Solução Encontre a transposição da 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 as etapas, consulte Calculadora de transposição de matriz ).
Multiplique a matriz por sua transposição: 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 ver as etapas, consulte calculadora de multiplicação de matrizes ).
Agora, encontre os valores e vetores próprios de W W W (para ver as etapas, consulte calculadora de valores e vetores próprios ).
Eigenvalue (valor próprio): 8 8 8 , eigenvector: [ 1 2 1 ] \left[\begin{array}{c}1\\2\\1\end{array}\right] ⎣ ⎡ 1 2 1 ⎦ ⎤ .
Eigenvalue (valor próprio): 2 2 2 , eigenvector: [ 1 − 1 1 ] \left[\begin{array}{c}1\\-1\\1\end{array}\right] ⎣ ⎡ 1 − 1 1 ⎦ ⎤ .
Eigenvalue (valor próprio): 0 0 0 , eigenvector: [ − 1 0 1 ] \left[\begin{array}{c}-1\\0\\1\end{array}\right] ⎣ ⎡ − 1 0 1 ⎦ ⎤ .
Encontre as raízes quadradas dos valores próprios não 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
A matriz Σ \Sigma Σ é uma matriz zero com σ i \sigma_{i} σ i em sua 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 ⎦ ⎤ .
As colunas da matriz U U U são os vetores normalizados (unitários): 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 saber as etapas para encontrar um vetor unitário, consulte calculadora de vetor unitário ).
Agora, 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 ver as etapas, consulte calculadora de multiplicação escalar de matrizes e calculadora de multiplicação de matrizes ).
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 ver as etapas, consulte calculadora de multiplicação escalar de matrizes e calculadora de multiplicação de matrizes ).
Como ficamos sem nenhum σ i \sigma_{i} σ i diferente de zero e precisamos de mais um vetor, encontre o vetor ortogonal a todos os vetores encontrados encontrando o espaço nulo da matriz cujas linhas são os vetores encontrados: [ 2 − 1 1 ] \left[\begin{array}{c}\sqrt{2}\\-1\\1\end{array}\right] ⎣ ⎡ 2 − 1 1 ⎦ ⎤ (para ver as etapas, consulte Calculadora de espaço nulo ).
Normalize o vetor: ele se torna [ 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 as etapas, consulte calculadora de vetor unitário ).
Portanto, 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 ⎦ ⎤ .
As matrizes U U U , Σ \Sigma Σ e V V V são tais que a 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 .
Resposta 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