Homepage › Solution manuals › Sheldon Axler › Linear Algebra Done Right › Exercise 7.B.1

Exercise 7.B.1

Answers

Proof. False. Define $T \in L (ℝ^{3})$ by

\begin{aligned} T e_{1} & = e_{1} \\ T e_{2} & = 2 e_{2} \\ T (e_{1} + e_{2} + e_{3}) & = 3 e_{1} + 3 e_{2} + 3 e_{3}, \end{aligned}

where $e_{1}, e_{2}, e_{3}$ is the standard basis of $ℝ^{3}$ . Clearly, $e_{1}, e_{2}, e_{1} + e_{2} + e_{3}$ is a basis of $ℝ^{3}$ consisting of eigenvectors of $T$ . However, the matrix of $T$ with respect to the standard basis (which is orthonormal) is

(\begin{matrix} 1 & 0 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{matrix}),

which does not equal its transpose (the matrix of $T^{∗}$ , by 7.10). Therefore $T$ is not self-adjoint. □

celiopassos

2017-10-06 00:00

Exercise 7.B.1

Answers

Comments

Add answer