v_1$ or $w=-v_2$, because then we end up with $(0,v_1)$ or $(v_1,0)$.

Here is a depiction that I drew to think about the problem.

Are there any other $w$'s?

Answer 1

There can't be a geometric interpretation in two dimensions, as every possible $w$ is in the span of $v_1$ and $v_2$. In three dimensions, two linearly independent vectors define a plane through the origin, then adding a $w$ to each that does not lie in the plane will give each vector an equal component in a direction perpendicular to the plane. Now for the sum of the new vectors to be zero, the sum of those components and hence the coefficients of the $w$'s must be zero - but that means the sum of the components of the $w$'s in the plane must also be zero, so the sum of the components of the new vectors in the plane is a linear combination of $v_1$ and $v_2$ - which we know can't be zero. I'm not sure if this is a geometrical interpretation, or just an elaborate way of describing the algebra.

Answer 2

You should strengthen the theorem to actually recognize the geometric interpretation. Specifically, we can show the following stronger theorem:

Suppose $v_1, v_2, \ldots, v_n \in V$ are linearly independent. Then $w \in V$ such that $v_1+w, v_2+w, \ldots, v_n+w$ is linearly dependent if and only if $w \in \operatorname{aff span}(-v_1, -v_2, \ldots, -v_n).$

In $V = \mathbb{R}^m$ for $m\geq 2$ and $n=2$, this means that $w$ lies on the line through $-v_1, -v_2$.

Answer 3

Consider the contrapositive (assuming $\{v_1,\cdots,v_n\}$ is LI), generalized:

If $w\not\in\mathrm{span}\{v_1,\cdots,v_n\}$ and $(a_1,\cdots,a_n)\ne\vec{0}$ then $\{v_1+a_1w,\cdots,v_n+a_nw_n\}$ is LI.

$\bullet$ Assumptions: WLOG $V=\mathrm{span}\{v_1,\cdots,v_n,w\}$ and we use the inner product where $\{v_1,\cdots,v_n,w\}$ is orthonormal. Then the spans of $\{v_1,\cdots,v_n\}$ and $\{v_1+a_1w,\cdots,v_n+a_n\}$ are hyperplanes (codimension $1$ subspaces).

$\bullet$ 2D: For the sake of discovering a geometric interpretation, let's start by considering $n=1$ and $\dim V=2$, in particular we might as well pick the vectors $v=[\begin{smallmatrix}1\\0\end{smallmatrix}]$ and $w=[\begin{smallmatrix}0\\1\end{smallmatrix}]$. Then $\mathrm{span}\{v\}$ is the $x$-axis, $\mathrm{span}\{w\}$ is the $y$-axis, and $\mathrm{span}\{v+aw\}$ is a line which is the graph of the linear function $y=ax$. This line is diagonal between the coordinate axes, and the parameter $a$ specifies just how much it is "tilted."

$\bullet$ 3D: Now consider $n=2$ and $\dim V=3$. Let $v_1,v_2,w$ be unit vectors along the $x$-, $y$-, $z$-axes, so $\mathrm{span}\{v_1,v_2\}$ is just the $xy$-plane. If we pick $a_1,a_2=1$ then the new plane $\mathrm{span}\{v_1+w,v_2+w\}$ has (non-orthogonal) axes through $v_1+w$ and $v_2+w$ which are diagonal in the $xz$-plane and $yz$-plane respectively. It is a plane which has been "tilted" - indeed, its normal has been tilted in the opposite direction of $(1,1,0)=a_1v_1+a_2v_2$. (Draw a picture!) In fact, the normal has been sheared (displaced parallel to $\mathrm{span}\{v_1,v_2\}$) ...

$\bullet$ Conclusion: going from $\mathrm{span}\{v_1,\cdots,v_n\}$ to $\mathrm{span}\{v_1+a_1w,\cdots,v_2+a_nw\}$ has the effect of shearing the normal vector from $w$ to the displacement $w-(a_1v_1+\cdots+a_nv_n)$. (Exercise!)

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More generally, given any subspace $V$ of an inner product space, we can write the inner product space as an orthogonal direct sum $V\oplus V^\perp$ (where $V^\perp$ is $V$'s orthogonal complement), and interpret this set-theoretically as a kind of Cartesian product. Then a "generic" subspace $W$ with the same dimension as $V$ is graph of a linear transformation $A:V\to V^\perp$, i.e. is of the form $W=\{v+Av\mid v\in V\}$ for a unique $A$. The transformation $A$ encodes geometrically how $V$ is "tilted" within $V\oplus V^\perp$. I like to call this a "linearized" version of Goursat's lemma from group theory (which classifies subgroups of a direct product $H\times K$ as "graphs" of isomorphisms between subquotients; $A$ is an isomorphism from its cokernel to its image).

(When I say "generic" subspace, I mean if we parametrize all subspaces in some fashion (i.e. the Grassmanian manifold / variety) the set of exceptions form a submanifold / subvariety with measure zero / positive codimension. Because the Grassmanian is very nice, it makes sense to speak in probabilistic terms of choosing a subspace "uniformly at random," and then we can replace "generic" with "almost surely.")