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Chapter 5: Problem 15

If \(A\) and \(B\) are matrices and the columns of \(A B\) are independent, show that the columns of \(B\) are independent.

Short Answer

Expert verified
The columns of matrix \(B\) are independent because the rank of \(B\) equals the number of its columns.

Step by step solution

01

Understanding Matrix Independence

The columns of a matrix are considered to be independent if no column can be written as a linear combination of the others. When matrix multiplication is involved, it's important to analyze how the transformation from one matrix affects another. In this case, we are given that the columns of the product matrix \(AB\) are independent, and we need to determine what this implies for the columns of matrix \(B\).
02

Analyzing the Transformation

Let the matrix \(A\) be an \(m \times n\) matrix and \(B\) be an \(n \times p\) matrix. The resulting product \(AB\) is an \(m \times p\) matrix. Given that the columns of \(AB\) are independent, it implies that the rank of \(AB\) is equal to the number of its columns, \(p\). This is a crucial fact in determining the characteristics of matrix \(B\).
03

Determining the Rank of B

Since \(AB\) has full column rank, it implies that each column of \(AB\) is linearly independent. Since matrix multiplication involving \(A\) does not affect the independence of matrix \(B\)'s columns, it follows that the rank of \(B\) must also be \(p\), the same as the number of columns in \(B\).
04

Concluding Independence of B's Columns

Since the rank of \(B\) equals the number of its columns, it shows that the columns of \(B\) are linearly independent. Therefore, if multiplying by \(A\) does not change this independence, it directly demonstrates that the columns of \(B\) were independent to begin with.

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Key Concepts

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

Matrix Multiplication
Matrix multiplication involves combining two matrices to produce a new matrix. This is a fundamental operation in linear algebra and is denoted as \(AB\), where \(A\) and \(B\) are the matrices being multiplied. The product is possible only if the number of columns in \(A\) matches the number of rows in \(B\).
  • Matrix \(A\), with dimensions \(m \times n\), and matrix \(B\), with dimensions \(n \times p\), produce a matrix \(AB\) with dimensions \(m \times p\).
  • Each element in the resulting matrix is calculated by taking the dot product of the corresponding row in \(A\) and column in \(B\).
Matrix multiplication is not commutative, meaning \(AB eq BA\).
However, it is associative, so \((AB)C = A(BC)\), and it is distributive over addition, \((A + B)C = AC + BC\). This operation is central to transformations and linear mappings in linear algebra.
Rank of a Matrix
The rank of a matrix is a measure of the maximum number of linearly independent column vectors in the matrix.
This is equivalent to the dimension of the column space or the row space of the matrix.
  • The rank helps in determining the consistency of linear equations associated with the matrix.
  • If the rank is equal to the number of columns, the columns are linearly independent.
In our given problem, since matrix \(AB\) has independent columns, its rank is equal to the number of columns \(p\).
This indicates full column rank and therefore implies constraints on the rank of matrix \(B\). Since \(AB\) reflects the transformation of \(B\) by \(A\),
the independence in \(AB\)'s columns ensures the rank of \(B\) must also be \(p\), confirming \(B\) has independent columns.
Linear Independence
Linear independence is an important concept determining the uniqueness of the vectors in a matrix. When columns are linearly independent,
no column can be expressed as a linear combination of the others.
  • This means for a set \([v_1, v_2, ..., v_n]\), the equation \(c_1v_1 + c_2v_2 + ... + c_nv_n = 0\) only has the trivial solution \(c_1 = c_2 = ... = c_n = 0\).
  • This property is crucial for solving systems of equations, as it ensures uniqueness and solvability.
In the given exercise, the columns of the matrix \(AB\) are independent. This results in each column of \(B\) maintaining its independence through the transformation by \(A\).
By understanding that the independence in the product \(AB\) is contingent on \(B\)'s columns being independently transformed into \(AB\),
we conclude the columns of \(B\) are, indeed, independent before the transformation.

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Most popular questions from this chapter

Let \(A\) denote an \(n \times n\) upper triangular matrix. a. If all the main diagonal entries of \(A\) are distinct, show that \(A\) is diagonalizable. b. If all the main diagonal entries of \(A\) are equal, show that \(A\) is diagonalizable only if it is already diagonal. c. Show that \(\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{array}\right]\) is diagonalizable but that \(\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{array}\right]\) is not diagonalizable.

Let \(A\) denote an \(m \times n\) matrix. a. Show that null \(A=\) null \((U A)\) for every invertible \(m \times m\) matrix \(U\) b. Show that \(\operatorname{dim}(\operatorname{null} A)=\operatorname{dim}(\) null \((A V))\) for every invertible \(n \times n\) matrix \(V\). [Hint: If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) is a basis of null \(A,\) show that \(\left\\{V^{-1} \mathbf{x}_{1}, V^{-1} \mathbf{x}_{2}, \ldots, V^{-1} \mathbf{x}_{k}\right\\}\) is a basis of \(\operatorname{null}(A V) .]\)

In each case, decide whether the matrix \(A\) is diagonalizable. If so, find \(P\) such that \(P^{-1} A P\) is diagonal. $$ \text { a. }\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{array}\right] \quad \text { b. }\left[\begin{array}{rrr} 3 & 0 & 6 \\ 0 & -3 & 0 \\ 5 & 0 & 2 \end{array}\right] $$ c. \(\left[\begin{array}{rrr}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{array}\right]\) d. \(\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 2 & 2 \\ 2 & 3 & 1\end{array}\right]\)

Show that \(\left[\begin{array}{rrrr}1 & 2 & -1 & 0 \\ 2 & 0 & 1 & 1 \\ 1 & 1 & 0 & -1 \\ 4 & 3 & 0 & 0\end{array}\right]\) and \(\left[\begin{array}{rrrr}1 & -1 & 3 & 0 \\ -1 & 0 & 1 & 1 \\ 0 & -1 & 4 & 1 \\\ 5 & -1 & -1 & -4\end{array}\right]\) are not similar.

A matrix obtained from \(A\) by deleting rows and columns is called a submatrix of \(A .\) If \(A\) has an invertible \(k \times k\) submatrix, show that rank \(A \geq k\). [Hint: Show that row and column operations carry \(A \rightarrow\left[\begin{array}{rr}I_{k} & P \\ 0 & Q\end{array}\right]\) in block form.] Remark: It can be shown that rank \(A\) is the largest integer \(r\) such that \(A\) has an invertible \(r \times r\) submatrix.
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