Linear Dependence and Independence

Aug 31, 2020
3 min read
Nov 24, 2020 05:10 UTC
A short piece on linearly dependent and independent sets of vectors.

The following material on linear dependence and independence, although short, is of fundamental importance in this class.

Definitions

A set (collection) of column vectors $\boldsymbol{a}_1, \cdots, \boldsymbol{a}_k$ of the same size are linearly dependent (LD) if there exist real numbers $x_1, \cdots, x_k$, not all zero, such that

$$ \sum_{i=1}^k x_i \boldsymbol{a}_i = \boldsymbol{0} $$

If no such scalars exist, the set is called linearly independent (LIN). The empty set is considered to be linearly independent.

Example 1: $\boldsymbol{a}_1 = (1, 1)^\prime$, $\boldsymbol{a}_2 = (1, 2)^\prime$. We set up a linear combination

$$ \begin{gathered} x_1\boldsymbol{a}_1 + x_2\boldsymbol{a}_2 = \boldsymbol{0} \\
x_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + x_2 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\
\begin{cases} x_1 + x_2 = 0 \\
x_1 + 2x_2 = 0 \end{cases} \Rightarrow x_1 = 0,, x_2 = 0 \end{gathered} $$

Since both coefficients are 0, $\boldsymbol{a}_1$ and $\boldsymbol{a}_2$ are LIN.

Example 2: $\boldsymbol{a}_1 = (1, 0, -1)^\prime$, $\boldsymbol{a}_2 = (2, 0, 1)^\prime$, $\boldsymbol{a}_3 = (1, 1, 2)^\prime$.

$$ \begin{gathered} x_1\boldsymbol{a}_1 + x_2\boldsymbol{a}_2 + x_3\boldsymbol{a}_3 = \boldsymbol{0} \\
\begin{pmatrix} x_1 + 2x_2 + x_3 \\
x_3 \\
-x_1 + x_2 + 2x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\
\Rightarrow x_1 = x_2 = x_3 = 0 \end{gathered} $$

So again the set of vectors are linearly independent.

Example 3: $\boldsymbol{a}_1 = (1, -1, 1)^\prime$, $\boldsymbol{a}_2 = (2, 0, 0)^\prime$, $\boldsymbol{a}_3 = (2, 1, -1)^\prime$.

We can express $\boldsymbol{a}_3$ as $-\boldsymbol{a}_1 + \frac{3}{2}\boldsymbol{a}_2$, so the set of vectors are linearly dependent.

Important facts

Lemma 1: Any set of vectors that include a zero vector is linearly dependent. This can be easily shown by setting all other coefficients to 0 and the coefficient for the zero vector to a non-zero value.

Lemma 2: A set of vectors $\{\boldsymbol{a}_1, \cdots, \boldsymbol{a}_n\}$ is linearly dependent if and only if there exists $\boldsymbol{a}_i$ such that $\boldsymbol{a}_i$ can be expressed as a linear combination of the rest. This is saying that we have some redundancy in the set of vectors.

To prove this, we first show the “if” ($\Leftarrow$) part. Assume that $\boldsymbol{a}_1, \cdots, \boldsymbol{a}_n$ are LD. By definition, there exists $x_1, \cdots, x_n$ that are not all zeros such that

$$ \sum_{i=1}^n x_i \boldsymbol{a}_i = \boldsymbol{0} $$

Let $x_k$ be a non-zero number among $x_1, \cdots, x_n$. We then have

$$ x_1\boldsymbol{a}_1 + x_2 \boldsymbol{a}_2 + \cdots + x_k\boldsymbol{a}_k + \cdots + x_n\boldsymbol{a}_n = \boldsymbol{0} $$

If we divide both sides by $x_k$ and isolate the $k^{th}$ term, we get:

$$ \begin{aligned} \boldsymbol{a}_k &= -\left( \frac{x_1}{x_k} \boldsymbol{a}_1 + \cdots + \frac{x_{k-1}}{x_k} \boldsymbol{a}_{k-1} + \frac{x_{k+1}}{x_k}\boldsymbol{a}_{k+1} + \cdots + \frac{x_n}{x_k}\boldsymbol{a}_n \right) \\
&= y_1\boldsymbol{a}_1 + \cdots + y_{k-1}\boldsymbol{a}_{k-1} + y_{k+1}\boldsymbol{a}_{k+1} + \cdots + y_n\boldsymbol{a}_n, \quad y_i \triangleq -\frac{x_i}{x_k} \end{aligned} $$

Now we show the “only if” part. Suppose we can express some $\boldsymbol{a}_k$ as a linear combination of the rest:

$$ \boldsymbol{a}_k = b_1 \boldsymbol{a}_1 + \cdots + b_{k-1}\boldsymbol{a}_{k-1} + b_{k+1}\boldsymbol{a}_{k+1} + \cdots + b_n\boldsymbol{a}_n $$

Moving the RHS to the LHS,

$$ -b_1 \boldsymbol{a}_1 - \cdots - b_{k-1}\boldsymbol{a}_{k-1} + \boldsymbol{a}_k - b_{k+1}\boldsymbol{a}_{k+1} - \cdots - b_n\boldsymbol{a}_n = \boldsymbol{0} $$

The coefficient for $\boldsymbol{a}_k$ is not zero, thus the set of vectors are LD.

Lemma 3: $\{\boldsymbol{a}_1, \cdots, \boldsymbol{a}_n\}$ are LIN if we can’t express any vector as a linear combination of the others.

Lemma 4: Any set containing a linearly dependent subset is linearly dependent.

Lemma 5: Any subset of a linearly independent set is linearly independent.


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