[SOLVED]Uniquely Determined Linear Operator

Sudharaka

Well-known member
MHB Math Helper
Hi everyone, Here's a problem that I want to confirm my answer. Note that for the second part of the question it states, "prove that $$T$$ is bonded by the above claim". I used a different method and couldn't find a method that relates the first part to prove the second.

Problem:

Suppose $$X$$ is a n-dimensional linear vector space. Prove that any linear operator $$T$$ on $$X$$ is uniquely determined by $$\{Tx_i\}_{i=1}^{n}$$ with $$\{x_i\}_{i=1}^{n}$$ a basis for $$X$$. Moreover, prove that $$T$$ is bounded by the above claim.

My Ideas:

Suppose there are two representations of $$Tv$$ where $$v\in V$$. That is,

$Tv=a_1 Tx_1+\cdots+a_n Tx_n=b_1 Tx_1+\cdots+b_n Tx_n$

$(a_1 Tx_1+\cdots+a_n Tx_n)-(b_1 Tx_1+\cdots+b_n Tx_n)=0$

Since $$T$$ is linear,

$T((a_1-b_1) x_1+(a_2-b_2) x_2+\cdots+(a_n-b_n) x_n)=0$

$(a_1-b_1) x_1+(a_2-b_2) x_2+\cdots+(a_n-b_n) x_n=0$

Since $$\{x_i\}_{i=1}^{n}$$ is linearly independent,

$a_i=b_i\mbox{ for all }i=1,\,2,\,\cdots,\,n$

That is, $$T$$ is uniquely determined by $$\{Tx\}_{i=1}^{n}$$.

Now we shall show that $$T$$ is bounded. Since $$X$$ is n-dimensional, $$X$$ is topologically isomorphic. That is there exist two positive constants $$c_1$$ and $$c_2$$ such that,

$c_1\|x\|\leq \|Tx\|\leq c_2 \|x\|$

for all $$x\in X$$. Hence it's obvious that $$T$$ is bounded.

Deveno

Well-known member
MHB Math Scholar
There is a problem with your proof:

One cannot, in general, deduce from $Tu = 0$ that $u = 0$.

I will give an example in $\Bbb R^3$, to illustrate:

Suppose $T(a,b,c) = (a,b,0)$. We have (for example), $T(1,1,1) = T(1,1,0)$, but clearly it is not the case that:

$(1,1,1) = (1,1,0)$.

In other words, $T(0,0,1) = (0,0,0)$, but $(0,0,1)$ is not the 0-vector.

What you need to show is not that $\{Tx_i\}$ is linearly independent (it may NOT be), but rather that it spans the image space. Thus we can choose some SUBSET of $\{Tx_i\}$ to be a basis for $T(X)$. We are unconcerned with elements of $X - T(X)$ since those HAVE no representation as linear combinations of the $Tx_i$.

The second half of your proof is also a bit confusing: what is it you are claiming $T(X)$ is topologically isomorphic TO? It certainly is NOT $T(X)$, unless $T$ is bijective.

Sudharaka

Well-known member
MHB Math Helper
There is a problem with your proof:

One cannot, in general, deduce from $Tu = 0$ that $u = 0$.

I will give an example in $\Bbb R^3$, to illustrate:

Suppose $T(a,b,c) = (a,b,0)$. We have (for example), $T(1,1,1) = T(1,1,0)$, but clearly it is not the case that:

$(1,1,1) = (1,1,0)$.

In other words, $T(0,0,1) = (0,0,0)$, but $(0,0,1)$ is not the 0-vector.

What you need to show is not that $\{Tx_i\}$ is linearly independent (it may NOT be), but rather that it spans the image space. Thus we can choose some SUBSET of $\{Tx_i\}$ to be a basis for $T(X)$. We are unconcerned with elements of $X - T(X)$ since those HAVE no representation as linear combinations of the $Tx_i$.
Thanks so much for the reply. I understand perfectly about the fault in proof. I was rather confused by the term "uniquely determined" and thought that I should show that it's linearly independent. But then why don't they just use the word "span" instead of "uniquely determined"? Is there a difference between the two? Hope you can explain this to me. Then the first part of the problem becomes quite easy, I guess. Take any vector $$v\in X$$. Since $$\{x_i\}_{i=1}^{n}$$ is a basis of $$X$$ we have,

$T(v)=T(a_1 x_1+\cdots+a_n x_n)=a_1 T(x_1)+\cdots+a_n T(x_n)$

Therefore $$\{Tx_i\}_{i=1}^{n}$$ spans $$\mbox{Img }T$$. Am I correct?

The second half of your proof is also a bit confusing: what is it you are claiming $T(X)$ is topologically isomorphic TO? It certainly is NOT $T(X)$, unless $T$ is bijective.
When I am writing the answer I took $$T$$ maps $$(X,\,\|\cdot\|_X)$$ into $$(Y,\,\|\cdot\|)_Y$$. And since both are $$n-dimensional$$ $$X$$ and $$Y$$ are topologically isomorphic. But there is a mistake in my original post. The last inequality should be,

$c_1\|x\|_X\leq \|Tx\|_Y\leq c_2 \|x\|_X$

I also figure out now that there's a obvious mistake here. We don't know whether $$Y$$ is n-dimensional or not. I have to start from the beginning. Edit: I think I got it. There's another theorem;

Let $$(X,\,\|\cdot\|_X)$$ and $$(Y,\,\|\cdot\|)_Y$$ be two normed linear spaces over $$F$$. If $$X$$ is finite dimensional, then any linear transformation $$T:\, X\rightarrow Y$$ is bounded.

We can use this theorem to show that $$T$$ is bounded. Am I correct? Deveno

Well-known member
MHB Math Scholar
I'd like to address the first part of the problem, and then we can move on to the second part.

It's not generally true for a function $f:X \to Y$ that the value of $f$ on a finite subset of $X$ completely determines $f$. For example, when $X = Y = \Bbb R$ (and these are both vector spaces), there are an infinite number of functions with:

$f(0) = a$
$f(1) = b$

no matter how we choose $a$ and $b$.

So linear functions are very special in this regard: for a vector space of dimension 2, for example, we only need the value of a linear function at 2 (linearly independent) points to know which one we have. This is essentially saying that a linear function $T: X \to X$ is completely determined by its matrix relative to any basis for $X$.

You seem to be focusing on the values of $T$ and arguing these have only "one representation". This isn't what you want to do (and it's not TRUE). What you want to do is THIS:

Suppose that $S,T$ are two linear functions with:

$S(x_i) = T(x_i),\ i = 1,2,\dots,n$.

Then, for any $v \in X$:

$S(v) = S(c_1x_1 + \cdots c_nx_n) = c_1S(x_1) + \cdots + c_nS(x_n)$

$= c_1T(x_1) + \cdots + c_nT(x_n) = T(c_1x_1 + \cdots + c_nx_n) = T(v)$

so we conclude $S = T$.

Before we move on to part 2, make sure you understand this. Also, the notion of boundedness generally requires a norm...what norm are we using for $X$ (as it is not given by the problem)?

Sudharaka

Well-known member
MHB Math Helper
I'd like to address the first part of the problem, and then we can move on to the second part.

It's not generally true for a function $f:X \to Y$ that the value of $f$ on a finite subset of $X$ completely determines $f$. For example, when $X = Y = \Bbb R$ (and these are both vector spaces), there are an infinite number of functions with:

$f(0) = a$
$f(1) = b$

no matter how we choose $a$ and $b$.

So linear functions are very special in this regard: for a vector space of dimension 2, for example, we only need the value of a linear function at 2 (linearly independent) points to know which one we have. This is essentially saying that a linear function $T: X \to X$ is completely determined by its matrix relative to any basis for $X$.

You seem to be focusing on the values of $T$ and arguing these have only "one representation". This isn't what you want to do (and it's not TRUE). What you want to do is THIS:

Suppose that $S,T$ are two linear functions with:

$S(x_i) = T(x_i),\ i = 1,2,\dots,n$.

Then, for any $v \in X$:

$S(v) = S(c_1x_1 + \cdots c_nx_n) = c_1S(x_1) + \cdots + c_nS(x_n)$

$= c_1T(x_1) + \cdots + c_nT(x_n) = T(c_1x_1 + \cdots + c_nx_n) = T(v)$

so we conclude $S = T$.
Thanks for the detailed explanation. I think I am getting the intuition behind the phrase "uniquely determined" after reading your post. However there's one little point I want to clarify.

So given the set of values $$\{Tx_i\}_{i=1}^{n}$$, only one linear transformation $$T$$ exists with these values as it's image with respect to the basis $$\{x_i\}_{i=1}^{n}$$. Is this statement correct? In your proof you have taken two linear transformations $$S$$ and $$T$$ and shown that they are the same. However isn't it more appropriate to let, $$\{T(x_i):\,i=1,\,\cdots,\,n\}=\{S(x_j):\, j=1,\,\cdots,\,n\}$$. ? I mean, for example, it could be that $$Tx_1=Sx_2,\,Tx_2=Sx_1$$ and so on. Or is there a necessity that $$Tx_i=Sx_i$$ for all $$i$$. ? Did you get what I meant? Before we move on to part 2, make sure you understand this. Also, the notion of boundedness generally requires a norm...what norm are we using for $X$ (as it is not given by the problem)?
Yes, I thought about this too. That why I took a general norm without specifying anything in particular. But then after your question I tried to find a norm for this vector space and I came up with the following.

$\|x\|=\sum_{i=1}^{n}|a_i|$

where $$x=\sum_{i=1}^{n}a_{i}x_i$$. How about that, is it correct? Deveno

Well-known member
MHB Math Scholar
Um, no...the indices have to match. Take a simple example:

Let $X = \Bbb R^2$ with:

$T = I = \text{id}_X$ (that is, $Tv = v$ for all $v \in \Bbb R^2$), and let:

$S(x,y) = (y,x)$. For the basis $B = \{(1,0),(0,1)\}$ we certainly have:

$T(1,0) = S(0,1)$
$T(0,1) = S(1,0)$, but these are clearly not the same linear transformation.

As to the norm question, there are MANY possible norms on a given vector space. It's very likely that the "usual" or "p-2 norm" (aka the Euclidean norm) is what is intended:

For $x = (x_1,\dots,x_n)$, we have:

$\displaystyle \|x\| = \left(\sum_{i = 1}^n x_i^2 \right)^{\frac{1}{2}}$

The norm you have given is also known as the "p-1 norm" or "taxi-cab norm" (or sometimes the "Manhattan norm"). We also get a norm for any real number $p \geq 1$ by taking:

$\displaystyle \|x\| = \left(\sum_{i = 1}^n x_i^p \right)^{\frac{1}{p}}$

and if we take the limit as $p \to \infty$, we get the "maximum norm":

$\|x\| = \text{max}(|x_1|,\dots,|x_n|)$.

It is not clear which of these norms is intended, of if the problem is asking you to show that $T$ is (locally) bounded for ANY norm, which is a slightly different question.

Sudharaka

Well-known member
MHB Math Helper
Um, no...the indices have to match. Take a simple example:

Let $X = \Bbb R^2$ with:

$T = I = \text{id}_X$ (that is, $Tv = v$ for all $v \in \Bbb R^2$), and let:

$S(x,y) = (y,x)$. For the basis $B = \{(1,0),(0,1)\}$ we certainly have:

$T(1,0) = S(0,1)$
$T(0,1) = S(1,0)$, but these are clearly not the same linear transformation.

As to the norm question, there are MANY possible norms on a given vector space. It's very likely that the "usual" or "p-2 norm" (aka the Euclidean norm) is what is intended:

For $x = (x_1,\dots,x_n)$, we have:

$\displaystyle \|x\| = \left(\sum_{i = 1}^n x_i^2 \right)^{\frac{1}{2}}$

The norm you have given is also known as the "p-1 norm" or "taxi-cab norm" (or sometimes the "Manhattan norm"). We also get a norm for any real number $p \geq 1$ by taking:

$\displaystyle \|x\| = \left(\sum_{i = 1}^n x_i^p \right)^{\frac{1}{p}}$

and if we take the limit as $p \to \infty$, we get the "maximum norm":

$\|x\| = \text{max}(|x_1|,\dots,|x_n|)$.

It is not clear which of these norms is intended, of if the problem is asking you to show that $T$ is (locally) bounded for ANY norm, which is a slightly different question.
Thanks so much for all your help. I think I understand this completely now. I think the question expects us to show that the linear map is bounded for any norm which is in fact a theorem given in our recommended text book. Deveno

Well-known member
MHB Math Scholar
For FINITE-DIMENSIONAL vector spaces, (local) boundedness is a consequence of continuity (at 0, and thus everywhere). I remark in passing that the definition of continuity is also dependent upon the norm (one uses the norm to establish a metric, and then uses the definition of continuity in a metric space...otherwise you lack quantities to be less than your epsilons and deltas).

In particular, by choosing a delta-ball centered at the origin that is small enough radius, we can ensure that $\|T(x)\| \leq 1$ for all $x$ with $\|x\| \leq \delta$ (in other words, we pick an epsilon equal to 1).

Then for ANY $v \in X$, we have:

$\displaystyle \|T(v)\| = \left\|\frac{\|v\|}{\delta}T\left(\delta \frac{v}{\|v\|}\right)\right\|$

$\displaystyle = \frac{\|v\|}{\delta}\left\|T\left(\delta \frac{v}{\|v\|}\right)\right\|$

$\displaystyle \leq \frac{1}{\delta}\|v\|$

so we may take $M = \dfrac{1}{\delta}$, which established local boundedness, and depends only on the generic property every norm has, namely:

$\|\alpha v\| = |\alpha|\cdot \|v\|$