Problem 26 - Differential Equations and Linear Algebra

Question

determine a linearly independent set of vectors that spans the same subspace of \(V\) as that spanned by the original set of vectors. $$V=\mathbb{R}^{3},\\{(3,1,5),(0,0,0),(1,2,-1),(-1,2,3)\\}$$.

Step-by-Step Solution

Answer

A linearly independent set of vectors that spans the same subspace of \(V\) as the original set is \(\{ (3,1,5),(1,2,-1) \}\). We found this by performing Gaussian elimination on the matrix formed by the given vectors and identifying the linearly independent rows, which correspond to the new spanning set.

Step 1: Write the Given Vectors as a Matrix

Let's write the given set of vectors as a matrix. $$ A = \begin{pmatrix} 3 & 1 & 5 \\ 0 & 0 & 0 \\ 1 & 2 & -1 \\ -1 & 2 & 3 \end{pmatrix} $$

Step 2: Perform Gaussian Elimination

Now we'll perform Gaussian elimination to reduce the matrix \(A\) to its reduced row-echelon form: $$ \begin{pmatrix} 3 & 1 & 5 \\ 0 & 0 & 0 \\ 1 & 2 & -1 \\ -1 & 2 & 3 \end{pmatrix} \rightarrow \begin{pmatrix} 3 & 1 & 5 \\ 1 & 2 & -1 \\ 0 & 0 & 0 \\ 0 & 3 & 8 \end{pmatrix} \rightarrow \begin{pmatrix} 3 & 1 & 5 \\ 0 & 3 & 8 \\ 1 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 8/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$

Step 3: Identify Linearly Independent Vectors

Now that we have our reduced row-echelon form, we can identify which rows correspond to linearly independent vectors: $$ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 8/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ The first and second rows are linearly independent, and they correspond to the first and third vectors in our original set.

Step 4: Construct the New Spanning Set with Linearly Independent Vectors

Now we can construct our new spanning set with the linearly independent vectors we found: $$\{ (3,1,5),(1,2,-1) \}$$ This new set of vectors is linearly independent and spans the same subspace of \(V\) as the original set.

Key Concepts

Linearly Independent Vectors

Linearly independent vectors are essential in linear algebra. They are vectors that cannot be written as a combination of other vectors in the set. If you tried to express one vector using others from the same set, you wouldn't succeed without using the vector itself. Here's why they matter:

• They form the basis of a vector space, meaning they can span the entire space without redundancy.
• In our exercise, we looked for a set of such vectors to span the same subspace of \( V \).
• This involves eliminating any vectors that can be expressed as a combination of others (like the zero vector, which offers no information).

By using linearly independent vectors, we ensure our representation of the vector space is efficient, meaning you use the fewest number of vectors required without losing any dimensionality or information about the space itself.

Gaussian Elimination

Gaussian elimination is a systematic method for simplifying a matrix to find solutions to systems of linear equations or to identify key properties, like linear independence. Here's how it works:

• Involves row operations to transform the matrix into row-echelon form and further into reduced row-echelon form.
• In our exercise, it helped in simplifying the matrix to easily spot the independent vectors through row-echelon form.
• Operations include swapping rows, multiplying rows by non-zero constants, and adding/subtracting multiples of rows.

For the given vectors in \( V \), Gaussian elimination allowed us to easily identify which vectors were redundant and which ones truly defined the subspace. It simplifies complex systems and helps in drawing clear conclusions about the characteristics of the vector set.

Subspace

A subspace is a subset of a vector space that is also a vector space itself. Understanding subspaces helps in identifying the essential parts of a vector space. Here’s what you need to know:

• Must satisfy conditions like containing the zero vector, and being closed under vector addition and scalar multiplication.
• In our exercise, once we identified linearly independent vectors within \( V \), we effectively found a subspace.
• Linearly independent vectors form a basis, which spans this subspace, meaning any vector in the subspace can be expressed as a combination of these basis vectors.

By identifying a subspace, you’re capturing an inherent structure within the larger space \( V \). It lays the groundwork for deeper exploration and understanding of vector spaces and their dimensions.