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Chapter 6: Problem 8

An \(n \times n\) matrix \(A\) is said to be idempotent if \(A^{2}=A .\) Show that if \(\lambda\) is an eigenvalue of an idempotent matrix, then \(\lambda\) must be either 0 or 1

Short Answer

Expert verified
Recalling the definition of eigenvalue and eigenvector, we use the idempotent property of the matrix \(A\) to substitute the eigenvector equation into the idempotent equation. We find an expression for \(A^2 v\) and use the given properties to show that the eigenvalue \(\lambda\) must be either 0 or 1.

Step by step solution

01

Recall the definition of eigenvalue and eigenvector

An eigenvalue \(\lambda\) with its associated eigenvector \(v\) for a given matrix \(A\) satisfies the equation: \[Av = \lambda v,\] where \(v\) is a non-zero vector.
02

Use the idempotent property of the matrix A

Since \(A\) is idempotent, we have the equation \(A^2 = A\). In order to demonstrate that the eigenvalue \(\lambda\) is either 0 or 1, we can substitute the eigenvector equation (from Step 1) into the idempotent equation. First, let's find an expression for \(A^2 v\): \[A^2 v = A(Av).\] We know that \(Av = \lambda v\), so we can substitute this into the previous equation: \[A^2 v = A (\lambda v).\] Now, use the idempotent property of the matrix \(A\) to write: \[Av = A(\lambda v).\]
03

Show that 位 must be 0 or 1

From Step 2, we have: \[Av = A(\lambda v).\] Since \(Av = \lambda v\), we can write: \[\lambda v = A(\lambda v).\] Now, let's factor \(\lambda\) from the right-hand side of the equation: \[\lambda v = \lambda (Av).\] We know that \(Av = \lambda v\), so we can substitute this into the previous equation: \[\lambda v = \lambda (\lambda v).\] Now there are two possibilities: 1. If \(\lambda = 0\), then we have \(\lambda v = 0\). 2. If \(\lambda \neq 0\), we can divide both sides by \(\lambda\). Doing this yields: \[v = \lambda v.\] In this case, since \(v\) is a non-zero vector, we must have \(\lambda = 1\). Hence, the eigenvalue \(\lambda\) must be either 0 or 1.

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Most popular questions from this chapter

Let \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\\}\) be an orthonormal basis for \(\mathbb{C}^{2},\) and let \(\mathbf{z}=(4+2 i) \mathbf{u}_{1}+(6-5 i) \mathbf{u}_{2}\) (a) What are the values of \(\mathbf{u}_{1}^{H} \mathbf{z}, \mathbf{z}^{H} \mathbf{u}_{1}, \mathbf{u}_{2}^{H} \mathbf{z},\) and \(\mathbf{z}^{H} \mathbf{u}_{2} ?\) (b) Determine the value of \(\|\mathbf{z}\|\)

Given that \\[ A=\left(\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 1 & i \\ 0 & -i & 1 \end{array}\right) \\] find a matrix \(B\) such that \(B^{H} B=A\)

For each of the following, find all possible values of the scalar \(\alpha\) that make the matrix defective or show that no such values exist: (a) \(\left(\begin{array}{lll}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & \alpha\end{array}\right)\) (b) \(\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (c) \(\left(\begin{array}{rrr}1 & 2 & 0 \\ 2 & 1 & 0 \\ 2 & -1 & \alpha\end{array}\right)\) \((\mathbf{d})\left(\begin{array}{rrr}4 & 6 & -2 \\ -1 & -1 & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (e) \(\left(\begin{array}{ccc}3 \alpha & 1 & 0 \\ 0 & \alpha & 0 \\ 0 & 0 & \alpha\end{array}\right)\) (f) \(\left(\begin{array}{ccc}3 \alpha & 0 & 0 \\ 0 & \alpha & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (g) \(\left(\begin{array}{ccc}\alpha+2 & 1 & 0 \\ 0 & \alpha+2 & 0 \\ 0 & 0 & 2 \alpha\end{array}\right)\) (h) \(\left(\begin{array}{ccc}\alpha+2 & 0 & 0 \\ 0 & \alpha+2 & 1 \\ 0 & 0 & 2 \alpha\end{array}\right)\)

Show that if \(A\) is a symmetric positive definite matrix, then \(A\) is nonsingular and \(A^{-1}\) is also positive definite.

Let \(\left\\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{n}\right\\}\) be an orthonormal basis for a complex inner product space \(V\), and let \\[ \begin{array}{l} \mathbf{z}=a_{1} \mathbf{u}_{1}+a_{2} \mathbf{u}_{2}+\cdots+a_{n} \mathbf{u}_{n} \\ \mathbf{w}=b_{1} \mathbf{u}_{1}+b_{2} \mathbf{u}_{2}+\cdots+b_{n} \mathbf{u}_{n} \end{array} \\] Show that \\[ \langle\mathbf{z}, \mathbf{w}\rangle=\sum_{i=1}^{n} \bar{b}_{i} a_{i} \\]
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