I have the linear transforation $f: W \to C^\infty(\mathbb{R})$ given by $f(x(t)=x''(t)-x'(t)$

$W$ is $span\{t,e^t,e^{-t}\}$ and the function $\frac{1}{2}e^{-t}+t \in W$

I found the image of f to be $Im(f)=e^{-t}-1$

"What is the dimension of the image f(W)? Determine the kernel for f"

I know that $dim(W)=dim(ker(f))+dim(f(W))$, but I'm confused as to what the dimension of the image is? What meaning does it have to the dimension of an image of a linear transformation if there's a constant in it?


Image of a set W under a transformation f , denoted by f(W) is the set of images of elements of W.Dimension of image of W means dim (f(W)) .


First of all your notation $Im(f)=e^{-t}-1$ is not valid because on LHS you have a set of function while the RHS is a function...

We see that $\{t, e^t, e^{-t}\}$ is a linearly independent set of functions, so we get $dim(W)=3$. (I hope this is clear)

Since $f$ is a linear map, it is sufficient to consider $$ f(t)=-1,~f(e^t)=0,~f(e^{-t})=e^{-t}-(-e^{-t})=2e^{-t} $$ because each linear map is determined by the way how it maps the base. You can compute from it the image and the kernel.

So your image is given by \begin{align} f(W)&=\{f(at+be^{t}+ce^{-t})~:~a,b,c\in\mathbb R\} =\{af(t)+bf(e^t)+cf(e^{-t})~:~a,b,c\in\mathbb R\} \\&=\{a(-1)+2ce^{-t}~:~c\in\mathbb R\}=span\{-1,e^{-1}\} \end{align} Since $\{-1,e^{-t}\}$ is independent, you get $\dim(f(W))=2$.

You can also see directly the kernel of $f$, but you should do it yourself.

  • $\begingroup$ Hi Mundron. Thanks for your thorough answer - I accidently wrote the wrong function - I edited it now (Included "+t") - Is my image still wrong? - Sorry for the inconvenience $\endgroup$ – Alex5207 Nov 12 '17 at 13:05
  • $\begingroup$ @Alex5207 I made a mistake myself and rewrote my answer and added why your notation is not valid and confusing. If you follow my computation of $f(W)$ you can see how the dimension of $f(W)$ becomes clear. (I hope so) $\endgroup$ – Mundron Schmidt Nov 12 '17 at 13:13
  • $\begingroup$ Thanks this helped a lot.. Also, I get what you're saying about the notation. Is the correct approach for calculating ker(f) to put up a matrix with the linear transformations of the base and the 0-vector like this: $$\left( \begin{matrix} -1 & 0 \\ 0 & 0 \\ 2 & 0 \end{matrix} \right) \to \left( \begin{matrix} -1 & 0 \\ 0 & 0 \\ 0 & 0\end{matrix} \right)$$ And thereby concluding that the only set in ker(f) is 0? @Mundron Schmidt $\endgroup$ – Alex5207 Nov 12 '17 at 13:51
  • $\begingroup$ @Alex5207 By the formula $\dim W=\dim(ker(f))+\dim(f(W))$ you get $\dim(ker(f))=\dim W-\dim(f(W))=3-2=1$. So you see that the kernel of $f$ can't be trivial! Further $f(e^t)=0$ so $e^t\in ker(f)$. Together you get $ker(f)=span\{e^t\}$. $\endgroup$ – Mundron Schmidt Nov 12 '17 at 17:32

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