← 한양대학교 2025년
행렬 A=(11−111−1)A= \left (\begin{matrix} 1 & 1 & -1 \\ 1 & 1 & -1 \end{matrix} \right)A=(1111−1−1) 의 특이값 분해(singular value decomposition)가
A=(u11u12u21−12)(6σ12σ13σ210σ23)(13v12v13v21−12v23v31026)TA= \left (\begin{matrix} u _ {11} & u _ {12} \\ u _ {21} & - \frac{1} {\sqrt {2}} \end{matrix} \right) \left (\begin{matrix} \sqrt {6} & \sigma _ {12} & \sigma _ {13} \\ \sigma _ {21} & 0 & \sigma _ {23} \end{matrix} \right) \left (\begin{matrix} \frac{1} {\sqrt {3}} & v _ {12} & v _ {13} \\ v _ {21} & - \frac{1} {\sqrt {2}} & v _ {23} \\ v _ {31} & 0 & \frac{2} {\sqrt {6}} \end{matrix} \right) ^ {T}A=(u11u21u12−21)(6σ21σ120σ13σ23)31v21v31v12−210v13v2362T
일 때, u112+(σ122+σ132+σ212+σ232)+(v122+v232+v312)u_ {11} ^ {2} + \left (\sigma _ {12} ^ {2} + \sigma _ {13} ^ {2} + \sigma _ {21} ^ {2} + \sigma _ {23} ^ {2} \right) + \left (v_ {12} ^ {2} +v_ {23} ^ {2} +v_ {31} ^ {2} \right)u112+(σ122+σ132+σ212+σ232)+(v122+v232+v312) 의 값은?