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Chapter 4: Problem 105

For \(k=1,2, \ldots, 5\), find the number \(n_{k}\) of linearly independent subsets consisting of \(k\) columns for each of the following matrices: (a) \(A=\left[\begin{array}{lllll}1 & 1 & 0 & 2 & 3 \\ 1 & 2 & 0 & 2 & 5 \\ 1 & 3 & 0 & 2 & 7\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{lllll}1 & 2 & 1 & 0 & 2 \\ 1 & 2 & 3 & 0 & 4 \\\ 1 & 1 & 5 & 0 & 6\end{array}\right]\)

Short Answer

Expert verified
For matrix \(A\), there are 2 linearly independent 1 column subsets and 1 linearly independent 2 column subset. There are no linearly independent subsets with more than 2 columns. For matrix \(B\), there are 2 linearly independent 1 column subsets and 1 linearly independent 2 column subset. There are no linearly independent subsets with more than 2 columns.

Step by step solution

01

1 - column subsets

2 linearly independent 1 column subsets: Columns 1 and 2
02

2 - column subsets

1 linearly independent 2 column subset: Columns 1&2
03

k > 2 column subsets

There are no linearly independent subsets with more than 2 columns as there are only 2 linearly independent columns in matrix A. Now let's analyze Matrix B: \(B=\left[\begin{array}{lllll}1 & 2 & 1 & 0 & 2 \\\ 1 & 2 & 3 & 0 & 4 \\\ 1 & 1 & 5 & 0 & 6\end{array}\right]\) We'll perform row reduction for Matrix B: #Row Reduction:# 1. Subtract first row from second and third rows: \(B=\left[\begin{array}{lllll}1 & 2 & 1 & 0 & 2 \\\ 0 & 0 & 2 & 0 & 2 \\\ 0 & -1 & 4 & 0 & 4\end{array}\right]\) 2. Subtract -1 times the second row from the third row: \(B=\left[\begin{array}{lllll}1 & 2 & 1 & 0 & 2 \\\ 0 & 0 & 2 & 0 & 2 \\\ 0 & 0 & 6 & 0 & 6\end{array}\right]\) 3. Subtract 3 times the second row from the third row: \(B=\left[\begin{array}{lllll}1 & 2 & 1 & 0 & 2 \\\ 0 & 0 & 2 & 0 & 2 \\\ 0 & 0 & 0 & 0 & 0\end{array}\right]\) Now, just like with Matrix A, let's identify the pivot columns for Matrix B: #Pivot Columns:# By looking at the pivot positions (1,1 and 2,3), we can identify the pivot columns to be the first and third columns of Matrix B: \[\left(\begin{array}{l}1\\\\ 1\\\\ 1\end{array}\right), \left(\begin{array}{l}1\\\\ 3\\\\ 5\end{array}\right)\] The number of linearly independent subsets for Matrix B:
04

1 - column subsets

2 linearly independent 1 column subsets: Columns 1 and 3
05

2 - column subsets

1 linearly independent 2 column subset: Columns 1&3
06

k > 2 column subsets

There are no linearly independent subsets with more than 2 columns as there are only 2 linearly independent columns in matrix B.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Row Reduction
Matrix row reduction is a fundamental process used to simplify a matrix while preserving its essential properties. This technique is essential for determining key attributes like the rank of a matrix and identifying linearly independent columns.
To perform row reduction, we apply elementary row operations which include:
  • Swapping two rows
  • Multiplying a row by a non-zero scalar
  • Adding or subtracting a multiple of one row from another row

Through these steps, a matrix can be transformed into its row-echelon form or even further into reduced row-echelon form (RREF). When the matrix is in this form, it becomes easier to identify pivot positions and determine linear independence among column subsets.
Pivot Columns
Pivot columns are the columns of a matrix that contain pivot positions, which are the leading non-zero entries in each row when the matrix is in row-echelon form. These positions help us determine the linear independence among the columns.
Identifying pivot columns is crucial because they represent the minimum number of columns needed to span the column space of the matrix. For example, in our matrix B, row reduction reveals that the first and third columns are pivot columns. These correspond to columns where the leading entries reside, indicating which columns are linearly independent.
Aligning the knowledge of pivot columns allows us to find out which column subsets form a basis for the column space.
Linear Independence
Linear independence is a concept that describes whether a set of vectors (or columns, in the context of matrices) are related or if they independently contribute to the vector space. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others.
Testing for linear independence often involves row reduction to see if one of the vectors can be represented by a combination of others. If so, that set of vectors (or matrix columns) is not linearly independent.
  • In Matrix A, only the first two columns are independent, as after simplification, no goods can be created from the others.
  • For Matrix B, by recognizing the pivot columns, we determined the first and third columns as independent subsets.

Understanding linear independence is critical in many analyses, such as solving systems of linear equations and in vector space theory.
Column Subsets
Column subsets involve selecting various groups of columns from a matrix for analysis of their linear relationships. It plays a role in determining which combinations of columns can form a basis or remain independent.
When working with matrix A, different subsets of columns are examined, starting from single columns to pairs, to determine how many linearly independent subset combinations exist.
  • Single column subsets that are independent can easily be identified as the pivot columns themselves.
  • Two-column subsets require verification to ensure no column is a combination of the other.

For both matrices A and B, the maximum linearly independent column subset determined was two, which corresponds to the rank of each matrix. By evaluating different column subsets, we understand how various column combinations interact and contribute to the matrix's structural properties.

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Most popular questions from this chapter

Let \(V\) be the vector space of \(n\)-square matrices over a field \(K\). Show that \(W\) is a subspace of \(V\) if \(W\) consists of all matrices \(A=\left[a_{i j}\right]\) that are (a) symmetric \(\left(A^{T}=A\right.\) or \(\left.a_{i j}=a_{j i}\right)\), (b) (upper) triangular, (c) diagonal, (d) scalar.

Let \(V\) be the vector space of functions \(f: \mathbf{R} \rightarrow \mathbf{R}\). Show that \(W\) is a subspace of \(V\), where (a) \(W=\\{f(x): f(1)=0\\}\), all functions whose value at 1 is 0 . (b) \(W=\\{f(x): f(3)=f(1)\\}\), all functions assigning the same value to 3 and 1 . (c) \(W=\\{f(t): f(-x)=-f(x)\\}\), all odd functions. Let 0 denote the zero function, so \(\hat{0}(x)=0\) for every value of \(x\). (a) \(\hat{0} \in W\), because \(\hat{0}(1)=0\). Suppose \(f, g \in W\). Then \(f(1)=0\) and \(g(1)=0\). Also, for scalars \(a\) and \(b\), we have $$ (a f+b g)(1)=a f(1)+b g(1)=a 0+b 0=0 $$ Thus, \(a f+b g \in W\), and hence \(W\) is a subspace. (b) \(\hat{0} \in W\), because \(\hat{0}(3)=0=\hat{0}(1)\). Suppose \(f, g \in W\). Then \(f(3)=f(1)\) and \(g(3)=g(1)\). Thus, for any scalars \(a\) and \(b\), we have $$ (a f+b g)(3)=a f(3)+b g(3)=a f(1)+b g(1)=(a f+b g)(1) $$ Thus, \(a f+b g \in W\), and hence \(W\) is a subspace. (c) \(\hat{0} \in W\), because \(\hat{0}(-x)=0=-0=-\hat{0}(x)\). Suppose \(f, g \in W\). Then \(f(-x)=-f(x)\) and \(g(-x)=-g(x)\). Also, for scalars \(a\) and \(b\), $$ (a f+b g)(-x)=a f(-x)+b g(-x)=-a f(x)-b g(x)=-(a f+b g)(x) $$ Thus, \(a b+g f \in W\), and hence \(W\) is a subspace of \(V\).

Prove Lemma 4.13: Suppose \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) spans \(V\), and suppose \(\left\\{w_{1}, w_{2}, \ldots, w_{m}\right\\}\) is linearly independent. Then \(m \leq n\), and \(V\) is spanned by a set of the form $$ \left\\{w_{1}, w_{2}, \ldots, w_{m}, v_{i_{1}}, v_{i_{2}}, \ldots, v_{i_{n-w}}\right\\} $$ Thus, any \(n+1\) or more vectors in \(V\) are linearly dependent. It suffices to prove the lemma in the case that the \(v_{i}\) are all not \(0 .\) (Prove!) Because \(\left\\{v_{i}\right\\}\) spans \(V\), we have by Problem \(4.34\) that $$ \left\\{w_{1}, v_{1}, \ldots, v_{n}\right\\} $$ is linearly dependent and also spans \(V\). By Lemma \(4.10\), one of the vectors in (1) is a linear combination of the preceding vectors. This vector cannot be \(w_{1}\), so it must be one of the \(v^{\prime}\) s, say \(v_{j}\). Thus by Problem \(4.34\), we can delete \(v_{j}\) from the spanning set (1) and obtain the spanning set $$ \left\\{w_{1}, v_{1}, \ldots, v_{j-1}, \quad v_{j+1} \ldots, v_{n}\right\\} $$ Now we repeat the argument with the vector \(w_{2}\). That is, because (2) spans \(V\), the set $$ \left\\{w_{1}, w_{2}, v_{1}, \ldots, v_{j-1}, \quad v_{j+1}, \ldots, v_{n}\right\\} $$ is linearly dependent and also spans \(V\). Again by Lemma \(4.10\), one of the vectors in (3) is a linear combination of the preceding vectors. We emphasize that this vector cannot be \(w_{1}\) or \(w_{2}\), because \(\left\\{w_{1}, \ldots, w_{m}\right\\}\) is independent; hence, it must be one of the \(v^{\prime}\) s, say \(v_{k} .\) Thus, by Problem \(4.34\), we can delete \(v_{k}\) from the spanning set (3) and obtain the spanning set $$ \left\\{w_{1}, w_{2}, v_{1}, \ldots, v_{j-1}, \quad v_{j+1}, \ldots, v_{k-1}, \quad v_{k+1}, \ldots, v_{n}\right\\} $$ We repeat the argument with \(w_{3}\), and so forth. At each step, we are able to add one of the \(w\) 's and delete one of the \(v\) 's in the spanning set. If \(m \leq n\), then we finally obtain a spanning set of the required form: $$ \left\\{w_{1}, \ldots, w_{m}, v_{i_{1}}, \ldots, v_{i_{n-\infty}}\right\\} $$ Finally, we show that \(m>n\) is not possible. Otherwise, after \(n\) of the above steps, we obtain the spanning set \(\left\\{w_{1}, \ldots, w_{n}\right\\} .\) This implies that \(w_{n+1}\) is a linear combination of \(w_{1}, \ldots, w_{n}\), which contradicts the hypothesis that \(\left\\{w_{i}\right\\}\) is linearly independent.

Find the dimension and a basis of the subspace \(W\) of \(\mathbf{M}=\mathbf{M}_{2,3}\) spanned by $$ A=\left[\begin{array}{lll} 1 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{lll} 2 & 4 & 3 \\ 7 & 5 & 6 \end{array}\right], \quad C=\left[\begin{array}{lll} 1 & 2 & 3 \\ 5 & 7 & 6 \end{array}\right] $$

Suppose \(S=\left\\{v_{1}, v_{2}, \ldots, v_{m}\right\\}\) spans a vector space \(V\) (a) If \(w \in V,\) then \(\left\\{w, v_{1}, \ldots, v_{m}\right\\}\) is linearly dependent and spans \(V\) (b) If \(v_{i}\) is a linear combination of \(v_{1}, \ldots, v_{i-1},\) then \(S\) without \(v_{i}\) spans \(V\)
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