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[SOLVED] Linear Function on a Vector Space

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Jan 25, 2013
236
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
Feb 5, 2012
1,621
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
Feb 15, 2012
1,967
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).

What is your conclusion?

(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).

johng's comments about "finite-dimensional" raises an interesting point:

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
Jan 12, 2014
1
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
Feb 5, 2012
1,621
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? :)
 
Last edited:

Deveno

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MHB Math Scholar
Feb 15, 2012
1,967
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 $F$:

Closure under vector addition:

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
Feb 5, 2012
1,621
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. :)