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math2009 · #4 November 3rd, 2009, 06:46 PM

generally, steps of calculating kernel & image are getting rref(matrix) at first

Image:

$rref\left[ \begin{array}{cccc} -1 & 0 & 0 & 1 \\1 & -1 & 0 & 0 \\-1 & 0 & 1 & 0 \\0 & 1 & 0 & -1 \\0 & 0 & -1 & 1 \\0 & -1 & 1 & 0 \end{array} \right]=\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 \end{array} \right]$
find columns including leading 1's in rref(matrix), these columns are linearly independent.

$im(A)=span(column1, \cdots , column4)$

Kernel:

$rref\left[ \begin{array}{cccccc} -1 & -1 & 0 & -1 & 0 & 0 \\1 & 0 & -1 & 0 & -1 & 0 \\0 & 1 & 1 & 0 & 0 & -1 \\0 & 0 & 0 & 1 & 1 & 1 \end{array} \right]=\left[ \begin{array}{cccccc} 1 & 0 & -1 & 0 & -1 & 0 \\0 & 1 & 1 & 0 & 0 & -1 \\0 & 0 & 0 & 1 & 1 & 1 \\0 & 0 & 0 & 0 & 0 & 0 \\\uparrow &\uparrow & &\uparrow & & \end{array} \right]$

Obviously, there are only 3 columns which are linearly independent.
The remainder 3 columns will spanned by linearly independent columns.
$\vec{v}_3=-\vec{v}_1+\vec{v}_2 \longrightarrow \vec{v}_1-\vec{v}_2+\vec{v}_3=0 \longrightarrow A\left[ \begin{array}{cccccc} 1\\-1\\1\\0\\0\\0 \end{array} \right]=0$

find another 2 vectors by the same method,

$ker(A)=span(\begin{bmatrix} 1\\-1\\1\\0\\0\\0 \end{bmatrix},\begin{bmatrix} 1\\0\\0\\-1\\1\\0 \end{bmatrix},\begin{bmatrix} 0\\1\\0\\-1\\0\\1 \end{bmatrix})$

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