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Let \(A\) be an \(m \times n\) matrix of \(\operatorname{rank} r\) Thinking of \(\mathbb{R}^{n}\) as rows, define \(V=\left\\{\mathbf{x}\right.\) in \(\left.\mathbb{R}^{m} \mid \mathbf{x} A=\mathbf{0}\right\\}\). Show that \(\operatorname{dim} V=m-r\).

Short Answer

Expert verified
The dimension of \( V \) is \( m - r \).

Step by step solution

01

Understand the Null Space Row Concept

First, we recognize that the set \( V \) described is essentially the null space of the matrix \( A \) when viewed in terms of row space. Specifically, this means we are looking at all vectors \( \mathbf{x} \in \mathbb{R}^m \) such that \( \mathbf{x}A = \mathbf{0} \) results in the zero vector. This is equivalent to finding the null space of \( A^T \), the transpose of \( A \).
02

Apply Rank-Nullity Theorem

The Rank-Nullity Theorem is foundational here. It states for any matrix \( B \) with \( n \) columns, \( \operatorname{rank}(B) + \operatorname{nullity}(B) = n \). Since we are considering \( A^T \) (an \( n \times m \) matrix), the theorem becomes: \[ \operatorname{rank}(A^T) + \operatorname{nullity}(A^T) = m \].
03

Calculate Rank and Nullity

The rank of \( A^T \) is the same as the rank of \( A \), which is given as \( r \). Thus, \( \operatorname{rank}(A^T) = r \). Applying the Rank-Nullity Theorem gives us: \[ r + \operatorname{nullity}(A^T) = m \].
04

Derive the Dimension of Null Space

Rearranging the equation from the Rank-Nullity Theorem, we obtain \( \operatorname{nullity}(A^T) = m - r \). Hence, the dimension of the null space of \( A^T \), which is our set \( V \), is \( m - r \).
05

Conclusion

We have shown through application of the Rank-Nullity Theorem that the dimension of \( V \), which is the null space of \( A^T \), is \( m - r \). This confirms that \( \operatorname{dim} V = m - r \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

null space
The concept of the null space is crucial when working with matrices. Think of the null space of a matrix as the set of all vectors that, when multiplied by the matrix, result in the zero vector. In mathematical terms, for a matrix \( A \), its null space consists of all vectors \( \mathbf{x} \) such that \( A\mathbf{x} = \mathbf{0} \).
  • **Importance**: The null space tells us about solutions to the homogeneous equation \( A\mathbf{x} = \mathbf{0} \). If the null space contains only the zero vector, the matrix is said to have full rank.
  • **Relation to Linear Independence**: A larger null space indicates more dimensions where solutions exist, meaning the columns of the matrix are not linearly independent.
Remember, understanding null space helps in solving systems of linear equations and is directly tied to concepts like the rank of a matrix and linear transformations.
matrix transpose
The transpose of a matrix is a fundamental operation in linear algebra. To transpose a matrix, you flip it over its diagonal. Essentially, the rows become the columns and vice-versa. If \( A \) is an \( m \times n \) matrix, its transpose \( A^T \) becomes an \( n \times m \) matrix.
  • **Properties**: - The transpose of the transpose brings you back to the original matrix: \( (A^T)^T = A \). - The transpose of a product of matrices follows \( (AB)^T = B^T A^T \).
  • **Uses**: Transposing is not just a mathematical exercise; it has practical applications such as changing coordinate systems, optimizing matrix multiplication, and solving linear equations where calculating the null space of \( A^T \) directly relates to solutions.
Understanding matrix transpose is crucial for delving into more advanced topics like matrix diagonalization and eigensystems.
vector space dimension
The dimension of a vector space is a measure of its size in terms of the number of independent directions it encompasses. In simpler words, it's the number of vectors in a basis of the space, which are the minimal set of vectors needed to span the entire space.
  • **Importance**: Dimension is a fundamental characteristic of a vector space and affects everything from mapping transformations to describing vector sets. - For an \( n \)-dimensional space, any set of more than \( n \) vectors must be linearly dependent. - Dimension reflects the maximum number of linearly independent vectors.
  • **Relation to Matrices and Null Spaces**: The dimension of the null space (or kernel) of a transformation gives insight into the solutions of associated linear systems, directly linking to the rank-nullity theorem.
By understanding vector space dimensions, students can build a deeper intuition about the structure and capabilities of different mathematical spaces they're working with.

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Most popular questions from this chapter

Suppose \(T: V \rightarrow V\) is a linear operator with the property that \(T[T(\mathbf{v})]=\mathbf{v}\) for all \(\mathbf{v}\) in \(V\). (For example, transposition in \(\mathbf{M}_{n n}\) or conjugation in \(\mathbb{C} .\) ) If \(\mathbf{v} \neq \mathbf{0}\) in \(V\), show that \(\\{\mathbf{v}, T(\mathbf{v})\\}\) is linearly independent if and only if \(T(\mathbf{v}) \neq \mathbf{v}\) and \(T(\mathbf{v}) \neq-\mathbf{v}\).

Let \(S\) and \(T\) be linear transformations \(V \rightarrow W,\) where \(\operatorname{dim} V=n\) and \(\operatorname{dim} W=m\) a. Show that \(\operatorname{ker} S=\operatorname{ker} T\) if and only if \(T=R S\) for some isomorphism \(R: W \rightarrow W\). [Hint: Let \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{r}, \ldots, \mathbf{e}_{n}\right\\}\) be a basis of \(V\) such that \(\left\\{\mathbf{e}_{r+1}, \ldots, \mathbf{e}_{n}\right\\}\) is a basis of \(\operatorname{ker} S=\operatorname{ker} T\). Use Theorem 7.2 .5 to extend \(\left\\{S\left(\mathbf{e}_{1}\right), \ldots, S\left(\mathbf{e}_{r}\right)\right\\}\) and \(\left\\{T\left(\mathbf{e}_{1}\right), \ldots, T\left(\mathbf{e}_{r}\right)\right\\}\) to bases of \(\left.W .\right]\) b. Show that \(\operatorname{im} S=\operatorname{im} T\) if and only if \(T=S R\) for some isomorphism \(R: V \rightarrow V\). [Hint: Show that \(\operatorname{dim}(\operatorname{ker} S)=\operatorname{dim}(\) ker \(T)\) and choose bases \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{r}, \ldots, \mathbf{e}_{n}\right\\}\) and \(\left\\{\mathbf{f}_{1}, \ldots, \mathbf{f}_{r}, \ldots, \mathbf{f}_{n}\right\\}\) of \(V\) where \(\left\\{\mathbf{e}_{r+1}, \ldots, \mathbf{e}_{n}\right\\}\) and \(\left\\{\mathbf{f}_{r+1}, \ldots, \mathbf{f}_{n}\right\\}\) are bases of ker \(S\) and ker \(T,\) respectively. If \(1 \leq i \leq r,\) show that \(S\left(\mathbf{e}_{i}\right)=T\left(\mathbf{g}_{i}\right)\) for some \(\mathbf{g}_{i}\) in \(V,\) and prove that \(\left\\{\mathbf{g}_{1}, \ldots, \mathbf{g}_{r}, \mathbf{f}_{r+1}, \ldots, \mathbf{f}_{n}\right\\}\) is a basis of \(\left.V .\right]\)

Determine whether each of the following transformations \(T\) has an inverse and, if so, determine the action of \(T^{-1}\). a. \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) \(\quad T(x, y, z)=(x+y, y+z, z+x)\) b. \(T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}\) \(\quad T(x, y, z, t)=(x+y, y+z, z+t, t+x)\) c. \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\) \(\quad T\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}a-c & b-d \\ 2 a-c & 2 b-d\end{array}\right]\) d. \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\) \(T\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}a+2 c & b+2 d \\ 3 c-a & 3 d-b\end{array}\right]\) e. \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{3} ; T\left(a+b x+c x^{2}\right)=(a-c, 2 b, a+c)\) f. \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{3} ; T(p)=[p(0), p(1), p(-1)]\)

In each case, find a basis for the space \(V\) of all sequences \(\left[x_{n}\right)\) satisfying the recurrence, and use it to find \(x_{n}\) if \(x_{0}=1, x_{1}=-1,\) and \(x_{2}=1\). a. \(x_{n+3}=x_{n}+x_{n+1}-x_{n+2}\) b. \(x_{n+3}=-2 x_{n}+3 x_{n+1}\) c. \(x_{n+3}=-4 x_{n}+3 x_{n+2}\) d. \(x_{n+3}=x_{n}-3 x_{n+1}+3 x_{n+2}\) e. \(x_{n+3}=8 x_{n}-12 x_{n+1}+6 x_{n+2}\)

Show that every matrix \(X\) in \(\mathbf{M}_{n n}\) has the form \(X=A^{T}-2 A\) for some matrix \(A\) in \(\mathbf{M}_{n n}\). [Hint: The dimension theorem.]

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