Let S= {v1.., vg} be a set of k vectors in R", with k

Assume A be a m×n matrix. Consider the following statement as given below: For each b in Rm, the…

Answered: Let S= (v1. Vk) be a set of k vectors…

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Let S= (v1. Vk) be a set of k vectors in R", with k

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Linear Algebra

In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied under linear algebra. The first-degree equations involving algebraic expressions are called linear equations. That is, if you represent it by drawing a graph you will get a straight line.

Question

Let S= {v1.., vg} be a set of k vectors in R", with k<n. Use a theorem about the matrix equation Ax = b to explain why S cannot be a basis for R".
equations, this statement is equivalent to what other statements? Choose all that apply.
A. The matrix A has a pivot position in each column.
B. Each b in Rm is a linear combination of the columns of A.
C. The columns of A span Rm.
D. The matrix A has a pivot position in each row.
E. The rows of A span R".
Let A be the nxk matrix [v, v2 .-- Vk] whose columns are the vectors of S. Since A has fewer
of A.
than
there cannot be a pivot position in each
Therefore, by the above theorem, the
of A do not span
Why is the proof complete?
O A. The proof is complete because the subspace spanned by S is RK.
B. The proof is complete by the Spanning Set Theorem.
C. The proof is complete because a basis for a subspace must consist of linearly independent vectors.
O D. The proof is complete because a basis for a subspace must span that subspace.

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