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If \(A\) and \(B\) are \(n \times n\) matrices, show that they have the same column space if and only if \(A=B U\) for some invertible matrix \(U\).

Short Answer

Expert verified
The column spaces of \(A\) and \(B\) are the same if and only if \(A=BU\) for some invertible \(U\).

Step by step solution

01

Understand the Problem

We need to prove the equivalence: The column space of matrix \(A\) equals the column space of matrix \(B\) if and only if there exists an invertible matrix \(U\) such that \(A = BU\). This means we have to prove the statement in both directions.
02

Prove \(A=B U\) implies \(Col(A)=Col(B)\)

Assume \(A = BU\) where \(U\) is invertible. Every column of \(A\) is a linear combination of the columns of \(B\) (since \(U\) is a matrix of linear combinations). Therefore, the column space of \(A\) is a subset of the column space of \(B\). The invertibility of \(U\) implies that every combination is possible, ensuring that the column spaces are identical.
03

Prove \(Col(A)=Col(B)\) implies \(A=B U\) for some \(U\)

Given \(Col(A) = Col(B)\), there must exist a transformation matrix that transforms \(B\) into \(A\), preserving the column space. Since \(B\) and \(A\) span the same space, each column of \(A\) can be expressed as a linear combination of the columns of \(B\). Thus, there exists matrix \(U\) such that \(A = BU\). The invertibility of \(U\) ensures that \(U\) includes all necessary transformations without losing the span.
04

Conclude with the Equivalence

We have shown that if \(A = BU\) for an invertible \(U\), then \(Col(A) = Col(B)\), and conversely, if \(Col(A) = Col(B)\), then there exists an invertible matrix \(U\) such that \(A = BU\). This completes the proof of the equivalence.

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Key Concepts

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

Invertible Matrix
An invertible matrix is a fundamental concept in linear algebra. It is a square matrix that has an inverse. This property allows us to reverse the matrix operations. If a matrix \( U \) is invertible, then there exists a matrix \( U^{-1} \) such that \( UU^{-1} = U^{-1}U = I \), where \( I \) is the identity matrix.

Invertibility is an important property when discussing transformations. In the context of the given problem, if a matrix \( U \) is invertible, it implies that the transformation it applies to another matrix \( B \) to give \( A = BU \), has no loss of dimensionality or information. Every operation made can be reversed by applying \( U^{-1} \). This ensures that the column spaces of \( A \) and \( B \) remain equivalent.

Some key points about invertible matrices:
  • Not all matrices are invertible. A matrix must be square (have the same number of rows and columns) to be potentially invertible.
  • An invertible matrix \( U \) has a non-zero determinant.
  • If a matrix is invertible, its row space and column space are full; they span the \( n \)-dimensional space for an \( n \times n \) matrix.
Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations, linear transformations, and their representations through matrices and vector spaces. It lies at the heart of mathematics and is key for various applications such as physics, computer science, and engineering.

The core components of linear algebra include vectors, matrices, and operations that can be performed on them. Vectors are quantities defined by a magnitude and direction, while matrices serve as arrays that can represent multiple vectors or linear transformations.

Linear transformations, which turn vectors into other vectors in a linear manner, are central in linear algebra. They can be represented by matrices, making problems like the one in our exercise possible to solve and conceptualize in different spaces.
  • Key concepts include vector spaces, basis, dimension, and transformations.
  • The operations within linear algebra rely on addition, scalar multiplication, and matrix multiplication.
  • Understanding linear algebra provides the foundation for more advanced mathematical topics and practical applications like computer graphics and machine learning.
Matrix Transformation
Matrix transformation involves using a matrix to perform a function or operation on vectors in a space. It is a way to execute linear transformations, which map input vectors to other vectors in the same or different space.

Every matrix can be thought of as a transformation. For example, in the expression \( A = BU \), the matrix \( U \) transforms the matrix \( B \) into matrix \( A \). This transformation is also described by the column spaces of the matrices, which tell us how vectors in those spaces are scaled and shifted.

The role of an invertible matrix in transformations is crucial. An invertible matrix ensures a one-to-one transformation, meaning no information or dimension is lost. This is particularly important in applications where precise consistency of transformations is needed, such as computer graphics or simulations.
  • Transformations can be visualized as moving, rotating, or scaling objects in space.
  • In our exercise, matrix transformations help establish the equivalence of column spaces, demonstrating how vectors are transformed from one space to another without loss.
  • Learning about transformation matrices is essential for solving systems of linear equations and understanding the structure of advanced mathematical models.

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Most popular questions from this chapter

Find the standard matrix of the rotation \(R\) about the line through the origin with direction vector \(\mathbf{d}=\left[\begin{array}{lll}2 & 3 & 6\end{array}\right]^{T} \cdot\left[\right.\) Hint: Consider \(\mathbf{f}=\left[\begin{array}{ccc}6 & 2 & -3\end{array}\right]^{T}\) and \(\left.\mathbf{g}=\left[\begin{array}{lll}3 & -6 & 2\end{array}\right]^{T} .\right]\)

Let \(A=\left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right]\) and consider \(T_{A}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) a. Show that the only eigenvalue of \(T_{A}\) is \(\lambda=0\). b. Show that \(\operatorname{ker}\left(T_{A}\right)=\mathbb{R}\left[\begin{array}{l}1 \\\ 0\end{array}\right]\) is the unique \(T_{A}\) -invariant subspace of \(\mathbb{R}^{2}\) (except for 0 and \(\mathbb{R}^{2}\) ).

Let \(V\) be a vector space. If \(f: V \rightarrow \mathbb{R}\) is a linear transformation and \(\mathrm{z}\) is a vector in \(V,\) define \(T_{f, \mathbf{z}}: V \rightarrow V\) by \(T_{f, \mathbf{z}}(\mathbf{v})=f(\mathbf{v}) \mathbf{z}\) for all \(\mathbf{v}\) in \(V\). Assume that \(f \neq 0\) and \(\mathbf{z} \neq \mathbf{0}\). a. Show that \(T_{f, z}\) is a linear operator of rank 1 . b. If \(f \neq 0\), show that \(T_{f, \mathrm{z}}\) is an idempotent if and only if \(f(\mathbf{z})=1\). (Recall that \(T: V \rightarrow V\) is called an idempotent if \(T^{2}=T\).) c. Show that every idempotent \(T: V \rightarrow V\) of rank 1 has the form \(T=T_{f, \mathbf{z}}\) for some \(f: V \rightarrow \mathbb{R}\) and some \(\mathbf{z}\) in \(V\) with \(f(\mathbf{z})=1\). [Hint: Write \(\operatorname{im} T=\mathbb{R} \mathbf{z}\) and show that \(T(\mathbf{z})=\mathbf{z}\). Then use \(\mathrm{Ex}\) - ercise \(9.3 .23 .]\)

In each case, find the matrix of the linear transformation \(T: V \rightarrow W\) corresponding to the bases \(B\) and \(D\) of \(V\) and \(W\), respectively. a. \(T: \mathbf{M}_{22} \rightarrow \mathbb{R}, T(A)=\operatorname{tr} A\) \(\begin{aligned} B=&\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\\}, \\ D=\\{1\\} \end{aligned}\) b. \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}, T(A)=A^{T}\) \(\left.\begin{array}{rl}B & =D \\ & =\left\\{\left[\begin{array}{ll}1 & 0 \\\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\\}\end{array}\right\\}\) c. \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{3}, T[p(x)]=x p(x) ; B=\left\\{1, x, x^{2}\right\\}\) and \(\quad D=\left\\{1, x, x^{2}, x^{3}\right\\}\) d. \(\quad T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}, T[p(x)]=p(x+1)\) \(\quad B=D=\left\\{1, x, x^{2}\right\\}\)

Let \(V\) be a vector space with ordered basis \(B=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}, \ldots, \mathbf{b}_{n}\right\\} .\) For each \(i=1,2, \ldots, m\) define \(S_{i}: \mathbb{R} \rightarrow V\) by \(S_{i}(r)=r \mathbf{b}_{i}\) for all \(r\) in \(\mathbb{R}\). a. Show that each \(S_{i}\) lies in \(\mathbf{L}(\mathbb{R}, V)\) and \(S_{i}(1)=\mathbf{b}_{i}\). b. Given \(T\) in \(\mathbf{L}(\mathbb{R}, V),\) let \(T(1)=a_{1} \mathbf{b}_{1}+a_{2} \mathbf{b}_{2}+\cdots+a_{n} \mathbf{b}_{n}, a_{i}\) in \(\mathbb{R} .\) Show $$\text { that } T=a_{1} S_{1}+a_{2} S_{2}+\cdots+a_{n} S_{n} \text { . }$$ c. Show that \(\left\\{S_{1}, S_{2}, \ldots, S_{n}\right\\}\) is a basis of \(\mathbf{L}(\mathbb{R}, V)\).

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