Solución Halla la transposición de la matriz: [ 0 1 1 2 2 0 0 1 1 ] T = [ 0 2 0 1 2 1 1 0 1 ] \left[\begin{array}{ccc}0 & 1 & 1\\\sqrt{2} & 2 & 0\\0 & 1 & 1\end{array}\right]^{T} = \left[\begin{array}{ccc}0 & \sqrt{2} & 0\\1 & 2 & 1\\1 & 0 & 1\end{array}\right] ⎣ ⎡ 0 2 0 1 2 1 1 0 1 ⎦ ⎤ T = ⎣ ⎡ 0 1 1 2 2 0 0 1 1 ⎦ ⎤ (para ver los pasos, consulta calculadora de transposición de matrices ).
Multiplicar la matriz por su transpuesto: W = [ 0 1 1 2 2 0 0 1 1 ] ⋅ [ 0 2 0 1 2 1 1 0 1 ] = [ 2 2 2 2 6 2 2 2 2 ] W = \left[\begin{array}{ccc}0 & 1 & 1\\\sqrt{2} & 2 & 0\\0 & 1 & 1\end{array}\right]\cdot \left[\begin{array}{ccc}0 & \sqrt{2} & 0\\1 & 2 & 1\\1 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}2 & 2 & 2\\2 & 6 & 2\\2 & 2 & 2\end{array}\right] W = ⎣ ⎡ 0 2 0 1 2 1 1 0 1 ⎦ ⎤ ⋅ ⎣ ⎡ 0 1 1 2 2 0 0 1 1 ⎦ ⎤ = ⎣ ⎡ 2 2 2 2 6 2 2 2 2 ⎦ ⎤ (para los pasos, véase calculadora de multiplicación de matrices ).
Ahora, encuentra los valores y vectores propios de W W W (para ver los pasos, consulta calculadora de valores y vectores propios ).
Valor propio: 8 8 8 , vector propio: [ 1 2 1 ] \left[\begin{array}{c}1\\2\\1\end{array}\right] ⎣ ⎡ 1 2 1 ⎦ ⎤ .
Valor propio: 2 2 2 , vector propio: [ 1 − 1 1 ] \left[\begin{array}{c}1\\-1\\1\end{array}\right] ⎣ ⎡ 1 − 1 1 ⎦ ⎤ .
Valor propio: 0 0 0 , vector propio: [ − 1 0 1 ] \left[\begin{array}{c}-1\\0\\1\end{array}\right] ⎣ ⎡ − 1 0 1 ⎦ ⎤ .
Hallar las raíces cuadradas de los valores propios no nulos (σ i \sigma_{i} σ i ):
σ 1 = 2 2 \sigma_{1} = 2 \sqrt{2} σ 1 = 2 2
σ 2 = 2 \sigma_{2} = \sqrt{2} σ 2 = 2
La matriz Σ \Sigma Σ es una matriz cero con σ i \sigma_{i} σ i en su diagonal: Σ = [ 2 2 0 0 0 2 0 0 0 0 ] \Sigma = \left[\begin{array}{ccc}2 \sqrt{2} & 0 & 0\\0 & \sqrt{2} & 0\\0 & 0 & 0\end{array}\right] Σ = ⎣ ⎡ 2 2 0 0 0 2 0 0 0 0 ⎦ ⎤ .
Las columnas de la matriz U U U son los vectores normalizados (unitarios): U = [ 6 6 3 3 − 2 2 6 3 − 3 3 0 6 6 3 3 2 2 ] U = \left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{3} & - \frac{\sqrt{2}}{2}\\\frac{\sqrt{6}}{3} & - \frac{\sqrt{3}}{3} & 0\\\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2}\end{array}\right] U = ⎣ ⎡ 6 6 3 6 6 6 3 3 − 3 3 3 3 − 2 2 0 2 2 ⎦ ⎤ (para los pasos para encontrar un vector unitario, véase calculadora de vectores unitarios ).
Ahora, v i = 1 σ i ⋅ [ 0 1 1 2 2 0 0 1 1 ] T ⋅ u i v_{i} = \frac{1}{\sigma_{i}}\cdot \left[\begin{array}{ccc}0 & 1 & 1\\\sqrt{2} & 2 & 0\\0 & 1 & 1\end{array}\right]^{T}\cdot u_{i} v i = σ i 1 ⋅ ⎣ ⎡ 0 2 0 1 2 1 1 0 1 ⎦ ⎤ T ⋅ u i :
v 1 = 1 σ 1 ⋅ [ 0 1 1 2 2 0 0 1 1 ] T ⋅ u 1 = 1 2 2 ⋅ [ 0 2 0 1 2 1 1 0 1 ] ⋅ [ 6 6 6 3 6 6 ] = [ 6 6 3 2 3 6 ] v_{1} = \frac{1}{\sigma_{1}}\cdot \left[\begin{array}{ccc}0 & 1 & 1\\\sqrt{2} & 2 & 0\\0 & 1 & 1\end{array}\right]^{T}\cdot u_{1} = \frac{1}{2 \sqrt{2}}\cdot \left[\begin{array}{ccc}0 & \sqrt{2} & 0\\1 & 2 & 1\\1 & 0 & 1\end{array}\right]\cdot \left[\begin{array}{c}\frac{\sqrt{6}}{6}\\\frac{\sqrt{6}}{3}\\\frac{\sqrt{6}}{6}\end{array}\right] = \left[\begin{array}{c}\frac{\sqrt{6}}{6}\\\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{6}\end{array}\right] v 1 = σ 1 1 ⋅ ⎣ ⎡ 0 2 0 1 2 1 1 0 1 ⎦ ⎤ T ⋅ u 1 = 2 2 1 ⋅ ⎣ ⎡ 0 1 1 2 2 0 0 1 1 ⎦ ⎤ ⋅ ⎣ ⎡ 6 6 3 6 6 6 ⎦ ⎤ = ⎣ ⎡ 6 6 2 3 6 3 ⎦ ⎤ (para los pasos, véase calculadora de multiplicación escalar de matrices y calculadora de multiplicación de matrices ).
v 2 = 1 σ 2 ⋅ [ 0 1 1 2 2 0 0 1 1 ] T ⋅ u 2 = 1 2 ⋅ [ 0 2 0 1 2 1 1 0 1 ] ⋅ [ 3 3 − 3 3 3 3 ] = [ − 3 3 0 6 3 ] v_{2} = \frac{1}{\sigma_{2}}\cdot \left[\begin{array}{ccc}0 & 1 & 1\\\sqrt{2} & 2 & 0\\0 & 1 & 1\end{array}\right]^{T}\cdot u_{2} = \frac{1}{\sqrt{2}}\cdot \left[\begin{array}{ccc}0 & \sqrt{2} & 0\\1 & 2 & 1\\1 & 0 & 1\end{array}\right]\cdot \left[\begin{array}{c}\frac{\sqrt{3}}{3}\\- \frac{\sqrt{3}}{3}\\\frac{\sqrt{3}}{3}\end{array}\right] = \left[\begin{array}{c}- \frac{\sqrt{3}}{3}\\0\\\frac{\sqrt{6}}{3}\end{array}\right] v 2 = σ 2 1 ⋅ ⎣ ⎡ 0 2 0 1 2 1 1 0 1 ⎦ ⎤ T ⋅ u 2 = 2 1 ⋅ ⎣ ⎡ 0 1 1 2 2 0 0 1 1 ⎦ ⎤ ⋅ ⎣ ⎡ 3 3 − 3 3 3 3 ⎦ ⎤ = ⎣ ⎡ − 3 3 0 3 6 ⎦ ⎤ (para los pasos, véase calculadora de multiplicación escalar de matrices y calculadora de multiplicación de matrices ).
Como nos hemos quedado sin σ i \sigma_{i} σ i distinto de cero y necesitamos un vector más, hallemos el vector ortogonal a todos los vectores encontrados hallando el espacio nulo de la matriz cuyas filas son los vectores encontrados: [ 2 − 1 1 ] \left[\begin{array}{c}\sqrt{2}\\-1\\1\end{array}\right] ⎣ ⎡ 2 − 1 1 ⎦ ⎤ (para los pasos, véase calculadora del espacio nulo ).
Normaliza el vector: se convierte en [ 2 2 − 1 2 1 2 ] \left[\begin{array}{c}\frac{\sqrt{2}}{2}\\- \frac{1}{2}\\\frac{1}{2}\end{array}\right] ⎣ ⎡ 2 2 − 2 1 2 1 ⎦ ⎤ , (para ver los pasos, consulta calculadora de vectores unitarios ).
Por lo tanto, V = [ 6 6 − 3 3 2 2 3 2 0 − 1 2 3 6 6 3 1 2 ] . V = \left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & - \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2}\\\frac{\sqrt{3}}{2} & 0 & - \frac{1}{2}\\\frac{\sqrt{3}}{6} & \frac{\sqrt{6}}{3} & \frac{1}{2}\end{array}\right]. V = ⎣ ⎡ 6 6 2 3 6 3 − 3 3 0 3 6 2 2 − 2 1 2 1 ⎦ ⎤ .
Las matrices U U U , Σ \Sigma Σ , y V V V son tales que la matriz inicial [ 0 1 1 2 2 0 0 1 1 ] = U Σ V T \left[\begin{array}{ccc}0 & 1 & 1\\\sqrt{2} & 2 & 0\\0 & 1 & 1\end{array}\right] = U \Sigma V^T ⎣ ⎡ 0 2 0 1 2 1 1 0 1 ⎦ ⎤ = U Σ V T .
Respuesta U = [ 6 6 3 3 − 2 2 6 3 − 3 3 0 6 6 3 3 2 2 ] ≈ [ 0.408248290463863 0.577350269189626 − 0.707106781186548 0.816496580927726 − 0.577350269189626 0 0.408248290463863 0.577350269189626 0.707106781186548 ] U = \left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{3} & - \frac{\sqrt{2}}{2}\\\frac{\sqrt{6}}{3} & - \frac{\sqrt{3}}{3} & 0\\\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2}\end{array}\right]\approx \left[\begin{array}{ccc}0.408248290463863 & 0.577350269189626 & -0.707106781186548\\0.816496580927726 & -0.577350269189626 & 0\\0.408248290463863 & 0.577350269189626 & 0.707106781186548\end{array}\right] U = ⎣ ⎡ 6 6 3 6 6 6 3 3 − 3 3 3 3 − 2 2 0 2 2 ⎦ ⎤ ≈ ⎣ ⎡ 0.408248290463863 0.816496580927726 0.408248290463863 0.577350269189626 − 0.577350269189626 0.577350269189626 − 0.707106781186548 0 0.707106781186548 ⎦ ⎤ A
Σ = [ 2 2 0 0 0 2 0 0 0 0 ] ≈ [ 2.82842712474619 0 0 0 1.414213562373095 0 0 0 0 ] \Sigma = \left[\begin{array}{ccc}2 \sqrt{2} & 0 & 0\\0 & \sqrt{2} & 0\\0 & 0 & 0\end{array}\right]\approx \left[\begin{array}{ccc}2.82842712474619 & 0 & 0\\0 & 1.414213562373095 & 0\\0 & 0 & 0\end{array}\right] Σ = ⎣ ⎡ 2 2 0 0 0 2 0 0 0 0 ⎦ ⎤ ≈ ⎣ ⎡ 2.82842712474619 0 0 0 1.414213562373095 0 0 0 0 ⎦ ⎤ A
V = [ 6 6 − 3 3 2 2 3 2 0 − 1 2 3 6 6 3 1 2 ] ≈ [ 0.408248290463863 − 0.577350269189626 0.707106781186548 0.866025403784439 0 − 0.5 0.288675134594813 0.816496580927726 0.5 ] V = \left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & - \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2}\\\frac{\sqrt{3}}{2} & 0 & - \frac{1}{2}\\\frac{\sqrt{3}}{6} & \frac{\sqrt{6}}{3} & \frac{1}{2}\end{array}\right]\approx \left[\begin{array}{ccc}0.408248290463863 & -0.577350269189626 & 0.707106781186548\\0.866025403784439 & 0 & -0.5\\0.288675134594813 & 0.816496580927726 & 0.5\end{array}\right] V = ⎣ ⎡ 6 6 2 3 6 3 − 3 3 0 3 6 2 2 − 2 1 2 1 ⎦ ⎤ ≈ ⎣ ⎡ 0.408248290463863 0.866025403784439 0.288675134594813 − 0.577350269189626 0 0.816496580927726 0.707106781186548 − 0.5 0.5 ⎦ ⎤ A