# [SOLVED]Linear Function on a Vector Space

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone,

Here's a problem that I need some help to continue. I would greatly appreciate if anybody could give me some hints as to how to solve this problem.

Problem:

Let $$f:V\rightarrow F$$ be a linear function, $$f\neq 0$$, on a vector space $$V$$ over a field $$F$$. Set $$U=\mbox{Ker } f$$. Prove the following.

a) $$U$$ is a maximal subspace, that is, not contained properly in any subspace, different from $$V$$.

b) $$V=U\oplus<a>$$, for any $$a\not\in U$$.

#### johng

##### Well-known member
MHB Math Helper
Don't be stingy with the adjective "finite dimensional". Assuming this is what you meant, you just need to show the kernel of any non-zero functional has co-dimension 1 (dim(V)-dim(kernel)=1). Then any such subspace is maximal and the 2nd part should follow easily.

#### Sudharaka

##### Well-known member
MHB Math Helper
Don't be stingy with the adjective "finite dimensional". Assuming this is what you meant, you just need to show the kernel of any non-zero functional has co-dimension 1 (dim(V)-dim(kernel)=1). Then any such subspace is maximal and the 2nd part should follow easily.
Thanks for your answer. By the Rank-Nullity theorem dim(V)=dim(Ker(f))+dim(Img(f)). So if the co-dimension of Ker(f) is 1, then that means, dim(Img(f))=1. Am I correct? But how to show that the dimension of the image of f is equal to one?

#### Deveno

##### Well-known member
MHB Math Scholar
You have only two possible choices for:

$\text{dim}_{\ F}(\text{im}(f))$,

either 1 (the maximum possible, if $f$ is surjective), or 0 (if $f$ is the 0-functional).

(and this is because $f$ is linear, so its image is a subspace of $F$).

This means linear functionals annihilate *most* of a vector space (perhaps even all).

Put another way, the null space of any linear functional on a finite-dimensional space is a hyperplane of that space.

Let me give you an example to show how this works out "geometrically".

A linear functional on $\Bbb R^3$ is of the form:

$f(x,y,z) = ax + by + cz$. For our purposes we will assume $c \neq 0$

(if $f \neq 0$, one of $a,b,c \neq 0$, you can easily see how the other cases work out).

The kernel of this linear functional is the plane $P$ spanned by:

$\{(c,0,-a),(0,c,-b)\}$

and the orthogonal complement to this plane (which is a line; namely, the line:

$L = \{t(a/c,b/c,1): t\in \Bbb R\}$ ) clearly satisfies:

$\Bbb R^3 = P \oplus L$.

It is "natural" to consider the projection onto the axes functionals:

$\pi_1(x,y,z) = x$
$\pi_2(x,y,z) = y$
$\pi_3(x,y,z) = z$

as a basis for $\Bbb R^{\ast}$, the hyperplanes we get in these examples are (respectively), the $yz$-plane, the $xz$-plane and the $xy$-plane, and the lines obtained are just the $x,y$ or $z$ axes.

The reason you are doing the present exercise in this form is to do it in a "basis-free" context (the above examples depend in an essential way on picking the standard basis $\{(1,0,0),(0,1,0),(0,0,1)\}$, but "vector spaces don't care which basis you use"). We also used inner product properties to define an orthogonal complement; we don't, in general, necessarily have such a non-degenerate bilinear form chosen for us (although we can impose one on a finite dimensional space by choosing a basis, and picking a matrix for the form using that basis).

What if $V$ is NOT finite-dimensional? In such a case, we can no longer appeal to the rank-nullity theorem. Let's still see if we can (under this restriction) prove $U$ is maximal.

First of all, note that if we have $v,v' \in V$ such that: $f(v) = f(v')$ then $v-v' \in U$.

Now suppose we have a subspace $W$ properly containing $U$, so that there is some $w \in W$ with $f(w) = a \neq 0$.

Let $v$ be any element of $V$. Now we have:

$f(v) = b = \frac{b}{a}a = \frac{b}{a}f(w) = f(\frac{b}{a}w)$

and $\frac{b}{a}w \in W$. Writing $w' = \frac{b}{a}w$ we see:

$v - w' \in U$.

Thus $v = w' + (v - w')$, that is: $V = W + U$. But since $W$ contains $U$:

$W + U$ = $W$, thus $V = W$. This shows $U$ is maximal.

Now $\langle w \rangle + U$ is clearly a subspace of $V$ properly containing $U$, and by the maximality of $U$, we have:

$V = \langle w\rangle + U$. The existence of $w$ is guaranteed by the fact that $f \neq 0$.

To show this sum is direct, it suffices to show:

$\langle w\rangle \cap U = \{0\}$.

Since $f(w) \neq 0$, for $f(\alpha w) = 0$, we must have $\alpha = 0$, whence:

$\alpha w = 0$, and we are done.

Thus we do not need the assumption of finite-dimensionality.

#### bns1357

##### New member
Let $$\displaystyle B = \{b_1, \ldots , b_n\}$$ be a basis of $V$

Suppose now that there is a set $M$ such that $U \subset M \subset V$ properly. We now select $b_i, b_j$ whereby $b_i \in M\setminus U$ and $b_j \in V \setminus M$. Now let

$$v = (f(b_i))^{-1}b_i - (f(b_j))^{-1}b_j.$$

Clearly $v \notin U$, however we have that

$$f(v) = f(b_i)(f(b_i))^{-1} - f(b_j)(f(b_j))^{-1} = 1 - 1 = 0$$

which is a contradiction. Hence, such a set $M$ cannot exists.

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#### Sudharaka

##### Well-known member
MHB Math Helper
You have only two possible choices for:

$\text{dim}_{\ F}(\text{im}(f))$,

either 1 (the maximum possible, if $f$ is surjective), or 0 (if $f$ is the 0-functional).
Thanks so much for the detailed reply Deveno. But I don't really understand the part quoted above. Simply, how do we know that the dimension of the image of $f$ should be equal to 1? I understand that it is 0 if $f$ is the 0-functional, but I don't understand how 1 is the only other possibility for the dimension of the image of $f$. Can you please elaborate this piece a little bit?

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#### Deveno

##### Well-known member
MHB Math Scholar
Thanks so much for the detailed reply Deveno. But I don't really understand the part quoted above. Simply, how do we know that the dimension of the image of $f$ should be equal to 1? I understand that it is 0 if $f$ is the 0-functional, but I don't understand how 1 is the only other possibility for the dimension of the image of $f$. Can you please elaborate this piece a little bit?
We are given that $f:V \to F$ is linear(equivalently, that $f \in V^{\ast}$).

It is easy to see that $\text{im}(f)$ is a subspace of $F$:

Suppose $a,b \in \text{im}(f)$. Thus $a = f(u),b = f(v)$ for some $u,v \in V$.

Hence $a+b = f(u) + f(v) = f(u+v)$ for $u+v \in V$, so $a+b \in \text{im}(f)$.

Closure under scalar multiplication:

Suppose $a \in \text{im}(f)$, and $c \in F$. Then:

$ca = c(f(v)) = f(cv)$ for $cv \in V$ so $ca \in \text{im}(f)$.

Finally, $0 = f(0)$ is clearly in the image.

Now $\text{dim}_{\ F}(F) = 1$. This is because $\{1\}$ is a basis (any non-zero field element is also a basis, of course. 1 is just convenient).

If we look at the power set of a set with only one element (in this case, the set $\{1\}$), we have only two elements:

1) the empty set = dimension 0 = the 0-subspace
2) the entire set = dimension 1 = the entire space (the field $F$).

This is why there is only 2 choices.

To re-iterate, for emphasis: when we are talking about a vector space $V$ over a field $F$ of finite dimension $n$, what we are talking about is isomorphic to $F^n$ (choosing a basis gives us an isomorphism). The case $n = 1$ should not be overlooked: a one-dimensional subspace of $V$ is isomorphic to a line (and by choosing "the right basis" we can think of this line as "the x-axis", for example). On the line itself, "scalar multiplication" (of a point on the line by a scalar) becomes "multiplication of scalars"-this is a consequence of the vector space axiom:

$\alpha(\beta v) = (\alpha\beta)v$ for all $\alpha,\beta \in F, v \in V$.

Let me explain how this works, in this particular example:

Suppose $f \neq 0$. This means that for SOME vector $v \in V$, we have:

$f(v) = a \neq 0$ (otherwise $f = 0$).

Since f is linear, we have:

$f(cv) = c(f(v))$ for EVERY $c \in F$.

In particular, given ANY $b \in F$, we may take:

$c = \frac{b}{a}$ (we can do this BECAUSE $a \neq 0$), leading to:

$f(cv) = f(\frac{b}{a}v) = \frac{b}{a}f(v) = \frac{b}{a}a = b$,

which shows that $f$ is thus surjective.

#### Sudharaka

##### Well-known member
MHB Math Helper
To Deveno, bns1357 and johng,

Thanks so much for all your valuable input on this question. I gained a lot of understanding about this question and many other related concepts through them. Thanks again.