Let [katex] A = \left( \begin{matrix} 1 & -1 & 2 \\ -1 & 0 & 1 \\ 2 & 1 & 1 \end{matrix} \right) [/katex] and let [katex] \left( \begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right) [/katex] be an eigenvector corresponding to the smallest eigenvalue of A, satisfying [katex] x_1^2+x_2^2+x_3^2 =1 [/katex]. Then the value of [katex] |x_1| +|x_2| +|x_3| [/katex] equals ____________________. (round off to 2 decimal points)
If [katex] \lambda [/katex] is an Eigen value of A then,
\left|A- \lambda I_3 \right| =0 \\ \implies (1-\lambda)(-\lambda(1-\lambda)-1) +1 (-1(1-\lambda) -2) \\ + 2(-1+2\lambda) =0 \\ \implies-(1-\lambda)(\lambda^2-\lambda-1) +\lambda-3+4\lambda-2=0 \\ \implies (1-\lambda)(\lambda-3)(\lambda+2)=0 \\ \implies \lambda=1,3,-2
Thus the smallest Eigen value is -2. Now we have,
A \underline{x} = -2 \underline x \\
\implies x_1-x_2+2x_3 +2x_1=0 \\
\hskip{1cm} -x_1+x_3 +2x_2 =0 \\
\hskip{1cm} 2x_1+x_2+x_3 + 2x_3 =0 \\
\implies 3x_1-x_2+2x_3 =0 \\
\hskip{1cm} -x_1+ 2x_2 +x_3 =0 \\
\hskip{1cm} 2x_1+x_2+3x_3 =0 \\\implies 3x_1-x_2+2x_3 =0 \\
\hskip{1cm} -x_1+ 2x_2 +x_3 =0 \\
\hskip{1cm} 2x_1+x_2+3x_3 =0 \\
(3x_1-x_2+2x_3)+3(-x_1+ 2x_2 +x_3) = 0 \\
\implies 0+5x_2+5x_3 = 0 \\
\implies x_2=-x_3 \\
(-x_1+ 2x_2 +x_3)-2(2x_1+x_2+3x_3)=0 \\
\implies -5x_1-5x_3=0 \\
\implies x_1=-x_3Thus,
x_1^2+x_2^2+x_3^2=1 \\
\implies x_1^2=x_2^2=x_3^2=\frac{1}{3} \\
|x_1|+|x_2|+|x_3|= 3|\frac{1}{\sqrt{3}}|=\sqrt{3}=1.732
Discovering Statistics 
Leave a comment