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Chapter 10: Problem 9

Prove that if \(P \in \mathcal{L}(V)\) satisfies \(P^{2}=P\), then trace \(P\) is a nonnegative integer.

Short Answer

Expert verified
Since a projector \(P\) satisfies \(P^2 = P\), we know its eigenvalues can only be \(0\) or \(1\). The trace of \(P\) is equal to the sum of its eigenvalues, which means it is equal to the number of eigenvalues equal to \(1\). Since the number of such eigenvalues must be a nonnegative integer, we conclude that the trace of \(P\) is a nonnegative integer.

Step by step solution

01

Identifying Complementary Subspaces of P

First, recall the definition of a projector: it is a linear transformation that is idempotent, meaning it satisfies \(P^2 = P\). This means that the range of \(P\) (\(\text{range}(P)\)) and the null space of \(P\) (\(\text{null}(P)\)) are complementary subspaces with respect to the domain, i.e., they are disjoint and their direct sum is the entire domain.
02

Eigenvectors and Eigenvalues of P

We know that the eigenvectors of a linear transformation lie in its range and null space. By the definition of a projector, all eigenvectors corresponding to eigenvalues of \(1\) will be in the range of \(P\), and all eigenvectors corresponding to eigenvalues of \(0\) will be in the null space of \(P\). In other words, each eigenvector in the range of \(P\) will satisfy: \[Pv = v\] And each eigenvector in the null space of \(P\) will satisfy: \[Pv = 0\] Since \(P\) is a projector, its eigenvalues can only be \(0\) or \(1\).
03

Determine the Trace of P

Now, let's denote the matrix representation of \(P\) as \(A\). The trace of \(P\) (i.e., the sum of the diagonal elements of \(A\)) is equal to the sum of its eigenvalues, as shown below: \[\text{trace}(P) = \sum_i \lambda_i\] Where \(\lambda_i\) is an eigenvalue of \(P\), and the sum runs over all eigenvalues. Since the eigenvalues of a projector can only be \(0\) or \(1\), the sum of the eigenvalues (and therefore the trace of \(P\)) will be equal to the number of eigenvalues equal to \(1\).
04

Conclude that the Trace of P is a Nonnegative Integer

As established in Steps 2 and 3, the trace of \(P\) is equal to the number of eigenvalues that are equal to \(1\). Since the number of such eigenvalues must be a nonnegative integer, the trace of \(P\) must also be a nonnegative integer. Thus, we have proven that if \(P \in \mathcal{L}(V)\) satisfies \(P^2 = P\), then the trace of \(P\) is a nonnegative integer.

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Most popular questions from this chapter

Suppose \(A\) is a block upper-triangular matrix \\[ A=\left[\begin{array}{ccc} A_{1} & & * \\ & \ddots & \\ 0 & & A_{m} \end{array}\right] \\] where each \(A_{j}\) along the diagonal is a square matrix. Prove that \\[ \operatorname{det} A=\left(\operatorname{det} A_{1}\right) \ldots\left(\operatorname{det} A_{m}\right) \\].

Suppose \(T \in \mathcal{L}\left(\mathbf{C}^{3}\right)\) is the operator whose matrix is \\[ \left[\begin{array}{ccc} 51 & -12 & -21 \\ 60 & -40 & -28 \\ 57 & -68 & 1 \end{array}\right] \\] Someone tells you (accurately) that -48 and 24 are eigenvalues of \(T .\) Without using a computer or writing anything down, find the third eigenvalue of \(T\).

Suppose \(T \in \mathcal{L}(V)\) and \(\left(\nu_{1}, \ldots, \nu_{n}\right)\) is a basis of \(V .\) Prove that \(\mathcal{M}\left(T,\left(\nu_{1}, \ldots, \nu_{n}\right)\right)\) is invertible if and only if \(T\) is invertible.

Suppose \(V\) is an inner-product space. Prove that if \(T \in \mathcal{L}(V)\) is a positive operator and trace \(T=0,\) then \(T=0\).

Suppose \(A\) is an \(n\) -by- \(n\) matrix with real entries. Let \(S \in \mathcal{L}\left(\mathbf{C}^{n}\right)\) denote the operator on \(\mathbf{C}^{n}\) whose matrix equals \(A,\) and let \(T \in\) \(\mathcal{L}\left(\mathbf{R}^{n}\right)\) denote the operator on \(\mathbf{R}^{n}\) whose matrix equals \(A\). Prove that trace \(S=\) trace \(T\) and \(\operatorname{det} S=\operatorname{det} T\).
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