Just starting linear algebra.

For every linear transformation $$f: V \longrightarrow W.$$

$\dim R(f) + \dim \ker(f) = \dim V$

Is this correct? $f(x)=2x$ The range of $f$, $R(f)= R_1$ dimension of $R_1=1$

The kernel of $f$, $\ker(f)=\{0\}$ dimension of $\{0\}$ is zero. $1+0$, so the dimensions of the vector is $1+0=1$


  • $\begingroup$ dimension $\{0\}$ is $0$ $\endgroup$ – Jlamprong Aug 4 '15 at 15:55
  • 1
    $\begingroup$ Yes, the dimension of the domain of a linear transformation is equal to the sum of the dimension of its range and the dimension of its kernel. $\endgroup$ – anakhro Aug 4 '15 at 16:02
  • $\begingroup$ Your post is very hard to read, please try to use [meta.math.stackexchange.com/questions/5020/… in your future posts. Also try to format you post better. And then you should provide come context, just giving $$f(x)=2x$$ does not say enough about the spaces you're looking at. $\endgroup$ – Hirshy Aug 4 '15 at 16:02

First of all: your statement is known as the rank-nullity-theorem. To point out what you are missing in your question:

Let $$f:\mathbb R^2\rightarrow \mathbb R,~f(\mathbf{x})=2x,$$ where $\mathbf{x}=\begin{pmatrix}x \\ y\end{pmatrix}\in\mathbb R^2$. Then $f$ is linear and we have $$\ker(f)=\left\{\begin{pmatrix} 0 \\ t \end{pmatrix}~|~t\in\mathbb R\right\}=\langle\begin{pmatrix}0\\1\end{pmatrix}\rangle,$$ thus the dimension of the kernel of $f$ is $1$.

Admittedly in this example I use $\mathbf{x}$ just to create some confusion, but it should show you that it is important to consider the domain when applying this theorem.


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