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Let $A$ be an orthogonal $3 \times 3$ matrix with real entries. Pick out the true statements: a. The determinant of $A$ is a rational number. b. $d(Ax,Ay) = d(x, y)$ for any two vectors $x, y \in \mathbb{R}^3$, where $d(u, v)$ denotes the usual Euclidean distance between vectors $u, v \in \mathbb{R}^3$. c. All the entries of $A$ are positive. d. All the eigenvalues of $A$ are real. |
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(a) True. $A$ being orthogonal means $A^tA = I$. Hence $\mathrm{det} (A^tA) = \mathrm{det}(I)$; i.e., $\mathrm{det}(A)^2 = 1$. So $\mathrm{det}(A) = \pm 1$. That is, its determinant is a rational number. (b) True: $d(Ax,Ay) = \vert\vert Ax - Ay \vert\vert = \vert\vert A(x-y) \vert\vert$, on the one hand. On the other hand, $d(x,y) = \vert\vert x-y \vert\vert$. So it's enough to prove that $\vert\vert Au \vert\vert = \vert\vert u \vert\vert$ for every vector $u$, right? And, since both guys in this last equation are positive numbers, it's enough to see that $\vert\vert Au \vert\vert^2 = \vert\vert u \vert\vert^2$. This fact can be deduced as follows: $$ \vert\vert Au \vert\vert^2 = \langle Au, Au \rangle = (Au)^tAu = u^tA^tAu = \langle u, A^tAu \rangle = \langle u,u \rangle = \vert\vert u \vert\vert^2 \ . $$ (c) False. For instance, $$ \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ is an orthogonal $3\times 3$ real matrix and not all of its entries are positive. (d) Shame on me. See Paul Z.'s comment. |
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