3.4: Espacio Nulo Izquierdo
- Page ID
- 113083
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Si uno entiende el concepto de un espacio nulo, el espacio nulo izquierdo es extremadamente fácil de entender.
El Espacio Nulo Izquierdo de una matriz es el espacio nulo de su transpuesta, es decir,
\[\mathcal{N}(A^T) = \{ \textbf{y} \in \mathbb{R}^{m} | A^{T} \textbf{y} = 0\} \nonumber\]
La palabra “izquierda” en este contexto deriva del hecho de que\(A^{T} \textbf{y} = 0\) equivale a\(\textbf{y}^{T} A = 0\) donde\(\textbf{y}\) “actúa” sobre A desde la izquierda.
Ejemplo
Como\(A_{red}\) fue la clave para identificar el espacio nulo de A, veremos que\(A^{T}_{red}\) es la clave del espacio nulo de\(A^T\). Si
\[A = \begin{pmatrix} {1}&{1}\\ {1}&{2}\\ {1}&{3} \end{pmatrix} \nonumber\]
entonces
\[A^{T} = \begin{pmatrix} {1}&{1}&{1}\\ {1}&{2}&{3} \end{pmatrix} \nonumber\]
y así
\[A^{T}_{red} = \begin{pmatrix} {1}&{1}&{1}\\ {0}&{1}&{2} \end{pmatrix} \nonumber\]
Resolvemos\(A^{T}_{red} = 0\) reconociendo que\(y_{1}\) y\(y_{2}\) son variables pivotes mientras que\(y_{3}\) es libre. Resolviendo\(A^{T}_{red} \textbf{y} = 0\) para el pivote en cuanto a lo libre que encontramos\(y_{2} = -(2y_{3})\) y\(y_{1} = y_{3}\) por lo tanto
\[\mathcal{N}(A^{T}) = \begin{equation} \left \{ y_{3} \begin{pmatrix} {1}\\ {-2}\\ {1} \end{pmatrix} | y_{3} \in \mathbb{R} \right \} \end{equation} \nonumber\]
Encontrar una base para el espacio nulo izquierdo
El procedimiento no es diferente al utilizado para calcular el espacio nulo de la propia A. De hecho
Supongamos que\(A^{T}\) es n-por-m con índices de pivote\(\{c_{j} | j = \{1, \cdots, r\}\}\) e índices libres\(\{c_{j} | j = \{r+1, \cdots, n\}\}\). Una base para se\(\mathcal{N}(A^T)\) puede construir de\(m-r\) vectores\(\{z^{1}, z^{2}, \cdots, z^{m-r}\}\) donde\(z^{k}\) y solo\(z^k\), posee un distinto de cero en su\(c_{r+k}\) componente.