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The matrices $$A = \begin{bmatrix}1&-2&9&5&4\\1&-1&6&5&-3\\-2&0&-6&1&-2\\4&1&9&1&-9\end{bmatrix}$$ $$B=\begin{bmatrix}1&-2&9&5&4\\0&1&-3&0&-7\\0&0&0&1&-2\\0&0&0&0&0\end{bmatrix}$$ are row equivalent. Find bases for Col $A$ and Nul $A$ and give the dimensions of these subspaces.

How exactly can I do this?

dantopa
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Yusha
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2 Answers2

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Let B be the 2nd matrix. To find a basis for col(A), take the columns of A corresponding to the leading 1's in B. To find a basis for nul(A), solve $Bx=0$ and find a basis for this solution space.

user84413
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Immediately: $$\mathbf{A}\in\mathbb{R}^{4\times 6}$$ The problem is to resolve a space of dimension $4$.

The row reduction identifies the fundamental columns: $$ \mathbf{A} = \left[ \begin{array}{crrcr} \boxed{1} & -2 & 9 & 5 & 4 \\ 0 & \boxed{1} & -3 & 0 & -7 \\ 0 & 0 & 0 & \boxed{1} & -2 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right] $$ The three fundamental columns identify the three vectors which span the $\color{blue}{range}$ space: $$ \left[ \begin{array}{rrrcr} \color{blue}{1} & \color{blue}{-2} & 9 & \color{blue}{5} & 4 \\ \color{blue}{1} & \color{blue}{-1} & 6 & \color{blue}{5} & -3 \\ \color{blue}{-2} & \color{blue}{0} & -6 & \color{blue}{1} & -2 \\ \color{blue}{4} & \color{blue}{1} & 9 & \color{blue}{1} & -9 \end{array} \right] $$ Formally, $$ \color{blue}{\mathcal{R}\left( \mathbf{A} \right)} = \text{span } \left\{ \, \color{blue}{\left[ \begin{array}{r} 1 \\ 1 \\ -2 \\ 4 \\ \end{array} \right]}, \, \color{blue}{\left[ \begin{array}{r} -2 \\ -1 \\ 0 \\ 1 \\ \end{array} \right]}, \, \color{blue}{\left[ \begin{array}{r} 5 \\ 5 \\ 1 \\ 1 \\ \end{array} \right]} \, \right\} $$ The matrices have rank $\rho = 3$. Therefore, the range space $\color{blue}{\mathcal{R}\left( \mathbf{A} \right)}$ has dimension $3$.
The Fundamental Theorem of Linear Algebra dictates that a matrix $\mathbb{A}\in\mathbb{C}^{m \times n}$ induces the four fundamental subspaces: $$ \begin{align} % \mathbb{C}^{n} = \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\ % \mathbb{C}^{m} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} % \end{align} $$

One vector is needed to complete the domain, the vector which will provide the span for $\color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}$.

The $\color{red}{null}$ space vectors are found through Gauss-Jordan elimination of the augmented matrix.

Clear column 1. $$ % \left[ \begin{array}{rrcc} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & -4 & 0 & 1 \\ \end{array} \right] % \left[ \begin{array}{c|c} \mathbf{A} & \mathbf{I}_{4} \end{array} \right] % = \left[ \begin{array}{crrrr|rrcc} \color{gray}{1} & -2 & 9 & 5 & 4 & 1 & 0 & 0 & 0 \\ \color{gray}{0} & 1 & -3 & 0 & -7 & -1 & 1 & 0 & 0 \\ \color{gray}{0} & -2 & 6 & 11 & -8 & 0 & 2 & 1 & 0 \\ \color{gray}{0} & 5 & -15 & -19 & 3 & 0 & -4 & 0 & 1 \\ \end{array} \right] % = \xi_{1} $$

Clear column 2. $$ \left[ \begin{array}{crcc} 1 & 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & -5 & 0 & 1 \\ \end{array} \right] % \xi_{1} = \left[ \begin{array}{ccrrr|rrcc} \color{gray}{1} & \color{gray}{0} & 3 & 5 & -10 & -1 & 2 & 0 & 0 \\ \color{gray}{0} & \color{gray}{1} & -3 & 0 & -7 & -1 & 1 & 0 & 0 \\ \color{gray}{0} & \color{gray}{0} & 0 & 11 & -22 & -2 & 4 & 1 & 0 \\ \color{gray}{0} & \color{gray}{0} & 0 & -19 & 38 & 5 & -9 & 0 & 1 \\ \end{array} \right] % = \xi_{2} % $$

Normalize row three, then clear column 3. $$ \left[ \begin{array}{ccrc} 1 & 0 & -5 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 19 & 1 \\ \end{array} \right] % \xi_{2} = \left[ \begin{array}{ccrcr|rrrc} \color{gray}{1} & \color{gray}{0} & 3 & \color{gray}{0} & 0 & -\frac{1}{11} & \frac{2}{11} & -\frac{5}{11} & 0 \\ \color{gray}{0} & \color{gray}{1} & -3 & \color{gray}{0} & -7 & -1\ & 1\ & 0\ & 0 \\ \color{gray}{0} & \color{gray}{0} & 0 & \color{gray}{1} & -2 & -\frac{2}{11} & \frac{4}{11} & \frac{1}{11} & 0 \\\hline \color{gray}{0} & \color{gray}{0} & 0 & \color{gray}{0} & 0 & \color{red}{\frac{17}{11}} & \color{red}{-\frac{23}{11}} & \color{red}{\frac{19}{11}} & \color{red}{1} \\ \end{array} \right] % $$ The red vector is $$ \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} = \text{span } \left\{ \, \color{red}{ \left[ \begin{array}{r} 17 \\ -23 \\ 19 \\ 11 \\ \end{array} \right]} \, \right\} $$

The decomposition of the image of $\mathbf{A}$ $$ \begin{align} \mathbb{C}^{m} &= \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} \\ % % range &= \text{span } \left\{ \, \color{blue}{\left[ \begin{array}{r} 1 \\ 1 \\ -2 \\ 4 \\ \end{array} \right]}, \, \color{blue}{\left[ \begin{array}{r} -2 \\ -1 \\ 0 \\ 1 \\ \end{array} \right]}, \, \color{blue}{\left[ \begin{array}{r} 5 \\ 5 \\ 1 \\ 1 \\ \end{array} \right]} \, \right\} % null \oplus \text{span } \left\{ \, \color{red}{ \left[ \begin{array}{r} 17 \\ -23 \\ 19 \\ 11 \\ \end{array} \right]} \, \right\} \end{align} $$ The dimension of the $\color{red}{null}$ space is $1$.
The rank-nullity theorem as applied to this matrix $\mathbf{A}\in\mathbf{R}^{4\times 6}_{3}$ $$ \begin{align} \text{rank } \mathbf{A} + \text{nullity } \mathbf{A} &= n \\ 3 + 1 & = 4 \end{align} $$
Another example. Here the matrix is symmetric $m=n$ and it is more challenging to keep track of vector spaces: Find base and dimension of given subspace
dantopa
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