Question

Find out whether the given vectors are dependent or independent; if they are dependent, find a linearly independent subset. In other words, find the dimension of the subspace spanned by the vectors, and a basis for it. Write each of the given vectors as a linear combination of the basis vectors. $$ (0,1,1),(-1,5,3),(1,0,2),(2,-15,1) $$

Short answer

The vectors are dependent. The independent subset is \((0,1,1), (-1,5,3), (1,0,2)\) with dimension 3.

Step-by-step solution

Write the vectors as a matrix

First, represent the given vectors as columns of a matrix. The matrix is: \(\begin{pmatrix} 0 & -1 & 1 & 2 \ 1 & 5 & 0 & -15 \ 1 & 3 & 2 & 1 \ \end{pmatrix}\)

Perform Gaussian Elimination

Use Gaussian elimination to transform the matrix to its row-echelon form. Perform row operations to simplify the matrix. \ \ \ \(\begin{pmatrix} 0 & -1 & 1 & 2 \ 1 & 5 & 0 & -15 \ 1 & 3 & 2 & 1 \ \end{pmatrix}\) becomes \ \ \(\begin{pmatrix} 1 & 5 & 0 & -15 \ 0 & -1 & 1 & 2 \ 0 & 0 & 1 & 8 \ \end{pmatrix}\) after appropriate row operations.

Identify the pivots in the row-echelon form

The matrix in row-echelon form shows the pivot positions in columns 1, 2, and 3. There is no pivot in column 4.

Determine Linearly Independent Vectors

The number of pivot columns (3) indicates that three vectors are linearly independent. The basis vectors are the ones corresponding to these pivot columns: \((0,1,1), (-1,5,3), (1,0,2)\). The dimension of the subspace is 3.

Express Each Vector as a Linear Combination

Rewrite each of the original vectors as a linear combination of the basis vectors: \((2,-15,1)\) can be written as: \ \((-2)*(0,1,1) + (-1)*(-1,5,3) + (2)*(1,0,2)\).

Key concepts

Gaussian Elimination

Gaussian elimination is a method used to solve systems of linear equations. It transforms a given matrix into its row-echelon form through a series of row operations. This form makes it easier to identify the pivot positions.

To transform the matrix, you perform operations such as:

• Rows swapping so the first non-zero entry in each row is a pivot.
• Multiplying rows by non-zero constants to make pivots equal to 1.
• Add or subtract multiples of rows to make zeros below and above pivots.

For the vectors \((0,1,1),(-1,5,3),(1,0,2),(2,-15,1)\), the matrix transformation gives us:

\[\begin{pmatrix} 1 & 5 & 0 & -15 \ 0 & -1 & 1 & 2 \ 0 & 0 & 1 & 8 \end{pmatrix} \].
This form helps us in identifying linearly independent vectors.

Basis Vectors

Basis vectors are crucial as they span a vector space and are linearly independent.

After performing Gaussian elimination, you determine the pivot columns. In our matrix:

\[\begin{pmatrix} 1 & 5 & 0 & -15 \ 0 & -1 & 1 & 2 \ 0 & 0 & 1 & 8 \end{pmatrix} \],
the pivots are in columns 1, 2, and 3. So, the basis vectors are
\((0,1,1),(-1,5,3),(1,0,2)\).

Basis vectors are important because they provide the simplest way to span the space. Any vector in this space can be written as a linear combination of these basis vectors, making calculations and vector manipulation much easier.

Subspace Dimension

The subspace dimension gives the number of basis vectors needed to span that space.

For our given problem, after transforming the matrix and identifying the pivot columns, we can determine the dimension of the subspace spanned by the vectors.

The pivot columns are 1, 2, and 3, relating to the basis vectors \((0,1,1),(-1,5,3),(1,0,2)\).
The count of these pivot columns is 3, thus the dimension of the subspace is 3.
This means any vector in this subspace can be represented using a combination of these three basis vectors. Dimension is fundamental in understanding how 'large' or 'complex' a subspace is.

Linear Combinations

A linear combination of vectors involves multiplying each by a constant and adding the results.

In our exercise, we found that the vectors \((0,1,1),(-1,5,3),(1,0,2)\) are linearly independent and form the basis.
The fourth vector \((2,-15,1)\) can be written as a linear combination of these basis vectors.

From the given solution, it is written as:

\( (2,-15,1) = -2(0,1,1) - 1(-1,5,3) + 2(1,0,2) \)
Meaning, we scaled and added the basis vectors to reconstruct the fourth vector.
This concept highlights the flexibility and power of working within a span and simplifies solving vector equations.