Writting Linear Functional as Direct Sum.

**qspeechc** · February 28th 2010, 01:57 AM

Hi everyone

Sorry about the goofy thread title, it should read "Linear Functionals and Direct Sums".

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.
Is this right?

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