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Chapter 3: Problem 34

Let \(B\) be an \(n \times m\) matrix whose columns are linearly independent. Show that \(B\) has a left inverse.

Short Answer

Expert verified
In order to show that matrix B has a left inverse, we need to understand the relationship between linearly independent columns and rank. Since B is an n x m matrix with linearly independent columns, it has a rank of m. A full column rank matrix always has a left inverse, A, with dimensions m x n. To find matrix A, we express each column of the identity matrix I as a linear combination of columns of B. Due to the linearly independent columns of B, we can form m linearly independent columns in I. Hence, B has a left inverse.

Step by step solution

01

Linearly independent columns and rank

A set of vectors is called linearly independent if no vector can be formed as a linear combination of the remaining vectors. In a matrix, the columns are linearly independent if and only if the rank of the matrix is equal to the number of columns. In our given exercise, B is an n x m matrix with linearly independent columns. So, the rank of B is m.
02

Left inverse of matrix B

Since the rank of B is m, it is a full column rank matrix. A full column rank matrix always has a left inverse. Given A * B = I, where A is the left inverse of B, the dimensions of A should be m x n, and I is the identity matrix of size m x m.
03

Using column-space

The column space of matrix B is the subspace formed by all linear combinations of columns of B. Since columns of B are linearly independent, the column space of B contains m linearly independent vectors. For a matrix A * B = I, each column of I can be expressed as a linear combination of columns of B. Since B has m linearly independent columns, we can form m linearly independent columns in I.
04

Finding matrix A

Let's find the left inverse matrix A of matrix B so that A * B = I. Let a_i be the rows of A and b_j be the columns of B. Then, we need to find a_i's such that: a_1 * b_1 + a_2 * b_2 + ... + a_m * b_m = 1 if i = j, a_1 * b_1 + a_2 * b_2 + ... + a_m * b_m = 0 if i 鈮� j. Since the columns of B are linearly independent, such a_i's exist for each i, and we can find the matrix A which will be the left inverse of B. Hence, the given n x m matrix B with linearly independent columns has a left inverse.

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Most popular questions from this chapter

Let \(A\) be an \(6 \times n\) matrix of rank \(r\) and let \(\mathbf{b}\) be a vector in \(\mathbb{R}^{6} .\) For each pair of values of \(r\) and \(n\) that follow, indicate the possibilities as to the number of solutions one could have for the linear system \(A \mathbf{x}=\mathbf{b} .\) Explain your answers. (a) \(n=7, r=5\) (b) \(n=7, r=6\) (c) \(n=5, r=5\) (d) \(n=5, r=4\)

Given the functions \(2 x\) and \(|x|,\) show that (a) these two vectors are linearly independent in \(C[-1,1]\) (b) the vectors are linearly dependent in \(C[0,1]\)

In each of the following, determine the subspace of \(\mathbb{R}^{2 \times 2}\) consisting of all matrices that commute with the given matrix: (a) \(\left(\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right)\) (b) \(\left(\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right)\) (c) \(\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\) (d) \(\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)\)

In \(\mathbb{R}^{4}\), let \(U\) be the subspace of all vectors of the form \(\left(u_{1}, u_{2}, 0,0\right)^{T},\) and let \(V\) be the subspace of all vectors of the form \(\left(0, v_{2}, v_{3}, 0\right)^{T}\). What are the dimensions of \(U, V, U \cap V, U+V ?\) Find a basis for each of these four subspaces. (See Exercises 20 and \(22 \text { of Section } 2 .)\)

Given \\[ \begin{aligned} \mathbf{x}_{1} &=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right], \quad \mathbf{x}_{2}=\left[\begin{array}{l} 3 \\ 4 \\ 2 \end{array}\right] \\ \mathbf{x} &=\left[\begin{array}{l} 2 \\ 6 \\ 6 \end{array}\right] , \quad \mathbf{y}=\left[\begin{array}{r} -9 \\ -2 \\ 5 \end{array}\right] \end{aligned} \\] (a) Is \(\mathbf{x} \in \operatorname{Span}\left(\mathbf{x}_{1}, \mathbf{x}_{2}\right) ?\) (b) Is \(\mathbf{y} \in \operatorname{Span}\left(\mathbf{x}_{1}, \mathbf{x}_{2}\right) ?\) Prove your answers.
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