Answer:
Step-by-step explanation:
From the given information:
[tex]T(\bar X) = \dfrac{1}{3}\left[\begin{array}{ccc}-1&-2&2\\-2&2&1\\2&1&2\end{array}\right][/tex]
For eigenvalues:
[tex]T(\bar X) = \left[\begin{array}{ccc}-1-\lambda &-2&2\\-2&2- \lambda &1\\2&1&2- \lambda \end{array}\right]=0[/tex]
[tex]\lambda^3 + 3 \lambda^2 + 9\lambda-27=0 \\ \\ -(\lambda+3)(\lambda-3)(\lambda-3) =0 \\ \\ \lambda = -3,3,3[/tex]
Let not forget that from the question; we are given [tex]\dfrac{1}{3}A[/tex] ; as such eigenvalues will be:
-1,1,1 since [tex]\dfrac{1}{3} (-3,3,3) = -1,1,1[/tex]
eigenvector for λ = -1
FInd [tex]|A - \lambda I |[/tex]
[tex]=|A+I| =\left[\begin{array}{ccc}\ \dfrac{2}{3}&-\dfrac{2}{3}&\dfrac{2}{3}\\ - \dfrac{2}{3}&\dfrac{5}{3}&\dfrac{1}{3}\\\dfrac{2}{3}&\dfrac{1}{3}&\dfrac{5}{3}\end{array}\right] \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{c}0\\0\\0\end{array}\right][/tex]
we have :
[tex]x_1 +2x_3 =0 \\ \\ x_2+x_3 = 0[/tex]
Thus;
[tex]x_1 = -2x_3 \\ \\ x_2= -x_3[/tex]
eigenvector for λ = -1 ⇒[tex]\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{c}-2x_3\\-x_3\\+x_3\end{array}\right][/tex][tex]=\left[\begin{array}{ccc}-2\\1\\1\end{array}\right] \ \ \ \ for \ x_3 = 1[/tex]
For λ = 1
[tex]|A - \lambda I |= |A-I|= \left[\begin{array}{ccc}\ - \dfrac{1}{3}&-\dfrac{2}{3}&\dfrac{2}{3}\\ - \dfrac{2}{3}& - \dfrac{1}{3}&\dfrac{1}{3}\\\dfrac{2}{3}&\dfrac{1}{3}&- \dfrac{1}{3}\end{array}\right] \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{c}0\\0\\0\end{array}\right][/tex]
[tex]x_1 + \dfrac{x_2}{2}- \dfrac{x_3}{3} =0[/tex]
[tex]x_1 = - \dfrac{x_2}{2}+\dfrac{1}{2}x_3[/tex]
Thus ;
[tex]\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{c} - \dfrac{1}{2} \\ 1 \\0\end{array}\right]x_2 + \left[\begin{array}{c} \dfrac{1}{2} \\ 0\\1\end{array}\right]x_3[/tex]
As such; [tex]\left[\begin{array}{c} - \dfrac{1}{2} \\ 1 \\0\end{array}\right] \ and \ \left[\begin{array}{c} \dfrac{1}{2} \\ 0\\1\end{array}\right] \ forms \ the \ basis \ of \ eigenpace[/tex]
Given that T is the reflection
Thus for [tex]T*T = I \ i.e \ T^2 = I[/tex]
we must have: [tex]\lambda^2 = 1 \to \lambda = \pm 1[/tex]
We can now conclude that the reflection pressures length and any other eigenvalue would mean reflection is stretching or shrinking a vector.
