- #1
dane502
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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 f^{\circ n}, where n\in\mathbb{N} and f:\mathbb{R}^4\to\mathbb{R}^4 where
f(\underline{x})=(\underline{x}\cdot\underline{a_1}) \underline{a_2} +<br /> (\underline{x}\cdot\underline{a_2}) \underline{a_3} +<br /> (\underline{x}\cdot\underline{a_3}) \underline{a_4}
and
<br /> \underline{a_1} =<br /> \begin{pmatrix}<br /> 1 \\<br /> 1 \\<br /> 0 \\<br /> 0 <br /> \end{pmatrix}<br /> ,<br /> \underline{a_2} =<br /> \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> 2 \\<br /> 0<br /> \end{pmatrix}<br /> ,<br /> \underline{a_3} =<br /> \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> 0 \\<br /> 1 <br /> \end{pmatrix}<br /> ,<br /> \underline{a_4} =<br /> \begin{pmatrix}<br /> 1 \\<br /> -1 \\<br /> 0 \\<br /> 0<br /> \end{pmatrix}<br />
The matrix representation of f with regards to the natural basis is
<br /> \underline{\underline{C}}=<br /> \begin{pmatrix}<br /> 0&0&0&1\\<br /> 0&0&0&-1\\<br /> 2&2&0&0\\<br /> 0&0&2&0<br /> \end{pmatrix}<br />
but with regards to the basis \mathcal{A} = (\underline{a_1},\ldots,\underline{a_4}) the matrix representation of f:\mathcal{A}\to\mathcal{A} is
<br /> \underline{\underline{A}} =<br /> \begin{pmatrix}<br /> 0 & 0 & 1 & 0 \\<br /> 2 & 0 & 0 & 0 \\<br /> 0 & 4 & 0 & 0 \\<br /> 0 & 0 & 1 & 0 \\<br /> \end{pmatrix}<br />
But the rank of a matrix equals the dimension of the image of the corresponding transformation. However, \text{Rk}(\underline{\underline{A}}^n) = \text{Rk}(\underline{\underline{C}}^n) only for n=1, 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 f^{\circ n}, where n\in\mathbb{N} and f:\mathbb{R}^4\to\mathbb{R}^4 where
f(\underline{x})=(\underline{x}\cdot\underline{a_1}) \underline{a_2} +<br /> (\underline{x}\cdot\underline{a_2}) \underline{a_3} +<br /> (\underline{x}\cdot\underline{a_3}) \underline{a_4}
and
<br /> \underline{a_1} =<br /> \begin{pmatrix}<br /> 1 \\<br /> 1 \\<br /> 0 \\<br /> 0 <br /> \end{pmatrix}<br /> ,<br /> \underline{a_2} =<br /> \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> 2 \\<br /> 0<br /> \end{pmatrix}<br /> ,<br /> \underline{a_3} =<br /> \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> 0 \\<br /> 1 <br /> \end{pmatrix}<br /> ,<br /> \underline{a_4} =<br /> \begin{pmatrix}<br /> 1 \\<br /> -1 \\<br /> 0 \\<br /> 0<br /> \end{pmatrix}<br />
Homework Equations
The matrix representation of f with regards to the natural basis is
<br /> \underline{\underline{C}}=<br /> \begin{pmatrix}<br /> 0&0&0&1\\<br /> 0&0&0&-1\\<br /> 2&2&0&0\\<br /> 0&0&2&0<br /> \end{pmatrix}<br />
but with regards to the basis \mathcal{A} = (\underline{a_1},\ldots,\underline{a_4}) the matrix representation of f:\mathcal{A}\to\mathcal{A} is
<br /> \underline{\underline{A}} =<br /> \begin{pmatrix}<br /> 0 & 0 & 1 & 0 \\<br /> 2 & 0 & 0 & 0 \\<br /> 0 & 4 & 0 & 0 \\<br /> 0 & 0 & 1 & 0 \\<br /> \end{pmatrix}<br />
The Attempt at a Solution
But the rank of a matrix equals the dimension of the image of the corresponding transformation. However, \text{Rk}(\underline{\underline{A}}^n) = \text{Rk}(\underline{\underline{C}}^n) only for n=1, 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.
