Let $T$ $\in M_{4}(\mathbb R)$ and consider $S= \{M \in M_{4\times1}|T.M = 0\}$. In the case $T=\begin{pmatrix}1&1&-1&-3\\0&1&0&-2\\-1&0&1&1\\-2&1&2&0\end{pmatrix}$, how can I find the base and the dimension of $S$?
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2How would you solve the matrix equation $Tx = 0$? – Ben Grossmann Sep 15 '14 at 15:23
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Let $X = \begin{pmatrix}a\b\c\d\end{pmatrix}$. I'd check the conditions for $a, b, c, d$ so that the 4 equations derived out of that product equal zero. It's a linear system 4x4. Then? – user71487 Sep 15 '14 at 15:42
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Fundamental Theorem of Linear Algebra
Given $\mathbf{A}\in\mathbb{C}^{m\times n}$, the four fundamental subspaces are $$ \begin{align} % \mathbf{C}^{n} = \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\ % \mathbf{C}^{m} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} % \end{align} $$
Resolve $\mathbf{C}^{m}$: column space $$ \mathbf{A} = \left[ \begin{array}{rcrr} 1 & 1 & -1 & -3 \\ 0 & 1 & 0 & -2 \\ -1 & 0 & 1 & 1 \\ -2 & 1 & 2 & 0 \\ \end{array} \right] $$ Row reduction of $\mathbf{A}$. Form augmented matrix. $$ \xi_{0} = \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I}_{4} \\ \end{array} \right] = \left[ \begin{array}{rcrr|cccc} 1 & 1 & -1 & -3 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 & 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ -2 & 1 & 2 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] $$
Reduction to row echelon form:
column 1
$$ \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 2 & 0 & 0 & 1 \\ \end{array} \right] % \xi_{0} % = % \left[ \begin{array}{ccrc|cccc} \boxed{\color{gray}{1}} & 1 & -1 & -3 & 1 & 0 & 0 & 0 \\ \color{gray}{0} & 1 & 0 & -2 & 0 & 1 & 0 & 0 \\ \color{gray}{0} & 1 & 0 & -2 & 1 & 0 & 1 & 0 \\ \color{gray}{0} & 3 & 0 & -6 & 2 & 0 & 0 & 1 \\ \end{array} \right] = \xi_{1} % $$ The boxed entry identifies the pivot in a fundamental column.
column 2
$$ \left[ \begin{array}{crcc} 1 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & -3 & 0 & 1 \\ \end{array} \right] % \xi_{1} % = % \left[ \begin{array}{ccrr|crcc} \boxed{\color{gray}{1}} & \color{gray}{0} & -1 & -1 & 1 & -1 & 0 & 0 \\ \color{gray}{0} & \boxed{\color{gray}{1}} & 0 & -2 & 0 & 1 & 0 & 0 \\\hline \color{gray}{0} & \color{gray}{0} & 0 & 0 & \color{red}{1} & \color{red}{-1} & \color{red}{1} & \color{red}{0} \\ \color{gray}{0} & \color{gray}{0} & 0 & 0 & \color{red}{2} & \color{red}{-3} & \color{red}{0} & \color{red}{1} \\ \end{array} \right] = \xi_{2} % $$ A span of the $\color{blue}{\mathcal{R} \left( \mathbf{A} \right)}$ is given by the fundamental columns of $\mathbf{A}$, columns 1 and 2. The red vectors form a span for $\color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}$.
$$ \begin{align} % \mathbf{C}^{m} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} % % range &= \text{span } \left\{ \, \color{blue}{\left[ \begin{array}{r} 1 \\ 0 \\ -1 \\ -2 \\ \end{array} \right]}, \, \color{blue}{\left[ \begin{array}{r} 1 \\ 1 \\ 0 \\ 1 \\ \end{array} \right]}, \, \right\} % null \oplus \text{span } \left\{ \, \color{red}{ \left[ \begin{array}{r} 1 \\ -1 \\ 1 \\ 0 \\ \end{array} \right], \, \left[ \begin{array}{r} 2 \\ -3 \\ 0 \\ 1 \\ \end{array} \right]} \, \right\}% \end{align} $$
Resolve $\mathbf{C}^{n}$: row space $$ \mathbf{A}^{T} = \left[ \begin{array}{rrrr} 1 & 0 & -1 & -2 \\ 1 & 1 & 0 & 1 \\ -1 & 0 & 1 & 2 \\ -3 & -2 & 1 & 0 \\ \end{array} \right] $$ Row reduction of $\mathbf{A}^{T}$. Form augmented matrix. $$ \xi_{0} = \left[ \begin{array}{c|c} \mathbf{A}^{T} & \mathbf{I}_{4} \\ \end{array} \right] = \left[ \begin{array}{rrrr|cccc} 1 & 0 & -1 & -2 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 2 & 0 & 0 & 1 & 0 \\ -3 & -2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right] $$
Reduction to row echelon form:
column 1
$$ \left[ \begin{array}{rccc} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 3 & 0 & 0 & 1 \\ \end{array} \right] % \xi_{0} % = % \left[ \begin{array}{crrr|rccc} \boxed{\color{gray}{1}} & 0 & -1 & -2 & 1 & 0 & 0 & 0 \\ \color{gray}{0} & 1 & 1 & 3 & -1 & 1 & 0 & 0 \\ \color{gray}{0} & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ \color{gray}{0} & -2 & -2 & -6 & 3 & 0 & 0 & 1 \\ \end{array} \right] = \xi_{1} % $$ The boxed entry identifies the pivot in a fundamental column.
column 2
$$ \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ \end{array} \right] % \xi_{1} % = % \left[ \begin{array}{ccrr|rccc} \boxed{\color{gray}{1}} & \color{gray}{0} & -1 & -2 & 1 & 0 & 0 & 0 \\ \color{gray}{0} & \boxed{\color{gray}{1}} & 1 & 3 & -1 & 1 & 0 & 0 \\\hline \color{gray}{0} & \color{gray}{0} & 0 & 0 & \color{red}{1} & \color{red}{0} & \color{red}{1} & \color{red}{0} \\ \color{gray}{0} & \color{gray}{0} & 0 & 0 & \color{red}{1} & \color{red}{2} & \color{red}{0} & \color{red}{1} \\ \end{array} \right] = \xi_{2} % $$ The red vector form a span for $\color{red} {\mathcal{N} \left( \mathbf{A} \right)}$.
$$ \begin{align} % \mathbf{C}^{m} = \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A} \right)} % % range &= \text{span } \left\{ \, \color{blue}{\left[ \begin{array}{r} 1 \\ 1 \\ -1 \\ -3 \\ \end{array} \right]}, \, \color{blue}{\left[ \begin{array}{r} 0 \\ 1 \\ 0 \\ -2 \\ \end{array} \right]}, \, \right\} % null \oplus \text{span } \left\{ \, \color{red}{ \left[ \begin{array}{r} 1 \\ 0 \\ 1 \\ 0 \\ \end{array} \right], \, \left[ \begin{array}{r} 1 \\ 2 \\ 0 \\ 1 \\ \end{array} \right]} \, \right\}% \end{align} $$
Another example is Given a matrix and its reduced row echelon form, resolve the image and the kernel.
dantopa
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I think you should check how many vectors are Linearly independent and dim=number of that vectors.
Panda
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