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This shows that \mathbf{w} is in the spanrange of the vectors \{L(\mathbf{v}_{k+1}), \ldots, L(\mathbf{v}_n)\}. The rank is . We can express this in homogeneous coordinates as:
After carrying out the matrix multiplication, the homogeneous component
w
c
{\displaystyle w_{c}}
will be equal to the value of
z
{\displaystyle z}
and the other three will not change.
However, we could have defined T
in this way:Given this definition, it is not at all obvious that T
is a matrix transformation, or what matrix it is associated to.
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Thus the nullspace is \{[x,0,0] \mid x \in \mathbb{R}\}, which is just a line with basis vector [1,0,0].
\]
Therefore, $S\circ T=T$. Although a translation is a non-linear transformation in a 2-D or 3-D Euclidean space described by Cartesian coordinates (i. ContinueThis 3Blue1Brown video provides some helpful animated illustrations of linear transformations:ContinueThe rank of a linear transformation from one vector space to another is the dimension of its range. However, this is not true when using perspective projections.
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To see that \{L(\mathbf{v}_{k+1}), \ldots, L(\mathbf{v}_n)\} a knockout post spanned by the range of L, note that if \mathbf{w} = L(\mathbf{v}) for some \mathbf{v}, then writing \mathbf{v} as a linear combination of \mathbf{v}_1, \ldots, \mathbf{v}_n, we have\begin{align*}\mathbf{w} = L(c_1 \mathbf{v}_1 + \cdots + c_n \mathbf{v}_n) = L(c_{k+1}\mathbf{v}_{k+1} + \cdots + c_n \mathbf{v}_n),\end{align*}by linearity of L. There is an ambiguity in this notation: one has to know from context that e
1
is meant to have n
entries.
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