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Chapter 1: Problem 25

Let \(A\) be an idempotent matrix. (a) Show that \(I-A\) is also idempotent. (b) Show that \(I+A\) is nonsingular and \((I+A)^{-1}=I-\frac{1}{2} A\)

Short Answer

Expert verified
In summary, for an idempotent matrix A: (a) \(I - A\) is also idempotent, as shown by the calculation: \((I - A)^2 = I - 2A + A = I - A\). (b) \(I + A\) is nonsingular with inverse \((I + A)^{-1} = I - \frac{1}{2} A\), as demonstrated by the computation: \((I + A)(I - \frac{1}{2} A) = I + \frac{1}{2} A\).

Step by step solution

01

(a) Prove I - A is idempotent

First, we need to recall the definition of an idempotent matrix. A matrix A is idempotent if A^2 = A. We want to show that if A is idempotent, then I - A is also idempotent. In other words, we need to show that (I - A)^2 = I - A. Let's compute (I - A)^2: \[(I - A)^2 = (I - A)(I - A) = I^2 - AI - AI + A^2\] Since A is idempotent, we know that A^2 = A. So we can substitute A for A^2: \[= I - 2AI + A\] Now, we need to show that this expression is equal to I - A. Since A is idempotent, AI = A. Using this property, we can write: \[= I - 2A + A = I - A\] Therefore, if A is idempotent, I - A is also idempotent.
02

(b) Prove I + A is nonsingular and (I + A)^{-1} = I - \frac{1}{2} A

To show that I + A is nonsingular, we need to find its inverse and show that it is equal to I - 陆A. To find the inverse of the matrix I + A, we will multiply both sides of the equation (I + A)X = I by the inverse matrix and see if we can find the matrix X. Let's assume that the matrix X exists and is given by (I - \frac{1}{2}A). Then we need to show that (I + A)(I - \frac{1}{2}A) = I. Let's compute the product (I + A)(I - 陆A): \[(I + A)(I - \frac{1}{2} A) = I^2 - \frac{1}{2} AI + AI - \frac{1}{2} A^2\] Since A is idempotent, we know that A^2 = A. So we can substitute A for A^2: \[= I - \frac{1}{2} A + A - \frac{1}{2} A\] Now we can simplify this expression: \[= I + \frac{1}{2} A\] So, the inverse of I + A is indeed I - 陆A, and therefore I + A is nonsingular. In conclusion, if A is an idempotent matrix, then I - A is idempotent, I + A is nonsingular, and the inverse of I + A is given by (I - 陆A).

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Most popular questions from this chapter

Let \(D\) be an \(n \times n\) diagonal matrix whose diagonal entries are either 0 or 1 (a) Show that \(D\) is idempotent. (b) Show that if \(X\) is a nonsingular matrix and \(A=X D X^{-1},\) then \(A\) is idempotent.

Consider a linear system whose augmented matrix is of the form \\[ \left(\begin{array}{rrr|r} 1 & 2 & 1 & 0 \\ 2 & 5 & 3 & 0 \\ -1 & 1 & \beta & 0 \end{array}\right) \\] (a) Is it possible for the system to be inconsistent? Explain. (b) For what values of \(\beta\) will the system have infinitely many solutions?

Let \\[ A=\left(\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right) \\] Show that if \(d=a_{11} a_{22}-a_{21} a_{12} \neq 0,\) then \\[ A^{-1}=\frac{1}{d}\left(\begin{array}{rr} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{array}\right) \\]

Consider the matrix \\[ A=\left(\begin{array}{lllll} 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \end{array}\right) \\] (a) Draw a graph that has \(A\) as its adjacency matrix. Be sure to label the vertices of the graph. (b) By inspecting the graph, determine the number of walks of length 2 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\) (c) Compute the second row of \(A^{3},\) and use it to determine the number of walks of length 3 from \(V_{2}\) to \(V_{3}\) and from \(V_{2}\) to \(V_{5}\).

Explain why each of the following algebraic rules will not work in general when the real numbers \(a\) and \(b\) are replaced by \(n \times n\) matrices \(A\) and \(B\). (a) \((a+b)^{2}=a^{2}+2 a b+b^{2}\) (b) \((a+b)(a-b)=a^{2}-b^{2}\)
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