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Chapter 4: Problem 71

Prove that row operations do not change the dependency relationships among the columns of an \(m \times n\) matrix.

Short Answer

Expert verified
Proof is via the premise that row operations, despite altering the rows of the matrix, do not change the linear combinations of the columns. If a column can be produced as a linear combination of other columns prior to the row operation, this combination remains true post-operation, preserving the column linear dependency.

Step by step solution

01

Define Column Dependency

Two or more columns in a matrix are said to be linearly dependent if one of the columns can be written as a linear combination of the other columns. If this is not possible, then the columns are linearly independent.
02

Understanding Row Operations

There are three main types of row operations: switching two rows, multiplying a row by a non-zero constant, and adding a multiple of one row to another row. Notice how all these operations alter only the rows of the matrix, not the columns.
03

Analyze Impact of Row Operations on Column Dependency

Even though row operations change the rows of the matrix, they do not change the linear combinations of the columns. In other words, if one column was written as a linear combination of other columns before the row operation, it could still be written as a linear combination of other columns after the row operation. Thus, the column linear dependency is preserved.
04

Example with the Row Operation

As an example, consider a 3脳3 matrix and suppose we obtain a new matrix by interchanging the first and second rows. Neither the linear dependency nor independency of the columns is affected by this row interchange, because the elements of each column are simply rearranged. The same is also true if we scale a row by a non-zero constant or if we add a multiple of one row to another row.

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Most popular questions from this chapter

Find the transition matrix from \(B\) to \(B^{\prime}\) by hand $$\begin{array}{l}B=\\{(1,0,0),(0,1,0),(0,0,1)\\} \\\B^{\prime}=\\{(1,0,0),(0,2,8),(6,0,12)\\}\end{array}$$

Test the given set of solutions for linear independence. $$\begin{array}{lll} \text { Differential Equation } & \text { Solutions } \\ y^{\prime \prime}+4 y^{\prime}+4 y=0 & \left\\{e^{-2 x}, x e^{-2 x}\right\\} \end{array}$$

Use a graphing utility or computer software program with matrix capabilities to find the transition matrix from \(B\) to \(B^{\prime}\) $$\begin{array}{l}B=\\{(1,3,3),(1,5,6),(1,4,5)\\} \\\B^{\prime}=\\{(1,0,0),(0,1,0),(0,0,1)\\}\end{array}$$

Test the given set of solutions for linear independence. $$\begin{array}{lll} \text { Differential Equation } & \text { Solutions } \\ y^{\prime \prime \prime \prime}+y^{\prime}=0 & \\{2,-1+2 \sin x, 1+\sin x\\} \end{array}$$

Identify and sketch the graph. $$y^{2}+8 x+6 y+25=0$$
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