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Chapter 3: Problem 8

Let \(A\) be an \(m \times n\) matrix. Prove that if \(c\) is any nonzero scalar, then \(\operatorname{rank}(c A)=\operatorname{rank}(A)\).

Short Answer

Expert verified
In summary, when a matrix A is multiplied by a non-zero scalar c, the linear independence of the rows or columns is preserved, and as a result, the rank of the matrix does not change. Therefore, we can conclude that \(\operatorname{rank}(cA) = \operatorname{rank}(A)\).

Step by step solution

01

Define the Problem

The goal here is to show that multiplying a matrix A by a non-zero scalar c will not change the rank of the matrix. This means showing that the maximum number of linearly independent columns (or rows) remains the same after multiplying by c.
02

Clarify Linear Independence

A set of vectors (or rows or columns of a matrix) are said to be linearly independent if no vector in the set can be represented as a linear combination of the other vectors in the set. Meaning any vector within the set cannot be formed by adding or subtracting multiples of other vectors within the set.
03

Apply the Non-Zero Scalar

Now consider the matrix \(cA\) for \(c \neq 0\). This represents multiplying every entry in the matrix by the scalar \(c\). When the entries of a vector are multiplied by a scalar, the vector is scaled (stretched or contracted), but its orientation remains the same, unless the scalar is negative, in which case, the vector is additionally reflected. However, the linear independence or dependence of the set of vectors will not be impacted by scalar multiplication. This is due to the distributive and associative property of scalar multiplication in matrices.
04

Understand the Rank of the Matrices

The rank of \(cA\) (denoted \(\operatorname{rank}(cA)\)) is the maximum number of linearly independent columns. Since the linear independence of the columns is not affected when every entry in the matrix is multiplied by \(c\), the rank of the matrix \(cA\) is the same as the rank of the matrix \(A\) (\(\operatorname{rank}(A)\)).
05

Conclude the Proof

To summarize, when a matrix is multiplied by a non-zero scalar, the linear independence of the rows or columns is preserved, and hence the rank of the matrix does not change. Hence, \(\operatorname{rank}(cA) = \(\operatorname{rank}(A)\).

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