/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Suppose that \(A \in M_{n \times... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 8

Suppose that \(A \in M_{n \times n}(F)\) has two distinct eigenvalues, \(\lambda_{1}\) and \(\lambda_{2}\), and that \(\operatorname{dim}\left(\mathrm{E}_{\lambda_{1}}\right)=n-1\). Prove that \(A\) is diagonalizable.

Short Answer

Expert verified
A is diagonalizable because the sum of dimensions of its eigenspaces, corresponding to the two distinct eigenvalues \( \lambda_1 \) and \( \lambda_2 \), equals the size of the matrix. Specifically, we are given dim(\( E_{\lambda_1} \)) = n-1, and we found that dim(\( E_{\lambda_2} \)) = 1. Since their sum (n-1) + 1 = n equals the size of the matrix A, A is diagonalizable.

Step by step solution

01

Find the dimension of eigenspace corresponding to each eigenvalue

We are given that the dimension of eigenspace corresponding to \( \lambda_1 \) is n-1. Let dim(\( E_{\lambda_2} \)) be x. We need to find the value of x.
02

Use the eigenspace equation to find the eigenspace dimensions

We know that the sum of dimensions of eigenspaces for a matrix A is equal to the size of the matrix (n, in this case) if and only if the matrix is diagonalizable. In other words: dim(\( E_{\lambda_1} \)) + dim(\( E_{\lambda_2} \)) = n We are given dim(\( E_{\lambda_1} \)) = n-1. Substituting this into the above equation, we get: (n-1) + x = n
03

Calculate the dimension of eigenspace corresponding to \( \lambda_2 \)

We have the equation (n-1) + x = n. Simplifying this, we get: x = n - (n - 1) x = 1 So, dim(\( E_{\lambda_2} \)) = 1.
04

Check if the sum of dimensions of eigenspaces equals the size of the matrix

Now we have the dimensions of eigenspaces for both eigenvalues: dim(\( E_{\lambda_1} \)) = n-1 and dim(\( E_{\lambda_2} \)) = 1. Let's check if the sum of these dimensions equals n: (n-1) + 1 = n n = n This equation holds true, so the sum of dimensions of eigenspaces equals the size of the matrix A.
05

Conclude that A is diagonalizable

Since the sum of dimensions of eigenspaces corresponding to \( \lambda_1 \) and \( \lambda_2 \) equals n, the size of the matrix A, we can conclude that A is diagonalizable.

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Most popular questions from this chapter

For each linear map \(F\) find a basis and the dimension of the kernel and the image of \(F\) (a) \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) defined by \(F(x, y, z)=(x+2 y-3 z, 2 x+5 y-4 z, x+4 y+z)\) (b) \(F: \mathbf{R}^{4} \rightarrow \mathbf{R}^{3}\) defined by \(F(x, y, z, t)=(x+2 y+3 z+2 t, 2 x+4 y+7 z+5 t, x+2 y+6 z+5 t)\)

Define \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) and \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) by \(F(x, y, z)=(2 x, y+z)\) and \(G(x, y, z)=(x-z, y)\) Find formulas defining the maps: (b) \(3 F\) (a) \(F+G\) (c) \(2 F-5 G\)

Suppose \(F_{1}, F_{2}, \ldots, F_{n}\) are linear maps from \(V\) into \(U .\) Show that, for any scalars \(a_{1}, a_{2}, \ldots, a_{n}\) and for any \(v \in V\) \\[ \left(a_{1} F_{1}+a_{2} F_{2}+\cdots+a_{n} F_{n}\right)(v)=a_{1} F_{1}(v)+a_{2} F_{2}(v)+\cdots+a_{n} F_{n}(v) \\]

Let \(F: \mathbf{R}^{4} \rightarrow \mathbf{R}^{3}\) be the linear mapping defined by \\[ F(x, y, z, t)=(x-y+z+t, \quad x+2 z-t, \quad x+y+3 z-3 t) \\] Find a basis and the dimension of (a) the image of \(F,\) (b) the kernel of \(F.\)

Consider the linear operator \(T\) on \(\mathbf{R}^{3}\) defined by \(T(x, y, z)=(2 x, 4 x-y, 2 x+3 y-z)\) (a) Show that \(T\) is invertible. Find formulas for (b) \(T^{-1}\), \((\mathrm{c}) T^{2},(d) T^{-2}\)
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