# Writting Linear Functional as Direct Sum.

• Feb 28th 2010, 01:57 AM
qspeechc
Linear Functionals and Direct Sums.
Hi everyone

Let $V$ be a finite-dimensional vector space over the field of scalars $F$, and $T$ a linear map from $V$ to $F$. Show that
$V = kerT\oplus span\{ u\}$
where $Tu \neq 0$
Well, I reasoned that if T is not the zero map, then the range of T is $F$ (simple to show), in which case the range of T is spanned by one vector. From dim(V)=dim(kerT)+dim(ranT) we must have that the dimension of the kernel is one less than that of V. So let $\{u_1,\ldots ,u_k \}$ be a basis for kerT. Extending it to a basis to V we only need add $u$, which is not in the kernel, because if it was it would contradict the linear independence of the base for the kernel.
So a basis for V is $\{u_1,\ldots ,u_k, u \}$. Obviously $kerT \cap span\{ u\} = 0$.
The fact that $\{u_1,\ldots ,u_k, u \}$ is a basis for $V$ shows that $V = kerT + span\{ u\}$. Thus, $V = kerT\oplus span\{ u\}$.
If $T$ is the zero map then the set of vectors not in the kerel is the empty set, and so $span\{ u\}=\{ 0\}$, and everything is all good.