Let U = -5 And A = 4 7 5. Is U In The Subset Of R3 Spanned By The Columns Of A? Why Or Why Not?-6 0 2
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Introduction
Understanding the concepts of vector spaces, subsets, and span is fundamental in linear algebra. When analyzing whether a particular vector belongs to a span generated by a set of vectors, we delve into the core of linear combinations, linear independence, and subspace theory. This article explores the specific question: Given the vector U = -5 and the matrix A with columns [4, 7, 5], is U in the subset of ℝ³ spanned by the columns of A? Why or why not? We will break down the problem systematically, providing detailed explanations of the necessary mathematical principles, step-by-step calculations, and interpretations to arrive at a comprehensive answer.
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Understanding the Problem
The Vectors and Matrices
- Vector U: U = -5 (Note: Since the context involves ℝ³, U should be interpreted as a vector in three dimensions. The given U = -5 suggests a scalar, so it’s important to clarify what this means. It could be that U is meant to be a vector with all components as -5, or a specific component. For this analysis, we will assume U is a vector in ℝ³, possibly U = [-5, -5, -5].)
- Matrix A: A is given with columns [4, 7, 5], [–6, 0, 2].
So, the matrix A in ℝ³×2 form is:
\[
A = \begin{bmatrix}
4 & -6 \\
7 & 0 \\
5 & 2
\end{bmatrix}
\]
The columns are vectors:
\[
\mathbf{a}_1 = \begin{bmatrix} 4 \\ 7 \\ 5 \end{bmatrix}, \quad
\mathbf{a}_2 = \begin{bmatrix} -6 \\ 0 \\ 2 \end{bmatrix}
\]
The Question
- Is U in the span of the columns of A?
In other words, does there exist scalars \( c1 \) and \( c2 \) such that:
\[
c1 \mathbf{a}1 + c2 \mathbf{a}2 = U
\]
- Why or why not?
The answer depends on whether the vector U can be expressed as a linear combination of the columns of A.
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Clarification of U and Its Components
Given the initial statement, U appears to be a scalar, but since the context involves ℝ³, U should be a 3-dimensional vector. To proceed, we interpret U as:
\[
U = \begin{bmatrix} -5 \\ -5 \\ -5 \end{bmatrix}
\]
This interpretation is consistent with the problem structure. If U is intended differently, adjustments may be necessary.
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Mathematical Approach to the Problem
Step 1: Formulate the Linear System
We seek scalars \( c1, c2 \) such that:
\[
c1 \begin{bmatrix} 4 \\ 7 \\ 5 \end{bmatrix} + c2 \begin{bmatrix} -6 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} -5 \\ -5 \\ -5 \end{bmatrix}
\]
This leads to the following system of equations:
\[
\begin{cases}
4c1 - 6c2 = -5 \\
7c1 + 0c2 = -5 \\
5c1 + 2c2 = -5
\end{cases}
\]
Step 2: Solve the System of Equations
- From the second equation:
\[
7c1 = -5 \Rightarrow c1 = -\frac{5}{7}
\]
- Substitute \( c_1 = -\frac{5}{7} \) into the first equation:
\[
4 \left(-\frac{5}{7}\right) - 6 c_2 = -5
\]
\[
-\frac{20}{7} - 6 c_2 = -5
\]
Multiply through by 7 to clear denominators:
\[
-20 - 42 c_2 = -35
\]
\[
-42 c_2 = -35 + 20 = -15
\]
\[
c_2 = \frac{-15}{-42} = \frac{15}{42} = \frac{5}{14}
\]
- Now, verify these values satisfy the third equation:
\[
5 c1 + 2 c2 = -5
\]
\[
5 \left(-\frac{5}{7}\right) + 2 \left(\frac{5}{14}\right) = -5
\]
Calculate each term:
\[
-\frac{25}{7} + \frac{10}{14}
\]
Note that \( \frac{10}{14} = \frac{5}{7} \):
\[
-\frac{25}{7} + \frac{5}{7} = -\frac{20}{7}
\]
Compare to -5:
\[
-\frac{20}{7} \neq -5
\]
Express -5 with denominator 7:
\[
-5 = -\frac{35}{7}
\]
Since \( -\frac{20}{7} \neq -\frac{35}{7} \), the third equation is not satisfied by the values of \( c1 \) and \( c2 \) found from the first two equations.
Step 3: Conclusion from the System
Because the third equation is not satisfied, the system is inconsistent, indicating that no solution exists for the scalars \( c1 \) and \( c2 \). Therefore:
> U is not in the span of the columns of A.
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Why Is U Not in the Span?
Linear Dependence and Span
- The columns of A form a subspace in ℝ³. The span of these columns is the set of all linear combinations of \( \mathbf{a}1 \) and \( \mathbf{a}2 \).
- Since the system has no solution, U cannot be written as a linear combination of the columns of A.
Geometric Interpretation
- The columns \( \mathbf{a}1 \) and \( \mathbf{a}2 \) span a subspace of at most dimension 2 (since there are only two vectors).
- The vector U, with all components -5, is not in this subspace because it lies outside the plane spanned by the two vectors.
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Additional Considerations
Matrix Rank and Column Space
- The rank of matrix A determines the dimension of the column space.
- Computing the rank:
\[
A = \begin{bmatrix}
4 & -6 \\
7 & 0 \\
5 & 2
\end{bmatrix}
\]
- The two columns are not scalar multiples, so they are linearly independent, and the rank of A is 2.
- The column space is a 2-dimensional subspace of ℝ³.
Implication for the Vector U
- Since U is not in the column space, the linear system has no solution, confirming that U is outside the subspace spanned by the columns.
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Summary and Final Verdict
| Aspect | Explanation | Result |
|---------|----------------|---------|
| Vector U | Interpreted as \([-5, -5, -5]\) | Given in ℝ³ |
| Columns of A | \(\mathbf{a}1 = [4,7,5]\), \(\mathbf{a}2 = [-6,0,2]\) | Generate a 2D subspace |
| System solution | No consistent solution found | U not in span |
| Geometric interpretation | U not in the plane spanned by columns | Confirmed |
Therefore, U is not in the subset of ℝ³ spanned by the columns of A because it cannot be expressed as a linear combination of those columns.
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Conclusion
In linear algebra, determining whether a vector belongs to a subspace generated by a set of vectors involves solving a system of linear equations. If the system is inconsistent, as in this case, the vector does not lie within the span. Our detailed analysis shows that the vector U = [-5, -5, -5] cannot be expressed as a linear combination of the columns of matrix A, which span a 2-dimensional subspace of ℝ³. Consequently, U is outside the subspace spanned by A's columns.
Understanding these principles is crucial for various applications, such as solving systems of linear equations, analyzing vector spaces, and performing projections in higher dimensions. This comprehensive approach demonstrates how to methodically analyze similar problems to determine vector inclusion in subspaces.
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Additional Resources
- Linear Algebra Textbooks: For foundational concepts on vector spaces, span, and linear independence.
- Online Calculators: Tools for solving systems of equations.
- Educational Videos: Visual explanations of span and subspace concepts.
