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Chapter 5: Problem 15

If \(A\) is nilpotent and not \(O\), show that \(A\) is not diagonalizable.

Short Answer

Expert verified
The matrix \(A\) is nilpotent and not equal to the zero matrix. All eigenvalues of \(A\) are 0, as shown in step 1. If \(A\) were diagonalizable, we would have a diagonal matrix \(D\) with all eigenvalues (0) on the diagonal. However, this would lead to a contradiction, as shown in step 3, where \(A\) would be equal to the zero matrix, which is not the case. Thus, \(A\) is not diagonalizable.

Step by step solution

01

1. Eigenvalues of a nilpotent matrix

Since \(A\) is a nilpotent matrix, there exists a positive integer \(k\) such that \(A^k=0\). We can prove that all eigenvalues of a nilpotent matrix must be 0. Let \(\lambda\) be an eigenvalue of \(A\), and \(v\) be the corresponding eigenvector, such that \(Av = \lambda v\). Now, we raise this equation to the \(k\)-th power, where \(k\) is the nilpotency index: $$(Av)^{k} = (\lambda v)^{k} \Rightarrow A^k v = \lambda^k v$$ Since \(A^k = 0\), we have \(0v = \lambda^k v\), and thus, \(\lambda^k = 0\). It follows that \(\lambda=0\).
02

2. The eigenvalues of a diagonalizable matrix

A diagonalizable matrix \(A\) has an invertible matrix \(P\) such that \(P^{-1}AP\) is a diagonal matrix. Let \(D=P^{-1}AP\). The diagonal entries of \(D\) are the eigenvalues of \(A\). Since \(A\) is nilpotent, we have found that all its eigenvalues are 0. Therefore, \(D\) is the zero matrix.
03

3. Diagonalizability contradiction

Now we have a contradiction. If \(A\) were diagonalizable, then we would have \(A=PDP^{-1}\) with \(D\) being the zero matrix. In this case, \(A=PD(P^{-1})P\), which is the zero matrix since \(PD=0\). But we know that \(A\) is not the zero matrix as given in the problem statement. Therefore, A cannot be diagonalizable. In conclusion, if \(A\) is nilpotent and not the zero matrix, then \(A\) is not diagonalizable.

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Key Concepts

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

Eigenvalues
Understanding eigenvalues is crucial when studying linear algebra, especially in the context of nilpotent matrices. An eigenvalue of a matrix is a scalar that, when the matrix is multiplied by an eigenvector associated with it, the result is simply a scalar multiple of that eigenvector. In mathematical terms, for a matrix \(A\) and eigenvector \(v\), the relationship \(Av = \lambda v\) holds, where \(\lambda\) represents the eigenvalue.

In the case of nilpotent matrices, a fascinating property emerges: all eigenvalues must be zero. Why is that so? Imagine repeatedly applying the matrix to its eigenvector. According to the definition of nilpotency, eventually, this repeated application will result in the zero vector. Now, since any power of zero is zero, the only possible eigenvalue for a nilpotent matrix is zero. This seemingly simple attribute is highly significant and forms the foundation for understanding why such matrices cannot be diagonalized if they are non-zero matrices themselves.
Diagonalizable Matrices
The concept of diagonalizable matrices ties directly into the heart of linear transformations and matrix representations. A matrix \(A\) is considered diagonalizable if it can be expressed in the form \(P^{-1}AP\), with \(P\) being an invertible matrix and \(D\) being a diagonal matrix. Diagonal matrices only have non-zero elements on their main diagonal, and these non-zero elements correspond to the eigenvalues of \(A\).

Why does this matter? Diagonal matrices are easier to raise to powers, compute exponentials, and solve differential equations, making diagonalizability a highly desirable property. However, in the context of nilpotent matrices that are non-zero, we encounter a conflict: as all eigenvalues are zero, diagonalizing such a matrix would imply converting it into the zero matrix, contradicting the fact that the matrix is non-zero. Therefore, non-zero nilpotent matrices stand apart as they cannot be diagonalized.
Nilpotency Index
Delving into the nilpotency index provides insight into the depth of nilpotence. The term refers to the smallest positive integer \(k\) such that applying the matrix to itself \(k\) times yields the zero matrix, i.e., \(A^k = 0\).

The nilpotency index is indicative of the matrix's inherent structure; it tells us how many times we must apply the matrix in succession to arrive at a state of 'nullity' or complete absence of effect. Coming back to eigenvalues, this ties into the fact that since \(A^k v = \lambda^k v = 0v\), and the eigenvector \(v\) is non-zero, the only logical conclusion is \(\lambda^k = 0\).

By understanding the nilpotency index, students not only grasp the inevitability of all eigenvalues being zero for nilpotent matrices but also why these matrices inherently contradict the nature of diagonalizable matrices, which rely on distinct eigenvalues to form a basis of eigenvectors.

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Most popular questions from this chapter

Let \(R\) be a ring containing a field \(K\) as a subfield with \(K\) in the center of \(R\). Assume that \(K\) is algebraically closed. Assume that \(R\) has no two- sided ideal other than 0 and \(R\). We also assume that \(R\) is of finite dimension \(>0\) over \(K\). Let \(L\) be a left ideal of \(R\), of smallest dimension \(>0\) over \(K\). (a) Prove that End \(_{R}(L)=K\) (t.e. the only \(R\) -linear maps of \(L\) consist of multiplication by elements of \(K\) ). [Hint: Cf. Schur's lemma and Exercise 23.] (b) Prove that \(R\) is ring-isomorphic to the ring of \(K\) -linear maps of \(L\) into itself. [Hint: Use Wedderburn-Rieffel.]

Assume that \(V\) is a simple \(S\) -space and that \(A B=B A\) for all \(B \in S\). Prove that either \(A\) is invertible or \(A\) is the zero map. Using the fact that \(V\) is finite dimensional and \(K\) algebraically closed, prove that there exists \(\alpha \in K\) such that \(A=x l .\)

Let \(M\) be an \(n \times n\) diagonal matrix with eigenvalues \(\lambda_{1} \ldots \ldots \hat{\lambda}_{r}\). Suppose that i. has multiplicity \(m_{i}\). Write down the minimal polynomial of \(M\), and also write down its characteristic polynomial.

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 \(f(u, w)=u-w\). \(]\)

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 \(R .]\)
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