- #1
dane502
- 21
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First of all I would like to wish a happy new year to all of you, who have helped us understand college math and physics. I really appreciate it.
Determine the dimension of the image of a linear transformations [tex]f^{\circ n}[/tex], where [tex]n\in\mathbb{N}[/tex] and [tex]f:\mathbb{R}^4\to\mathbb{R}^4[/tex] where
[tex]f(\underline{x})=(\underline{x}\cdot\underline{a_1}) \underline{a_2} +
(\underline{x}\cdot\underline{a_2}) \underline{a_3} +
(\underline{x}\cdot\underline{a_3}) \underline{a_4}[/tex]
and
[tex]
\underline{a_1} =
\begin{pmatrix}
1 \\
1 \\
0 \\
0
\end{pmatrix}
,
\underline{a_2} =
\begin{pmatrix}
0 \\
0 \\
2 \\
0
\end{pmatrix}
,
\underline{a_3} =
\begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix}
,
\underline{a_4} =
\begin{pmatrix}
1 \\
-1 \\
0 \\
0
\end{pmatrix}
[/tex]
The matrix representation of [tex]f[/tex] with regards to the natural basis is
[tex]
\underline{\underline{C}}=
\begin{pmatrix}
0&0&0&1\\
0&0&0&-1\\
2&2&0&0\\
0&0&2&0
\end{pmatrix}
[/tex]
but with regards to the basis [tex]\mathcal{A} = (\underline{a_1},\ldots,\underline{a_4})[/tex] the matrix representation of [tex]f:\mathcal{A}\to\mathcal{A}[/tex] is
[tex]
\underline{\underline{A}} =
\begin{pmatrix}
0 & 0 & 1 & 0 \\
2 & 0 & 0 & 0 \\
0 & 4 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{pmatrix}
[/tex]
But the rank of a matrix equals the dimension of the image of the corresponding transformation. However, [tex]\text{Rk}(\underline{\underline{A}}^n) = \text{Rk}(\underline{\underline{C}}^n)[/tex] only for [tex]n=1[/tex], which puzzles me for two reasons. Firstly, because I cannot solve problem. Secondly, because it implies that the dimension of the image of a linear transformation depends on the basis, which contradicts my "visualization" of changes of basis.
Homework Statement
Determine the dimension of the image of a linear transformations [tex]f^{\circ n}[/tex], where [tex]n\in\mathbb{N}[/tex] and [tex]f:\mathbb{R}^4\to\mathbb{R}^4[/tex] where
[tex]f(\underline{x})=(\underline{x}\cdot\underline{a_1}) \underline{a_2} +
(\underline{x}\cdot\underline{a_2}) \underline{a_3} +
(\underline{x}\cdot\underline{a_3}) \underline{a_4}[/tex]
and
[tex]
\underline{a_1} =
\begin{pmatrix}
1 \\
1 \\
0 \\
0
\end{pmatrix}
,
\underline{a_2} =
\begin{pmatrix}
0 \\
0 \\
2 \\
0
\end{pmatrix}
,
\underline{a_3} =
\begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix}
,
\underline{a_4} =
\begin{pmatrix}
1 \\
-1 \\
0 \\
0
\end{pmatrix}
[/tex]
Homework Equations
The matrix representation of [tex]f[/tex] with regards to the natural basis is
[tex]
\underline{\underline{C}}=
\begin{pmatrix}
0&0&0&1\\
0&0&0&-1\\
2&2&0&0\\
0&0&2&0
\end{pmatrix}
[/tex]
but with regards to the basis [tex]\mathcal{A} = (\underline{a_1},\ldots,\underline{a_4})[/tex] the matrix representation of [tex]f:\mathcal{A}\to\mathcal{A}[/tex] is
[tex]
\underline{\underline{A}} =
\begin{pmatrix}
0 & 0 & 1 & 0 \\
2 & 0 & 0 & 0 \\
0 & 4 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{pmatrix}
[/tex]
The Attempt at a Solution
But the rank of a matrix equals the dimension of the image of the corresponding transformation. However, [tex]\text{Rk}(\underline{\underline{A}}^n) = \text{Rk}(\underline{\underline{C}}^n)[/tex] only for [tex]n=1[/tex], which puzzles me for two reasons. Firstly, because I cannot solve problem. Secondly, because it implies that the dimension of the image of a linear transformation depends on the basis, which contradicts my "visualization" of changes of basis.