Dimension of the image of a linear transformation dependent on basis?

[dane502]

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.

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
$$\underline{a_1} =<br /> \begin{pmatrix}<br /> 1 \\<br /> 1 \\<br /> 0 \\<br /> 0 <br /> \end{pmatrix}<br /> ,<br /> \underline{a_2} =<br /> \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> 2 \\<br /> 0<br /> \end{pmatrix}<br /> ,<br /> \underline{a_3} =<br /> \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> 0 \\<br /> 1 <br /> \end{pmatrix}<br /> ,<br /> \underline{a_4} =<br /> \begin{pmatrix}<br /> 1 \\<br /> -1 \\<br /> 0 \\<br /> 0<br /> \end{pmatrix}$$

Homework Equations

The matrix representation of $$f$$ with regards to the natural basis is

$$\underline{\underline{C}}=<br /> \begin{pmatrix}<br /> 0&0&0&1\\<br /> 0&0&0&-1\\<br /> 2&2&0&0\\<br /> 0&0&2&0<br /> \end{pmatrix}$$

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

$$\underline{\underline{A}} =<br /> \begin{pmatrix}<br /> 0 & 0 & 1 & 0 \\<br /> 2 & 0 & 0 & 0 \\<br /> 0 & 4 & 0 & 0 \\<br /> 0 & 0 & 1 & 0 \\<br /> \end{pmatrix}$$

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.

 

[micromass]

I think the matrix A is incorrect. Specifically, I don't think that the third column is correct...

 

[dane502]

I think the matrix A is incorrect. Specifically, I don't think that the third column is correct...

Thanks for your fast response - Your right. The entry in column 3 row 1 should be a 0 and not a 1.
Which makes $$\text{Rk}(\underline{\underline{A}}^n) = \text{Rk}(\underline{\underline{C}}^n)$$ for all $$n\in\mathbb{N}$$.

Would someone care to comment on whether or not the dimension of the image of a linear transformation (or rank of its matrix representation) depends on the basis being used in general?

 

[micromass]

Well, the rank of a linear transformation does not depend on the basis. So in this case, you would know immediately that you made a mistake.
Also the kernel of a transformation does not depend on the basis...