91影视

In linear algebra, a symmetric matrix is a square matrix that stays unchanged when its transpose is calculated. In that instance, a symmetric matrix is one whose transpose equals the matrix itself.

Given that, A is any symmetric 3x3 matrix with eigenvalues -2,3 and 4 , and $\stackrel{鈫�}{u}$is a unit vector in ${\mathrm{鈩�}}^3$. From the spectral theorem, we know that there exists an orthonormal eigenbasis ${\stackrel{鈫�}{v}}_{3}forT$, with associated real eigenvalues ${位}_1=4,{位}_2=3$and ${位}_3=-2$(Arrange things so that $\left|{位}_1\right|鈮�\left|{位}_2\right|鈮�\left|{位}_3\right|$). Now consider unit vector as represented below:

$$\begin{gathered} \stackrel{鈫�}{u}=c_1{\stackrel{鈫�}{v}}_1+c_2{\stackrel{鈫�}{v}}_2+c_3{\stackrel{鈫�}{v}}_3 \\ 鈬�A\stackrel{鈫�}{u}={位}_1{c}_1{\stackrel{鈫�}{v}}_1+{位}_2{c}_2{\stackrel{鈫�}{v}}_2+{位}_3{c}_3{\stackrel{鈫�}{v}}_3 \\ 鈬�A\stackrel{鈫�}{u}=4c_1{\stackrel{鈫�}{v}}_1+3c_2{\stackrel{鈫�}{v}}_2+2c_3{\stackrel{鈫�}{v}}_3 \end{gathered}$$

Now evaluate as follows:

$$\begin{gathered} \stackrel{鈫�}{u}.A\stackrel{鈫�}{u}=\left(c_1{\stackrel{鈫�}{v}}_1+c_2{\stackrel{鈫�}{v}}_2+c_3{\stackrel{鈫�}{v}}_3\right).\left(4c_1{\stackrel{鈫�}{v}}_1+3c_2{\stackrel{鈫�}{v}}_2+2c_3{\stackrel{鈫�}{v}}_3\right) \\ 鈬�\stackrel{鈫�}{u}.A\stackrel{鈫�}{u}=4c_1^2+3c_2^2-2c_3^2\left(1\right) \end{gathered}$$

Since

$4c_1^2+3c_2^2-2c_3^2鈮�\left(4c_1^2+4c_2^2-4c_3^2=4\right)\left(2\right)$

and

$\left(-2c_1^2-2c_2^2-2c_3^2=-2\right)鈮�4c_1^2+3c_2^2-2c_3^2\left(3\right)$

From (1),(2) and (3) we can imply that the possible values of the dot product areas represented below:

$-2鈮�\stackrel{鈫�}{u}.A\stackrel{鈫�}{u}鈮�4$