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If the rows (or columns, equivalently) of a matrix ''A'' are [[Linear independence|linearly independent]], then the image of that transformation is the entire space it is applied to.
If the rows (or columns, equivalently) of a matrix ''A'' are [[Linear independence|linearly independent]], then the image of that transformation is the entire space it is applied to.
==Examples==
===Example 1===
Consider all the points (vectors) in the plane, ie, ''(x,y)'' acting under the transformation
<math>\mathbf{A} = \begin{pmatrix}
-1 & 1 \\
1 & 0 \end{pmatrix}</math>.
We can be sure that the image of this transformation is the entire plane, because for any point
<math>\begin{pmatrix}
x \\
y \end{pmatrix}</math>
in the plane, there is another vector in the plane
<math>\begin{pmatrix}
y \\
x+y \end{pmatrix}</math>
such that
<math>\begin{pmatrix} -1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix}y \\x+y \end{pmatrix} = \begin{pmatrix}x \\y \end{pmatrix}</math>.
===Example 2===
We continue to work in the plane, but now we examine the matrix
<math>\mathbf{A} = \begin{pmatrix}
3 & 1 \\
-6 & -2 \end{pmatrix}</math>.
Now if we examine how this matrix acts on an arbitary point ''(a,b)'', we find that point is carried to
<math>\mathbf{A} = \begin{pmatrix}3 & 1 \\-6 & -2 \end{pmatrix}\begin{pmatrix}a \\b \end{pmatrix} = \begin{pmatrix}3a+b \\-6a-2b \end{pmatrix} = \begin{pmatrix}1 \\-3 \end{pmatrix}(a+b)</math>,
in other words, all points in the plane are carried to the line <math>y=-3x \ </math>.
We write
<math>im(\mathbf{A}) = \begin{pmatrix}1 \\-3 \end{pmatrix}t, \forall t\in\mathbb{R} \ </math>.
[[Category:linear algebra]]
[[Category:linear algebra]]
[[Category:mathematics]]
[[Category:mathematics]]