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Let $T : C^{\infty}(\mathbb{R}) \rightarrow C^{\infty}(\mathbb{R})$ be a linear map, which is defined by $T = \frac{d^2}{d x^2} - 3\frac{d}{dx} + 2 \text{id}$, that is, a map which takes a function $f(x)$ and maps it to $$f''(x) - 3f'(x) + 2f(x)$$

Now my exercise asks me to find coordinate systems for the kernel of $T$ and for the nullity of $T$.

The kernel is the vectors in $C^{\infty}(\mathbb{R})$ which maps to $0$, and thus, it is exactly the functions on the form $\alpha e^{2x} + \beta e^{x}$, since that is the general solution to the differential equation $f''(x) - 3f'(x) + 2f(x) = 0$. To get a coordinate system for the kernel, I take the linear map from $\mathbb{R}^2 \rightarrow \text{ker} T$ associated with the matrix $$ \begin{pmatrix} e^{2x} & e^{x}\\ \end{pmatrix} $$

This map is both linear and bijective, thus an isomorphism of vector spaces.

My problem is with the second part. By my notes, the nullity of the map $T$ is the dimension of its kernel. In this case, the dimension of the kernel is 2. How do I define a coordinate system on the number 2? Have I misunderstood something?

The exact wording of the exercise:

Let $T : C^{\infty}(\mathbb{R}) \rightarrow C^{\infty}(\mathbb{R})$ be the linear map defined by $T = \frac{d^2}{d x^2} - 3\frac{d}{dx} + 2 \text{id}$. Find a co-ordinate system for $\text{ker} T$ and $\text{null} T$.

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    $\begingroup$ I'm with you. I thought, maybe "nullity" was supposed to be "nullspace", but then that's the same as the kernel, so it seems a bit redundant. $\endgroup$ Aug 20 '12 at 5:36
  • $\begingroup$ The kernel and nullspace of a linear transformation are exactly the same thing. $\endgroup$
    – user38268
    Aug 20 '12 at 5:40
  • $\begingroup$ The exact wording of the exercise has a greater-than symbol between the 2 and the id? $\endgroup$ Aug 20 '12 at 5:46
  • $\begingroup$ @Gerry No, that was introduced by the editor when trying to make a citation. $\endgroup$
    – utdiscant
    Aug 20 '12 at 5:48
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The intended meaning of the question is "Find a coordinate system for $\text{ker}\ T$, and find $\text{null}\ T$".

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  • $\begingroup$ So there is only one question remaining: what is null $T$? $\endgroup$
    – Fabian
    Aug 20 '12 at 6:31
  • $\begingroup$ I think you are right. @Fabian: null $T$ is the dimension of the kernel, thus 2. $\endgroup$
    – utdiscant
    Aug 20 '12 at 6:45

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