LetWbethesubspaceofR5 spannedbythefollowingvectors: u1¼ð1;2;1;3;2Þ; u2¼ð1;3;3;5;3Þ; u3¼ð3;8;7;13;8Þ u4¼ð1;4;6;9;7Þ; u5¼ð5;13;13;25;19Þ FindabasisofWconsistingof theoriginalgivenvectors,andfinddimW. FormthematrixMwhosecolumnsarethegivenvectors,andreduceMtoechelonform:

Given the problem, we need to:

1. Find a basis for the subspace \( W \) spanned by the given vectors.
2. Determine the dimension of \( W \), denoted as \( \dim(W) \).
3. Form a matrix \( M \) whose columns are the given vectors, and reduce \( M \) to echelon form to identify the linearly independent columns, which will help find the basis.

Let's break this down:

### Step 1: Form the Matrix \( M \)

We are given five vectors \( u_1, u_2, u_3, u_4, u_5 \) in \( \mathbb{R}^5 \):
\[
u_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \\ 3 \\ 2 \end{pmatrix}, 
u_2 = \begin{pmatrix} 1 \\ 3 \\ 3 \\ 5 \\ 3 \end{pmatrix}, 
u_3 = \begin{pmatrix} 3 \\ 8 \\ 7 \\ 13 \\ 8 \end{pmatrix}, 
u_4 = \begin{pmatrix} 1 \\ 4 \\ 6 \\ 9 \\ 7 \end{pmatrix}, 
u_5 = \begin{pmatrix} 5 \\ 13 \\ 13 \\ 25 \\ 19 \end{pmatrix}
\]
Matrix \( M \) will be formed by taking these vectors as columns:
\[
M = \begin{pmatrix}
1 & 1 & 3 & 1 & 5 \\
2 & 3 & 8 & 4 & 13 \\
1 & 3 & 7 & 6 & 13 \\
3 & 5 & 13 & 9 & 25 \\
2 & 3 & 8 & 7 & 19
\end{pmatrix}
\]

### Step 2: Perform Row Reduction on \( M \)

We now need to reduce this matrix to echelon form to identify the linearly independent columns. This will help us find the basis and dimension of the subspace.

I'll perform the row reduction now.The row-reduced echelon form of the matrix \( M \) is:

\[
\begin{pmatrix}
1 & 1 & 3 & 1 & 5 \\
0 & 1 & 2 & 2 & 3 \\
0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
\]

### Step 3: Identify the Basis and Dimension of \( W \)

From the row-reduced form, we can observe that there are three leading 1's (pivots) in columns 1, 2, and 4. This means that:

- The vectors \( u_1 \), \( u_2 \), and \( u_4 \) form a basis for the subspace \( W \) because their corresponding columns are linearly independent.
- The dimension of \( W \), \( \dim(W) \), is 3, since there are three linearly independent vectors.

### Final Answer:
- A basis for \( W \) is \( \{ u_1, u_2, u_4 \} \).
- The dimension of \( W \) is \( 3 \).

Would you like further details or have any questions?

Here are five follow-up questions to expand understanding:
1. How can we verify that \( u_1, u_2, u_4 \) are linearly independent?
2. What does the rank of a matrix represent in this context?
3. How would the process change if we had more vectors than the dimension of \( \mathbb{R}^5 \)?
4. Can we find the null space of \( M \), and what would that represent?
5. How does echelon form differ from reduced row echelon form?

**Tip**: Row reduction is a powerful tool for determining linear independence and finding bases for subspaces. Always check the pivot positions in the reduced matrix to identify the independent vectors.

Let W be the subspace of R^5 spanned by the following vectors: u1 = (1, 2, 1, 3, 2), u2 = (1, 3, 3, 5, 3), u3 = (3, 8, 7, 13, 8), u4 = (1, 4, 6, 9, 7), u5 = (5, 13, 13, 25, 19). Find a basis of W consisting of the original given vectors, and find dim(W). Form the matrix M whose columns are the given vectors, and reduce M to echelon form.

Explore how to find a basis and the dimension of a subspace W in R^5 spanned by given vectors using matrix row reduction. This problem involves concepts of linear independence, subspaces, and echelon form. Step-by-step solution provided, suitable for college-level students.

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