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Chapter 3: Problem 11

Let \(A\) be an \(m \times n\) matrix. Prove that \\[ \operatorname{rank}(A) \leq \min (m, n) \\]

Short Answer

Expert verified
To prove that the rank of an \(m \times n\) matrix A is less than or equal to the minimum of its dimensions, we can consider the column and row spaces of A. Since the dimension of the column space of A is a subspace of \(R^m\), it cannot be greater than m. Similarly, the dimension of the row space of A is a subspace of \(R^n\) and cannot be greater than n. By the rank-nullity theorem, we have \(\operatorname{rank}(A) = \operatorname{dimension}( \operatorname{Col}(A) ) = \operatorname{dimension}( \operatorname{Row}(A) )\). Combining these inequalities, we conclude that \(\operatorname{rank}(A) \leq \min(m, n)\).

Step by step solution

01

Definition of Rank

First, recall that the rank of a matrix is the maximum number of linearly independent rows or columns present in the matrix. The rank tells us about the dimension of the vector space spanned by the rows or columns.
02

Consider the column space

Let's look at the column space of the matrix A. The column space of A consists of all the linear combinations of the columns of A. It is a vector space whose dimension is equal to the number of linearly independent columns in A.
03

Relation between dimensions of column space and matrix

Since there are m elements in each column of the matrix A, the column space can be thought of as a subspace of \(R^m\). Therefore, by the properties of vector spaces, the dimension of the column space cannot be greater than the dimension of the larger space it is contained in, which in our case is m. This result tells us: \[ \operatorname{dimension}( \operatorname{Col} (A) ) \leq m\]
04

Consider the row space

Now let's look at the row space of the matrix A. The row space of A consists of all the linear combinations of the rows of A. It is a vector space whose dimension is equal to the number of linearly independent rows in A.
05

Relation between dimensions of row space and matrix

Each row of the matrix A has n elements. Hence, the row space can be thought of as a subspace of \(R^n\). Therefore, by the properties of vector spaces, the dimension of the row space cannot be greater than the dimension of the larger space it is contained in, which in our case is n. This result tells us: \[ \operatorname{dimension}( \operatorname{Row}(A) ) \leq n\]
06

Rank of the matrix

By the rank-nullity theorem, we know: \[ \operatorname{rank}(A) = \operatorname{dimension}( \operatorname{Col}(A) ) = \operatorname{dimension}( \operatorname{Row}(A) ) \]
07

Combining the inequalities

Now we combine our results from step 3 and step 5: \[ \operatorname{rank}(A) \leq \min(m, n) \] This completes the proof that the rank of an m x n matrix is less than or equal to the minimum of its dimensions.

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Most popular questions from this chapter

Let \(A\) be an \(m \times n\) matrix with rank equal to \(n\). Show that if \(\mathbf{x} \neq \mathbf{0}\) and \(\mathbf{y}=A \mathbf{x},\) then \(\mathbf{y} \neq \mathbf{0}\).

Let \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) be a spanning set for a vector space \(V\) (a) If we add another vector, \(\mathbf{x}_{k+1},\) to the set, will we still have a spanning set? Explain. (b) If we delete one of the vectors, say \(\mathbf{x}_{k},\) from the set, will we still have a spanning set? Explain.

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 \(C[0,1]\)

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Let \(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\) be linearly independent vectors in a vector space \(V .\) Show that \(\mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\) cannot \(\operatorname{span} V\)
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