1
$\begingroup$

I'm self-studying the book Introduction to Applied Linear Algebra – Vectors, Matrices, and Least Squares

Below is an excerpt from the book:

Independence-dimension inequality. If the $n$-vectors $\vec{a_1}, \cdots, \vec{a_k}$ are linearly independent, then $k\leq n$. In words:

A linearly independent collection of $n$-vectors can have at most $n$ elements.

Put another way:

Any collection of $n+1$ or more $n$-vectors is linearly dependent.

Proof of independence-dimension inequality. The proof is by induction on the dimension $n$.First consider a linearly independent collection $\vec{a_1}, \cdots, \vec{a_k}$ of $1$-vectors. We must have $\vec{a_1}\neq 0$. This means that every element $\vec{a_i}$ of the collection can be expressed a multiple $\vec{a_i}=(\vec{a_i}/\vec{a_1})\vec{a_1}$ of the first element $\vec{a_1}$. This contradicts linear independence unless $k=1$.

Next suppose $n\geq 2$ and the independence-dimension inequality holds for dimension $n-1$. Let $\vec{a_1}, \cdots, \vec{a_k}$ be a linearly independentg list of $n$-vectors. We need to show that $k\leq n$. We partition the vectors as

$$ \vec{a}_i= \begin{bmatrix} \vec{b_i} \\ \alpha_i \end{bmatrix}, \qquad i = 1, \ldots, k, $$ where $\vec{b_i}$ is an $(n-1)$-vector and $\alpha_i$ is a scalar.

First suppose that $\alpha_1 = \cdots = \alpha_k=0$. Then the vectors $\vec{b_1}, \cdots, \vec{b_k}$ are linearly independent: $\sum_{i=1}^k\beta_i\vec{b_i}=0$ holds if and only if $\sum_{i=1}^k\beta_i\vec{a_i}=0$, which is only possible for $\beta_1 = \ldots = \beta_k =0$ because the vectors $\vec{a_i}$ are linearly independent. The vectors $\vec{b_1}, \ldots, \vec{b_k}$ form a linearly independent collection of $(n-1)$-vectors. By the induction hypothesis we have $k\leq n-1$, so certainly $k \leq n$.

Next suppose that the scalars $\alpha_i$ are not all zero. Assume $\alpha_j\neq 0$. We define a collection of $k-1$ vectors $\vec{c_i}$ of length $n-1$ as follows:

$$ \vec{c_i} = \vec{b_i} - \frac{\alpha_i}{\alpha_j}\vec{b_j}, \qquad i = 1, \ldots, j-1, \qquad \vec{c_i}=\vec{b_{i+1}} - \frac{\alpha_{i+1}}{\alpha_j}\vec{b_j}, \qquad i = j, \ldots, k-1 $$

These $k-1$ vectors are linearly independent: If $\sum_{i=1}^{k-1} \beta_i \vec{c_i}=0$ then

\begin{equation} \tag{1}\label{eq:1} \sum_{i=1}^{j-1}\beta_i \begin{bmatrix} \vec{b_i} \\ \alpha_i \end{bmatrix} + \gamma \begin{bmatrix} \vec{b_j} \\ \alpha_j \end{bmatrix} + \sum_{i=j+1}^k \beta_{i-1} \begin{bmatrix} \vec{b_i} \\ \alpha_i \end{bmatrix} =0 \end{equation}

with $$ \gamma = -\frac{1}{\alpha_j}(\sum_{i=1}^{j-1}\beta_i\alpha_i + \sum_{i=j+1}^k \beta_{i-1}\alpha_i). $$

Since the vectors $\vec{a_i}=(\vec{b_i}, \alpha_i)$ are linearly independent, the equality $\eqref{eq:1}$ only hold when all the coefficients $\beta_i$ and $\gamma$ are all zero. This in turns implies that the vectors $\vec{c_i}, \cdots, \vec{c_{k-1}}$ are linearly independent. By the induction hypothesis $k-1 \leq n-1$, so we have established that $k \leq n$.

My Question:

After reading some times, the ideas of the proof in my understanding:

  • First prove Independence-dimension inequality holds for $n$-vectors when $n=1$
  • Then proves when $n>=2$ if Independence-dimension inequality holds for $n-1$-vectors, then it holds for $n$-vectors

I stuck with part 2. How the equation $\eqref{eq:1}$ comes from? Especially about the $\gamma$.

$\endgroup$

migrated from stats.stackexchange.com Apr 30 at 17:46

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

0
$\begingroup$

If $\sum_{i=1}^{k-1}\beta_i\vec{c}_i=0$, then we have

$$\sum_{i=1}^{j-1}\beta_i\vec{c}_i + \sum_{i=j}^{k-1}\beta_i \vec{c}_i=0$$

Then we have from definition of $\vec{c}_i$,

$$\sum_{i=1}^{j-1}\beta_i\left(\vec{b}_i-\frac{\alpha_i}{\alpha_j}\vec{b}_j\right) + \sum_{i=j}^{k-1}\beta_i \left(\vec{b}_{i+1}-\frac{\alpha_{i+1}}{\alpha_j}\vec{b}_j\right)=0$$

By shifting index by a place in the second sum, we have

$$\sum_{i=1}^{j-1}\beta_i\left(\vec{b}_i-\frac{\alpha_i}{\alpha_j}\vec{b}_j\right) + \sum_{i=j+1}^{k}\beta_{i-1} \left(\vec{b}_{i}-\frac{\alpha_{i}}{\alpha_j}\vec{b}_j\right)=0$$

Let's explicitly write out the $\vec{b}_j$ term clearly,

$$\sum_{i=1}^{j-1}\beta_i \vec{b}_i-\frac1{\alpha_j}\left(\sum_{i=1}^{j-1}\beta_i\alpha_i + \sum_{i=j+1}^{k}\beta_{i-1} \alpha_{i}\right)\vec{b}_j + \sum_{i=j+1}^{k}\beta_{i-1}\vec{b}_i=0$$

We let the coefficient of $\vec{b}_j$ be known as $\gamma$,

Hence $\gamma = -\frac1{\alpha_j}\left(\sum_{i=1}^{j-1}\beta_i\alpha_i + \sum_{i=j+1}^{k}\beta_{i-1} \alpha_{i}\right)$ and we have

$$\sum_{i=1}^{j-1}\beta_i \vec{b}_i+\gamma\vec{b}_j + \sum_{i=j+1}^{k}\beta_{i-1}\vec{b}_i=0\tag{2}$$

Also check that

\begin{align}&\sum_{i=1}^{j-1}\beta_i \alpha_i+\gamma\alpha_j + \sum_{i=j+1}^{k}\beta_{i-1}\alpha_i\\&=\sum_{i=1}^{j-1}\beta_i \alpha_i -\frac1{\alpha_j}\left(\sum_{i=1}^{j-1}\beta_i\alpha_i + \sum_{i=j+1}^{k}\beta_{i-1} \alpha_{i}\right)\alpha_j + \sum_{i=j+1}^{k}\beta_{i-1}\alpha_i=0 \tag{3}\end{align}

Equation $(1)$ is just concatenation of equation $(2)$ and equation $(3)$.

The expression might seems cryptic to you, what we usually do in developing proof is something called working backward, that is on a draft paper, we construct the expression of $\gamma$ from equation $(3)$ first.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.