Find a linearly independent set of vectors that spans the same subspace of R as that spanned by the vectors 日EA A linearly independent spanning set for the subspace is:
Find a linearly independent set of vectors that spans the same subspace of R as that spanned by the vectors 日EA A linearly independent spanning set for the subspace is:
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Problem Statement:**
Find a linearly independent set of vectors that spans the same subspace of \(\mathbb{R}^3\) as that spanned by the vectors
\[
\begin{bmatrix}
3 \\
1 \\
-1
\end{bmatrix}
, \quad
\begin{bmatrix}
3 \\
2 \\
0
\end{bmatrix}
, \quad
\begin{bmatrix}
0 \\
-1 \\
-1
\end{bmatrix}
.
\]
**Solution:**
A linearly independent spanning set for the subspace is:
\[
\left\{
\begin{bmatrix}
\boxed{} \\
\boxed{} \\
\boxed{}
\end{bmatrix}
,
\begin{bmatrix}
\boxed{} \\
\boxed{} \\
\boxed{}
\end{bmatrix}
\right\}
.
\]
**Explanation:**
The original set of vectors provided spans the same subspace of \(\mathbb{R}^3\). However, to find a linearly independent set that also spans this subspace, one must choose vectors such that none can be written as a linear combination of the others. The solution leaves room for finding these values, which should be determined through methods like the row reduction or Gram-Schmidt process.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2192a9f0-1ae9-4ed3-a35b-fc11f1172fc0%2F48151939-f742-4b9a-a309-2760042a399d%2Fujzqq1h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find a linearly independent set of vectors that spans the same subspace of \(\mathbb{R}^3\) as that spanned by the vectors
\[
\begin{bmatrix}
3 \\
1 \\
-1
\end{bmatrix}
, \quad
\begin{bmatrix}
3 \\
2 \\
0
\end{bmatrix}
, \quad
\begin{bmatrix}
0 \\
-1 \\
-1
\end{bmatrix}
.
\]
**Solution:**
A linearly independent spanning set for the subspace is:
\[
\left\{
\begin{bmatrix}
\boxed{} \\
\boxed{} \\
\boxed{}
\end{bmatrix}
,
\begin{bmatrix}
\boxed{} \\
\boxed{} \\
\boxed{}
\end{bmatrix}
\right\}
.
\]
**Explanation:**
The original set of vectors provided spans the same subspace of \(\mathbb{R}^3\). However, to find a linearly independent set that also spans this subspace, one must choose vectors such that none can be written as a linear combination of the others. The solution leaves room for finding these values, which should be determined through methods like the row reduction or Gram-Schmidt process.
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