Engineering Analysis/Linear Independence and Basis

Linear Independance

A set of vectors $V = {v_1, v_2, \cdots, v_n}$ are said to be linearly dependant on one another if any vector v from the set can be constructed from a linear combination of the other vectors in the set. Given the following linear equation:

a_1v_1 + a_2v_2 + \cdots + a_nv_n = 0

The set of vectors V is linearly independent only if all the a coefficients are zero. If we combine the v vectors together into a single row vector:

\hat{V} = [v_1 v_2 \cdots v_n]

And we combine all the a coefficients into a single column vector:

\hat{a} = [a_1 a_2 \cdots a_n]^T

We have the following linear equation:

\hat{V}\hat{a} = 0

We can show that this equation can only be satisifed for $\hat{a} = 0$ , the matrix $\hat{V}$ must be invertable:

\hat{V}^{-1}\hat{V}\hat{a} = \hat{V}^{-1}0

\hat{a} = 0

Remember that for the matrix to be invertable, the determinate must be non-zero.

Non-Square Matrix V

If the matrix $\hat{V}$ is not square, then the determinate can not be taken, and therefore the matrix is not invertable. To solve this problem, we can premultiply by the transpose matrix:

\hat{V}^T\hat{V}\hat{a} = 0

And then the square matrix $\hat{V}^T\hat{V}$ must be invertable:

(\hat{V}^T\hat{V})^{-1}\hat{V}^T\hat{V}\hat{a} = 0

\hat{a} = 0

Rank

The rank of a matrix is the largest number of linearly independent rows or columns in the matrix.

To determine the Rank, typically the matrix is reduced to row-echelon form. From the reduced form, the number of non-zero rows, or the number of non-zero columns (whichever is smaller) is the rank of the matrix.

If we multiply two matrices A and B, and the result is C:

AB = C

Then the rank of C is the minimum value between the ranks A and B:

\operatorname{Rank}(C) = \operatorname{min}[\operatorname{Rank}(A), \operatorname{Rank}(B)]

Span

A Span of a set of vectors V is the set of all vectors that can be created by a linear combination of the vectors.

Basis

A basis is a set of linearly-independent vectors that span the entire vector space.

Basis Expansion

If we have a vector $y \in V$ , and V has basis vectors ${v_1 v_2 \cdots v_n}$ , by definition, we can write y in terms of a linear combination of the basis vectors:

a_1v_1 + a_2v_2 + \cdots + a_nv_n = y

\hat{V}\hat{a} = y

If $\hat{V}$ is invertable, the answer is apparent, but if $\hat{V}$ is not invertable, then we can perform the following technique:

\hat{V}^T\hat{V}\hat{a} = \hat{V}^Ty

\hat{a} = (\hat{V}^T\hat{V})^{-1}\hat{V}^Ty

And we call the quantity $(\hat{V}^T\hat{V})^{-1}\hat{V}^T$ the left-pseudoinverse of $\hat{V}$ .

Change of Basis

Frequently, it is useful to change the basis vectors to a different set of vectors that span the set, but have different properties. If we have a space V, with basis vectors $\hat{V}$ and a vector in V called x, we can use the new basis vectors $\hat{W}$ to represent x:

x = \sum_{i = 0}^na_iv_i = \sum_{j = 1}^n b_jw_j

or,

x = \hat{V}\hat{a} = \hat{W}\hat{b}

If V is invertable, then the solution to this problem is simple.

Grahm-Schmidt Orthogonalization

If we have a set of basis vectors that are not orthogonal, we can use a process known as orthogonalization to produce a new set of basis vectors for the same space that are orthogonal:

Given:

\hat{V} = {x_1 v_2 \cdots v_n}

Find the new basis

\hat{W} = {w_1 w_2 \cdots w_n}

Such that

\langle w_i, w_j\rangle = 0\quad\forall i, j

We can define the vectors as follows:

$w_1 = v_1$
$w_m = v_m - \sum_{i = 1}^{m-1}\frac{\langle v_m, u_i\rangle }{\langle u_i, u_i\rangle }u_i$

Notice that the vectors produced by this technique are orthogonal to each other, but they are not necessarily orthonormal. To make the w vectors orthonormal, you must divide each one by its norm:

\bar{w} = \frac{w}{\|w\|}

Reciprocal Basis

A Reciprocal basis is a special type of basis that is related to the original basis. The reciprocal basis $\hat{W}$ can be defined as:

\hat{W} = [\hat{V}^T]^{-1}

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