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Chapter 5: Problem 9

Let \(B\) be another endomorphism of \(V\). Assume that \(A B=B A\) and both \(A, B\) are diagonalizable. Prove that \(A\) and \(B\) are simultaneously diagonalizable, that is \(V\) has a basis consisting of elements which are eigenvectors for both \(A\) and \(B\).

Short Answer

Expert verified
To prove that \(A\) and \(B\) are simultaneously diagonalizable, follow these steps: 1. Show that the eigenspaces of \(A\) are \(B\)-invariant: for any eigenvector \(v\) of \(A\) with eigenvalue \(\lambda\), \(Bv\) is also an eigenvector of \(A\) with the same eigenvalue \(\lambda\). 2. Restrict \(B\) to act on the eigenspaces of \(A\), and find a basis for each eigenspace consisting of eigenvectors of the restricted \(B\). 3. Combine the eigenvectors of the restricted \(B\) to form a basis for \(V\). 4. Show that the elements of this basis are eigenvectors of both \(A\) and \(B\), proving that they are simultaneously diagonalizable.

Step by step solution

01

Prove that the eigenspaces of \(A\) are \(B\)-invariant

Let \(\lambda\) be an eigenvalue of \(A\) and \(E_\lambda\) be its eigenspace, so for any \(v\in E_\lambda\), we have \(Av=\lambda v\). We need to show that for any \(v\in E_\lambda\), \(Bv\in E_\lambda\). Since \(AB=BA\), we can rewrite it as \(ABv=BAv\). Now, let \(v\in E_\lambda\), then \(ABv=BAv\Rightarrow A(Bv)=B(Av)=B(\lambda v) \Rightarrow A(Bv)= \lambda (Bv)\). This implies that \(Bv\) is also an eigenvector of \(A\) for the same eigenvalue \(\lambda\). Therefore, \(Bv\in E_\lambda\), and the eigenspace \(E_\lambda\) is \(B\)-invariant.
02

Use the diagonalizable property of \(B\) on the eigenspaces of \(A\)

Since \(B\) is diagonalizable and \(E_\lambda\) is \(B\)-invariant, we can restrict \(B\) to act on \(E_\lambda\). Now we can find a basis for each eigenspace \(E_\lambda\) consisting of eigenvectors of the restricted \(B|_{E_\lambda}\). Let's denote the eigenvectors of \(B|_{E_\lambda}\) as \(v_1,v_2,\dots,v_k\).
03

Combine the eigenvectors of \(B|_{E_\lambda}\) as a basis for \(V\)

As \(A\) is assumed to be diagonalizable, it has a basis consisting of eigenvectors. Let's say \(A\) has eigenvalues \(\lambda_1,\lambda_2,\dots,\lambda_n\) with corresponding eigenspaces \(E_{\lambda_1},E_{\lambda_2},\dots,E_{\lambda_n}\). Now, for each eigenspace \(E_{\lambda_i}\), we find the eigenvectors of \(B|_{E_{\lambda_i}}\) (as done in Step 2) and combine them to form a basis for \(V\). Let's call this basis set \(S\).
04

Prove the elements of \(S\) are eigenvectors of both \(A\) and \(B\)

Finally, we need to show that the elements of set \(S\) are eigenvectors of both \(A\) and \(B\). Let's take a vector \(v\) from set \(S\). Since \(v\) was chosen as an eigenvector of \(A\) from the eigenspace \(E_\lambda\), we know that \(Av=\lambda v\). And from the construction of the basis set in the previous step, we also have \(Bv=\mu v\) for some eigenvalue \(\mu\) of \(B\). Therefore, \(v\) is an eigenvector of both \(A\) and \(B\), and this is true for all elements in the basis set \(S\). This means \(A\) and \(B\) are simultaneously diagonalizable.

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Key Concepts

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

Endomorphism
An endomorphism is a special type of function that maps a vector space to itself. In simpler terms, it's a transformation that doesn't take vectors outside of their original space. Endomorphisms are really important because they help us understand how vectors behave under certain operations.

To picture an endomorphism, think about rotating or reflecting shapes within a plane. These operations keep the shapes within the same space, similar to how an endomorphism would act on vectors.
  • **Key Point:** Endomorphisms preserve the structure of the vector space.
  • They help analyze properties like symmetry and dynamics within the vector space.
Understanding endomorphisms is crucial for diving deeper into concepts like eigenvalues and diagonalizability.
Eigenvalue
An eigenvalue is like a special "tag" that tells us how a particular transformation affects a vector. When you have a matrix (let鈥檚 say representing a transformation), and it acts on a vector, the result can be a scalar (the eigenvalue) times the original vector.

Mathematically, if a vector \( \mathbf{v} \) is transformed by matrix \( A \) and we get \( A\mathbf{v} = \lambda\mathbf{v} \), then \( \lambda \) is the eigenvalue corresponding to \( \mathbf{v} \).
  • **Critical Concept:** Eigenvalues allow us to understand how stretches, compressions, or rotations occur in the vector space.
  • They are essential for simplifying complex systems like in systems of differential equations.
Figuring out eigenvalues can help us predict how a transformation affects different directions in the space, especially when dealing with endomorphisms.
Eigenspace
When we talk about eigenspace, we refer to all vectors that are affected in the same way by a transformation, characterized by the same eigenvalue. If you imagine each eigenvalue as defining a "direction," the eigenspace represents all possible vectors that travel in that direction.
  • **Why It Matters:** Eigenspaces help us understand the geometry and structure introduced by transformations.
  • They allow for decomposing a vector space into simpler, understandable components.
For an eigenvalue \( \lambda \), the eigenspace is the set of all vectors \( \mathbf{v} \) such that \( A\mathbf{v} = \lambda\mathbf{v} \). Studying the properties of these spaces is vital for understanding simultaneous diagonalizability.
Eigenvector
An eigenvector is a non-zero vector that remains on the same line, even after a linear transformation is applied. It might stretch or compress, but its direction stays constant. When you multiply a matrix by this vector, the result is the vector itself scaled by the eigenvalue.

This relationship is expressed as \( A\mathbf{v} = \lambda\mathbf{v} \), where \( A \) is a matrix, \( \mathbf{v} \) is an eigenvector, and \( \lambda \) is the eigenvalue. Fine-tuning of these vectors is pivotal to diagonalization.
  • **Importance of Eigenvectors:** They offer insight into the stable states or directions unaffected by transformations, making them key in many applications.
  • They are key in constructing diagonal matrices and understanding the structure of transformations over a vector space.
Eigenvectors are the building blocks that make understanding transformations through matrices much easier.
Diagonalizable Matrix
A diagonalizable matrix is one that can be expressed in a diagonal form through a basis of eigenvectors. This process simplifies complex matrix operations, such as raising a matrix to large powers or finding its exponential.

The crux of diagonalization is to transform a matrix \( A \) into a format \( D \) with eigenvalues along the diagonal, through similar matrices, where \( A = PDP^{-1} \) and \( P \) is a matrix whose columns are eigenvectors of \( A \).
  • **Why Diagonalizability is Useful:** It reduces computational complexity and eases finding derived properties like matrix powers.
  • Simplifies solving differential equations, making it crucial in applied mathematics and engineering.
In simultaneous diagonalizability, two matrices share a common eigenbasis, greatly refining the analysis and understanding of properties linked to both matrices concurrently.

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Most popular questions from this chapter

Show that the following vectors are linearly independent, over \(\mathbf{R}\) and over \(\mathbf{C}\). (a) \((1,1,1)\) and \((0,1,-1)\) (b) \((1,0)\) and \((1,1)\) (c) \((-1,1,0)\) and \((0,1,2)\) (d) \((2,-1)\) and \((1,0)\) (c) \((\pi, 0)\) and \((0,1)\) (f) \((1,2)\) and \((1,3)\) (g) \((1,1,0),(1,1,1)\) and \((0,1,-1)\) (h) \((0,1,1),(0,2,1)\) and \((1,5,3)\)

Let \(E\) be a free module over the ring \(R\), with basis \(\left\\{0_{1} \ldots, v_{n}\right\\}\). Let \(F\) be a module, and \(w_{1}, \ldots, w_{n}\) elements of \(F\). Show that there exists a unique homomorphism \(f: E \rightarrow F\) such that \(f\left(v_{i}\right)=w_{i}\) for \(i=1, \ldots, n\).

Let \(A\) be an invertible matrix. If \(\lambda\) is an cigenvalue of \(A\) show that \(\lambda \neq 0\) and that \(\lambda^{-1}\) is an eigenvalue of \(A^{-1}\).

Let \(R\) be a ring. Show that \(R\) can be viewed as a module over itself, and has one generator.

Express the given vector \(X\) as a linear combination of the given vectors \(A, B\) and find the coordinates of \(X\) with respect to \(A, B\). (a) \(X=(1,0), A=(1,1), B=(0,1)\) (b) \(X=(2,1), A=(1,-1), B=(1,1)\) (c) \(X=(1,1), A=(2,1), B=(-1,0)\) (d) \(X=(4,3), A=(2,1), B=(-1,0)\) (You may view the above vectors as elements of \(\mathbf{R}^{2}\) or \(\mathbf{C}^{2}\). The coordinates will be the same.)
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