Why This Question Came Up
While studying the nullspace in Section 3.2, an important question appeared:
Why does linear algebra test dependence by asking whether a weighted combination of vectors equals the zero vector?
The connection is
\[ x_1v_1+x_2v_2+\cdots+x_nv_n=0. \]This equation is not asking us to make vectors cancel for its own sake. It is a test for whether there is a hidden relationship or redundancy among the vectors.
First: Can One Vector Always Be Produced From the Others?
No. Arbitrary real or complex coefficients do not automatically allow a set of vectors to produce every other vector.
Consider three vectors in \(\mathbb R^3\):
\[ a_1=\begin{bmatrix}1\\0\\0\end{bmatrix},\qquad a_2=\begin{bmatrix}0\\1\\0\end{bmatrix},\qquad a_3=\begin{bmatrix}0\\0\\1\end{bmatrix}. \]Ask whether \(a_3\) can be produced from \(a_1\) and \(a_2\):
\[ c_1a_1+c_2a_2=a_3. \]The left side is always
\[ \begin{bmatrix}c_1\\c_2\\0\end{bmatrix}. \]But the required vector is
\[ \begin{bmatrix}0\\0\\1\end{bmatrix}. \]The third component would require
\[ 0=1, \]which is impossible.
Even if \(c_1\) and \(c_2\) are complex numbers, the third component of the combination remains zero. Therefore \(a_3\) cannot be produced by \(a_1\) and \(a_2\).
The Missing Idea Is Span
The combinations
\[ c_1a_1+c_2a_2 \]produce only vectors belonging to the span of \(a_1\) and \(a_2\):
\[ \operatorname{span}\{a_1,a_2\}. \]For the previous example, this span is the entire \(xy\)-plane:
\[ \left\{\begin{bmatrix}x\\y\\0\end{bmatrix}:x,y\in\mathbb R\right\}. \]There are infinitely many such vectors, but they still do not fill all of \(\mathbb R^3\). The vector \(a_3=(0,0,1)\) lies outside that plane.
This gives the important rule:
\[ \boxed{a_3\text{ can be produced from }a_1,a_2\iff a_3\in\operatorname{span}\{a_1,a_2\}.} \]Why Three Vectors in R2 Are Different
Suppose
\[ a_1=\begin{bmatrix}1\\0\end{bmatrix},\qquad a_2=\begin{bmatrix}0\\1\end{bmatrix}. \]These two vectors already span all of \(\mathbb R^2\), because
\[ c_1a_1+c_2a_2=\begin{bmatrix}c_1\\c_2\end{bmatrix}. \]Therefore any third vector
\[ a_3=\begin{bmatrix}p\\q\end{bmatrix} \]can be written as
\[ a_3=pa_1+qa_2. \]So once two independent vectors already span \(\mathbb R^2\), any third vector is redundant. This is why Strang states that three vectors in \(\mathbb R^2\) cannot be linearly independent.
What Does Redundant Mean Mathematically?
Suppose one vector can be produced from the others:
\[ a_3=2a_1+5a_2. \]Then move everything to one side:
\[ 2a_1+5a_2-a_3=0. \]This is the same as
\[ 2a_1+5a_2+(-1)a_3=0. \]The coefficient vector is
\[ \begin{bmatrix}2\\5\\-1\end{bmatrix}. \]These coefficients are not all zero. Therefore we have found a nontrivial zero combination.
This does not mean every weighted combination equals zero. It means:
There exists at least one choice of coefficients, not all zero, whose weighted vector sum is zero.
Why Use the Zero Vector?
The zero vector provides a standard way to expose a relationship among the vectors.
If
\[ a_3=2a_1+5a_2, \]then
\[ 2a_1+5a_2-a_3=0. \]The zero equation reveals that the vectors are not providing completely independent information.
More generally, suppose two different combinations produce the same vector \(b\):
\[ 2a_1+3a_2=b \]and
\[ 5a_1+a_2=b. \]Subtract the equations:
\[ (2a_1+3a_2)-(5a_1+a_2)=b-b, \]which gives
\[ -3a_1+2a_2=0. \]So a hidden relationship between different representations naturally becomes a zero combination.
Strang’s Formal Definition
Strang formally introduces this idea in Chapter 3, Section 3.4, Independence, Basis and Dimension.
A sequence of vectors is linearly independent when
\[ x_1v_1+x_2v_2+\cdots+x_nv_n=0 \]is possible only when
\[ x_1=x_2=\cdots=x_n=0. \]If the zero vector can be produced with coefficients that are not all zero, the vectors are linearly dependent.
Connection to the Nullspace
Now put the vectors into the columns of a matrix:
\[ A=\begin{bmatrix}|&|&|\\a_1&a_2&a_3\\|&|&|\end{bmatrix}. \]Let
\[ x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}. \]Then matrix multiplication means
\[ Ax=x_1a_1+x_2a_2+x_3a_3. \]Therefore
\[ Ax=0 \]is exactly the same question as
\[ x_1a_1+x_2a_2+x_3a_3=0. \]So the nullspace searches for coefficient vectors that reveal relationships among the columns.
If
\[ N(A)=\{0\}, \]then no nonzero coefficient vector produces zero, and the columns are independent.
If there is some
\[ x\neq0 \]with
\[ Ax=0, \]then the columns are dependent.
Thus:
\[ \boxed{N(A)=\{0\}\iff\text{columns of }A\text{ are linearly independent}} \]and
\[ \boxed{N(A)\text{ contains a nonzero vector}\iff\text{columns are linearly dependent}.} \]The Mental Picture
The most useful chain to remember is:
\[ \text{one vector can be produced from the others} \] \[ \Downarrow \] \[ \text{one vector is redundant} \] \[ \Downarrow \] \[ \text{there exists a nonzero coefficient combination giving }0 \] \[ \Downarrow \] \[ Ax=0\text{ has a nonzero solution} \] \[ \Downarrow \] \[ N(A)\neq\{0\} \] \[ \Downarrow \] \[ \text{columns are linearly dependent}. \]The crucial word is exists. Dependence does not mean every combination gives zero. It means at least one nontrivial combination gives zero.
Where This Fits in Our Study
We encountered this connection while studying Section 3.2 because nonzero solutions of \(Ax=0\) naturally raised the question of what those solutions mean for the columns of \(A\).
However, Strang formally develops linear independence later in Section 3.4. For now, the useful connection is simply:
\[ \boxed{\text{a nonzero nullspace vector records a dependence relation among the columns}.} \]Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Introduction to Linear Algebra, 5th Edition — Chapter 3, Section 3.4