Answer :
An orthonormal basis for the -eigenspace of [tex]\pi[/tex]2 is
B1=(V3=[tex]\left[\begin{array}{ccc}1/\sqrt{3}\\1\sqrt{3}\\1/\sqrt{3} \end{array}\right][/tex]
What is Spectral Theorem?
There exists an invertible matrix C such that C 1 M C C-1 MC C1MC is a diagonal matrix, according to the spectral theorem, which asserts that if M equals the transpose of M, then M is diagonalizable.
Concept of theorem
If A is a symmetric matrix, then eigenvectors of different eigenvalues are orthogonal
here if v3 is an eigenvectors for eigenvalues [tex]\pi[/tex]2 of matrix A,then by spectral theorem we have
v1.v3=0
-2x+0y+2z=0 ------------(1)
and v2.v3=0
-2x+y+z=0 ------------(2)
To find v3 we have solve system
-2x+0y+2z=0
-2x+y+z=0
[tex]\left[\begin{array}{ccc}-2&0&-1\\-2&1&1\end{array}\right][/tex] [tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}0\\0\\\end{array}\right][/tex]
Applying R1 = -1/2 X R1
R2= R2+2 X R2 We get
[tex]\left[\begin{array}{ccc}1&0&-1\\0&1&-1\\\end{array}\right][/tex][tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}0\\0\\\end{array}\right][/tex]
x-z=0
y-z=0
put z=t
x=t
y=t
[tex]\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex]=[tex]\left[\begin{array}{ccc}t\\t\\t\end{array}\right][/tex]=t[tex]\left[\begin{array}{ccc}1\\1\\1\end{array}\right][/tex]
thus eigenvector for eigenvalue [tex]\pi[/tex]2 is v3=[tex]\left[\begin{array}{ccc}1\\1\\1\end{array}\right][/tex]
and basis for eigenspace of [tex]\pi \\[/tex]2 is B=[tex]\left[\begin{array}{ccc}1\\1\\1\end{array}\right][/tex]
Next we will convert v3 into a unit vector
for this v3=[tex]\sqrt{(1.1)+(1.1)+(1.1)}[/tex] = [tex]\sqrt{v3.v3}[/tex]
v3=[tex]\sqrt{3}[/tex]
v3= v3/v3 =1/[tex]\sqrt{3}[/tex] (1 1 1)
an orthonormal basis for the -eigenspace is B1 = V3=[tex]\left[\begin{array}{ccc}1/\sqrt{3} \\1/\sqrt{3} \\1/\sqrt{3} \end{array}\right][/tex]
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