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Chapter 5: Problem 3

Let \(V\) be a finite dimensional vector space, and let \(A: V \rightarrow V\) be an endomorphism. Suppose \(A^{2}=A\). Show that there is a basis of \(V\) such that the matrix of \(A\) with respect to this basis is diagonal, with only 0 or 1 on the diagonal. Or, if you prefer, show that \(V=V_{0} \oplus V_{1}\) is a direct sum, where \(V_{0}=\) Ker \(A\) and \(V_{1}\) is the \((+1)\) -eigenspace of \(A\).

Short Answer

Expert verified
In summary, we showed that there exist eigenspaces V鈧 and V鈧 corresponding to eigenvalues 0 and 1, respectively, such that V is a direct sum of V鈧 and V鈧. By taking a basis B for V formed by the union of bases of V鈧 and V鈧, the matrix of endomorphism A with respect to basis B is diagonal, with only 0 or 1 on the diagonal.

Step by step solution

01

Find the eigenspaces of A

First, let's define the eigenspaces of A corresponding to the eigenvalues 0 and 1: V鈧 = {v 鈭 V | Av = 0} (the kernel of A) V鈧 = {v 鈭 V | Av = v} (the (+1)-eigenspace of A)
02

Prove that V is a direct sum of V鈧 and V鈧

To show that V is a direct sum of V鈧 and V鈧, we must show two things: 1. V鈧 鈭 V鈧 = {0} 2. V鈧 + V鈧 = V To prove that V鈧 鈭 V鈧 = {0}, consider any vector v in V鈧 鈭 V鈧. Then we have both Av = 0 and Av = v. Since Av = v = 0, v must be the zero vector. Thus, V鈧 鈭 V鈧 = {0}. Now, we need to show that V鈧 + V鈧 = V. For any vector v 鈭 V, consider the vector u = v - Av. Since A^2 = A, we have: Au = A(v - Av) = A(v) - A(Av) = Av - A^2v = Av - Av = 0, which means u 鈭 V鈧. Now, let w = Av. Then Aw = A(Av) = A^2v = Av = w, and hence w 鈭 V鈧. Since v = u + w, V鈧 + V鈧 = V. Thus, we have shown that V is a direct sum of V鈧 and V鈧.
03

Diagonalize A

Now, let's find a basis for V鈧 and V鈧. We can extend the bases of each of these eigenspaces to obtain a basis B鈧 for V鈧 and a basis B鈧 for V鈧. Then, we can form a basis B for V by taking the union of these bases, i.e., B = B鈧 鈭 B鈧. Now, let's see what the matrix of A looks like with respect to this basis B. For any basis vector v 鈭 B鈧, we have Av = 0, which means that the column of A corresponding to this basis vector will have a 0 in the diagonal entry and 0's in all other entries. For any basis vector v 鈭 B鈧, we have Av = v, which means that the column of A corresponding to this basis vector will have a 1 in the diagonal entry and 0's in all other entries. Therefore, the matrix representation of A with respect to basis B is diagonal, with only 0 or 1 on the diagonal, as desired.

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Key Concepts

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

Vector spaces
A vector space is a fundamental concept in algebra and is essential for understanding the diagonalization of matrices. It consists of a collection of objects called vectors, which can be added together and multiplied by scalars (numbers from a field) in a way that satisfies certain axioms, such as associativity, commutativity, and distributivity.

For our purposes, think of a vector space as a stage where the entire action takes place. The vectors themselves can represent anything from physical quantities like force, to abstract concepts like polynomials or functions. The only requirement is that they adhere to the rules of the vector space. When diagonalizing a matrix, what we are really doing is finding a basis鈥攁 set of vectors that spans the entire vector space and consists of eigenvectors, making the matrix representation relatively simple and, importantly, diagonal.
Endomorphism
An endomorphism is a special type of function in the realm of vector spaces. Specifically, an endomorphism is a linear transformation that maps a vector space to itself. This means that for an endomorphism, say, \( A \), if you take any vector \( v \) in the vector space \( V \), the product \( A(v) \) is also a vector in the same space.

Consider our original problem with the transformation \( A \) that satisfies \( A^{2} = A \). In the world of linear transformations, such a property hints that \( A \) might have special eigenvectors corresponding to eigenvalues 0 and 1 because \( A \) has a sort of 'idempotent' behavior鈥攊t doesn't change upon repeated applications.
Direct sum decomposition
Direct sum decomposition is an elegant way of breaking down a vector space into simpler, non-overlapping pieces. When we say that a vector space \( V \) is the direct sum of two subspaces, \( V_0 \) and \( V_1 \), we are stating that each vector in \( V \) can be uniquely written as the sum of vectors from each subspace.

This concept ties beautifully into the idea of splitting a vector space into eigenspaces when diagonalizing a matrix. By showing that the vector space is the direct sum of eigenspaces, we essentially demonstrate that every vector in our space is a clear-cut combination of eigenvectors, thus making the diagonalization process possible and relatively straightforward.
Eigenspaces
Eigenspaces are central to the process of diagonalizing a matrix. An eigenspace arises from an eigenvalue and consists of all vectors that, when transformed by the matrix (or linear operator), simply get scaled by the eigenvalue鈥攈ence, the term 'eigen,' which is German for 'own.' Each eigenspace corresponds to an eigenvalue, and the vectors in that eigenspace are called eigenvectors.

In our exercise, the endomorphism \( A \) has eigenspaces \( V_0 \) and \( V_1 \), corresponding to the eigenvalues 0 and 1, respectively. What's compelling here is that by organizing a vector space as a direct sum of eigenspaces, we can pick a basis that turns the transformation's matrix into a simple diagonal matrix where each entry represents the eigenvalue associated with each eigenspace.

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Most popular questions from this chapter

Let \(V\) be a finite dimensional vector space over \(K .\) Let \(W\) be a subspace. Let \(\left\\{w_{1}, \ldots, w_{m}\right\\}\) be a basis of \(W\), Show that there exist elements \(w_{m+1} \ldots \ldots w_{n}\) in \(V\) such that \(\left\\{w_{1}, \ldots, w_{n}\right\\}\) is a basis of \(V\).

Let \(K\) be a field and \(R=K[X]\) the polynomial ring over \(K\). Let \(f(X)\) be a polynomial of degree \(d>0\) in \(K[X]\). Let \(J\) be the ideal generated by \(f(X)\). What is the dimension of \(R / J\) over \(K\) ? Exhibit a basis of \(R / J\) over \(K\). Show that \(R / J\) is an integral ring if and only if \(f\) is irreducible.

Let \(V\) be a finite dimensional vector space over the field \(K\). Let \(R\) be the ring of \(K\) -linear maps of \(V\) into itself. Show that \(R\) has no two- sided ideals except \(\\{O\\}\) and \(R\) itself. [Hint: Let \(A \in R, A \neq O\). Let \(v_{1} \in V, v_{t} \neq 0\), and \(A v_{1} \neq 0\). Complete \(v_{1}\) to a basis \(\left\\{v_{1+\ldots, v_{n}}\right\\}\) of \(V\). Let \(\left\\{w_{1}, \ldots, w_{n}\right\\}\) be arbitrary elcments of \(V .\) For each \(i=1, \ldots, n\) there exists \(B_{i} \in R\) such that $$ B_{1} v_{i}=v_{1} \quad \text { and } \quad B_{i} v_{j}=0 \text { if } j \neq i_{,} $$ and there exists \(C_{i} \in R\) such that \(C_{1} A v_{1}=w_{i}\) (justify these two existence statements in detail). Let \(F=C_{1} A B_{1}+\cdots+C_{n} A B_{s^{+}}\) Show that \(F\left(v_{j}\right)=w_{i}\) for all \(i=1, \ldots, n .\) Conclude that the two-sided ideal generated by \(A\) is the whole ring \(R .]\)

Let \(E\) be a module over the ring \(R\), and let \(L\) be a left ideal of \(R\). Let \(L E\) be the set of all elements \(x_{1} v_{1}+\cdots+x_{n} t_{n}\) with \(x_{i} \in L\) and \(v_{j} \in E\), Show that \(L E\) is a submodule of \(E\).

Let \(U, W\) be subspaces of a vector space \(V\). (a) Show that \(U+W\) is a subspace. (b) Define \(U \times W\) to be the set of all pairs \((u, w)\) with \(u \in U\) and \(w \subset W\). Show how \(U \times W\) is a vector space. If \(U, W\) are finite dimensional, show that $$ \operatorname{dim}(U \times W)=\operatorname{dim} U+\operatorname{dim} W $$ (c) Prove that \(\operatorname{dim} U+\operatorname{dim} W=\operatorname{dim}(U+W)+\operatorname{dim}(U \cap W)\). [Hint: Con- sider the linear map \(f: U \times W \rightarrow U+W\) given by \(f(u, w)=u-w\). \(]\)
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