Problem 14 - Differential Equations and Linear Algebra

Question

Determine all values of the constant \(k\) for which the given set of vectors is linearly independent in \(\mathbb{R}^{4}\). \(\\{(1,0,1, k),(-1,0, k, 1),(2,0,1,3)\\}\)

Step-by-Step Solution

Answer

For the given set of vectors to be linearly independent in \(\mathbb{R}^{4}\), the constant \(k\) must satisfy the condition \(k \neq -1\). Therefore, all values of \(k\) except for \(-1\) will make the set linearly independent.

Step 1: Create matrix from the given vectors

We create a matrix A with given vectors as its columns. \[ A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 0 & 0 \\ 1 & k & 1 \\ k & 1 & 3 \end{pmatrix} \]

Step 2: Perform row reduction

Now, we'll row reduce this matrix to obtain its Row Echelon Form (REF). Row reduction: \(R_2 \leftrightarrow R_3\) and \(R_3 \leftrightarrow R_4\) \[ A = \begin{pmatrix} 1 & -1 & 2 \\ 1 & k & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix} \] \(R_2 \rightarrow R_2 - R_1\) \[ A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & k+1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix} \] \(R_1 \rightarrow R_1 + R_2\) \[ A = \begin{pmatrix} 1 & k & 1 \\ 0 & k+1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix} \]

Step 3: Determine conditions for linear independence

The matrix is now in Row Echelon Form. In order to have linearly independent vectors, there should be no zero rows after row reduction. So our condition is that there should be no zero rows (all rows have at least one nonzero element). We can analyze each row one by one: - Row 1: Always has a nonzero entry in the first element (1), so it is never a zero row. - Row 2: Can be a zero row if \(k = -1\). Since we don't want any zero rows for linear independence, we require \(k \neq -1\). - Row 3: Always a zero row. However, since it's below all other rows with pivots, this doesn't affect the linear independence of the matrix, so we don't have to consider it for the final result. - Row 4: Always has a nonzero entry in the second element (1), so it is never a zero row. We conclude that for the given set of vectors to be linearly independent, \(k\) must satisfy the following condition: \(k \neq -1\) Thus, all values of the constant \(k\) except for \(-1\) will make this set of vectors linearly independent in \(\mathbb{R}^{4}\).

Key Concepts

Row Reduction

Row reduction is a fundamental method used in linear algebra to simplify matrices and solve systems of linear equations. It involves performing a series of elementary row operations on a matrix to transform it into a simpler form. This process helps in identifying key characteristics of the matrix, such as the number and type of solutions a system has.

Here are the primary elementary row operations:

• Swapping two rows (\(R_i \leftrightarrow R_j\))
• Multiplying a row by a nonzero scalar (\(cR_i\) )
• Adding or subtracting the multiple of one row from another (\(R_i + cR_j\rightarrow R_i\) )

During the process of row reduction, you'll often work towards obtaining the Row Echelon Form, which is an intermediary step for interpreting linear dependence or solving systems. By systematically reducing the rows, the matrix becomes easier to work with and essential properties of the system are illuminated.

Row Echelon Form

Row Echelon Form (REF) is a special type of matrix form achieved through row reduction. It reveals readily the rank and possible pivots of a matrix, which are integral in determining solutions to linear systems. When a matrix is in row echelon form, each successive row contains more zeros than the previous one, with leading coefficients (or pivots) to the left of any below them.

A matrix is said to be in Row Echelon Form if:

• Any rows consisting entirely of zeros are at the bottom of the matrix.
• The leading entry (pivot) of a non-zero row is always to the right of the pivot in the row directly above it.
• All pivot positions are populated by 1s, and the columns containing a pivot have all zeros except in the pivot row.

These criteria make REF particularly useful for identifying linear independence. For vectors to be linearly independent, the matrix created from these vectors should be free from zero rows in its REF. Therefore, in our exercise, the vector set in \(\mathbb{R}^4\) was checked for zero rows after transforming the matrix to its row echelon form. Any zero rows would suggest a linear dependency.

Vector Spaces

Vector spaces are a foundational concept in linear algebra, consisting of a set of vectors that can be added together and multiplied by scalars while satisfying specific axioms. These axioms include closure under addition and scalar multiplication, associativity, distributivity, and existence of additive identity and inverses. Understanding vector spaces helps in comprehending the structure of solutions and the nature of linear combinations at a deeper level.

Notable properties of vector spaces include:

• Every vector in a vector space can be expressed as a linear combination of a basis set.
• The dimension of a vector space is defined by the number of vectors in its basis.
• Subspaces of vector spaces retain these properties, allowing for the exploration of more complex structures.

In solving exercises regarding linear independence, understanding the concepts of vector spaces is crucial. Linear independence is a property within vector spaces wherein no vector in the set is a linear combination of others. This is decisive as it determines if vectors span a particular space, or if they can form a basis. By examining whether a matrix has a full row of pivots at REF, we can ascertain the dimensionality of the span and thereby the linear independence of the set of vectors.