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Chapter 4: Problem 56

Prove that the nonzero row vectors of a matrix in row-echelon form are linearly independent.

Short Answer

Expert verified
The nonzero row vectors of a row-echelon form matrix are linearly independent as each nonzero row has its leading 1 in a unique column, the only way to make the sum equals to zero in this column is to multiply this row by zero. This implies that all coefficients of the linear combination must be zero, which is the definition of linear independence.

Step by step solution

01

Understand the Row-Echelon Form

Each nonzero row in a row-echelon matrix starts with a leading 1 (also known as a pivot), and the subsequent entries in this columns are zeros. Moreover, each leading 1 is to the right of the leading 1 in the previous row.
02

Consider a Linear Combination

Let's combine the row vectors into a single vector using coefficients \(a_1\), \(a_2\), ..., \(a_n\). This will result in a new row vector. If this new row vector is the zero vector, this will imply that all the coefficients \(a_i\) are zero, hence proving the original row vectors are linearly independent.
03

Make Observations

Since each nonzero row has its leading 1 in a unique column, the only way to make the sum equals to zero in this column is to multiply this row by zero (since that specific entry is the only non-zero entry in that column among all other row vectors, due to the echelon form). This will result in all the multiplying coefficients being zero, hence proving the linear independence of the row vectors.

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