0
$\begingroup$

Is the following Proof Correct?

Theorem. There exists a Linear map $T:\mathbf{R^4}\to\mathbf{R^4}$ such that $$\operatorname{null} T = \operatorname{range} T$$ Proof. We know that the following list of vectors constitutes a basis for $\mathbf{R^4}$. $$\beta_1 = (1,0,0,0),\beta_2 =(0,1,0,0,0),\beta_3=(0,0,1,0),\beta_4 =(0,0,0,1)$$ now let us define the the transformation $T$ as follows $$T(\beta_1)=T(\beta_2) = 0$$ $$T(\beta_3)=\beta_1,T(\beta_4)=\beta_2$$ We now show that $T$ as defined above is indeed a linear transform.

Let $v_1$ and $v_2$ be arbitrary vectors in $\mathbf{R^4}$ thus for some $a_1,a_2,a_3,a_4\in\mathbf{F}$ and $b_1,b_2,b_3,b_4\in\mathbf{F}$ we have $$v_1=\sum_{j=1}^{4}a_j\beta_j,v_2 = \sum_{j=1}^{4}b_j\beta_j$$ consequently $T(v_1+v_2) = T(\sum_{j=1}^{4}a_j\beta_j+\sum_{j=1}^{4}b_j\beta_j) =T(\sum_{j=1}^{4}(a_j+b_j)\beta_j) = (a_3+b_3)\beta_1+(a_4+b_4)\beta_2 = (a_3\beta_1+a_4\beta_2)+(b_3\beta_1+b_4\beta_2) = T(v_1)+T(v_2)$

Now Let $w$ be an arbitrary vector in $V$ and $\lambda$ be an arbitrary scalar in $\mathbf{F}$, thus for some $c_1,c_2,c_3,c_4\in\mathbf{F}$ we have $$w=\sum_{j=1}^{4}c_j\beta_j$$ consequently $T(\lambda w) = T(\lambda\sum_{j=1}^{4}c_j\beta_j) = T(\sum_{j=1}^{4}\lambda c_j\beta_j) = \lambda c_3\beta_1+\lambda c_4\beta_2 = \lambda(c_3\beta_1+c_4\beta_2) = \lambda T(w)$.

The above two results together with the fact that $\beta_1,\beta_2$ is a basis for $\operatorname{range}T$ implies that $T$ is well-defined and is indeed a linear transformation on $\mathbf{R^4}$.

$\endgroup$
  • $\begingroup$ A linear operator on $\mathbb{R}^4$ can only have equal null and range spaces if they're both $\mathbb{R}^2$, by the rank plus nullity theorem. It is very easy to construct a matrix with this property, such as that given by Robert Z below. $\endgroup$ – g.s Aug 26 '17 at 8:22
3
$\begingroup$

Your ideas are good. Your proof is bad.

Proof. We know that the following list of vectors constitutes a basis for $\mathbf{R^4}$. $$\beta_1 = (1,0,0,0),\beta_2 =(0,1,0,0,0),\beta_3=(0,0,1,0),\beta_4 =(0,0,0,1)$$

That part is fine, although not really needed, as I'll show below.

now let us define the the transformation $T$ as follows $$T(\beta_1)=T(\beta_2) = 0$$ $$T(\beta_3)=\beta_1,T(\beta_4)=\beta_2$$

At this point, you've defined $T$ on exactly four elements of the domain. You could write, as your next sentence, that for $x$ not equal to any of these four, $T(x) = 0$, and that would define a transformation $T$, albeit a useless one.

The problem here is that you're trying to define a linear transformation, but you didn't say that. If you had said that, then you'd need to show that there is a linear transformation that takes on these four values. There's a theorem that says that such a transformation exists (although might not be unique) whenever the domain-points used (in your case, the $\beta$s) are linearly independent, and that the transformation is unique if the domain-points used are actually a basis. To make your proof valid at this point, you'd need to cite that theorem.

We now show that $T$ as defined above is indeed a linear transform.

You can't show that, because the example I gave (extending $T$ to be zero everywhere else) is consistent with what you wrote, but it's not linear.

So the next several lines of your proof are both wrong and irrelevant. Let me suggest a simple correction: just write down $T$, i.e., say the following:

"Let $$T: \Bbb R^4 \to \Bbb R^4 : (x, y, z, w) \mapsto (0, 0, x, y)."$$

There. That's it. Nice and clean and simple.

I'm deleting the irrelevant parts of your proof, and moving on:

The above two results together with the fact that $\beta_1,\beta_2$ is a basis for $range\ T$

How do you know that these form a basis for the range? Even going back to your original definition, or perhaps a nicer form of it, namely $$ T(x \beta_1 + y \beta_2 + z \beta_3 + w \beta_4) = x \beta_3 + y \beta_4 $$ which implicitly shows linearity, etc., what you know (by looking at it) is that $\beta_3, \beta_4$ is a spanning set for the image; how do you know it's a basis? You need to say "it's linearly independent, because..." and then either say "it's a subset of a basis, and hence of linearly independent set" or give some other reason.

implies that $T$ is well-defined and is indeed a linear transformation on $\mathbf{R^4}$.

The arguments above (if they were correct) would imply that $T$ is well-defined and a linear transformation. The fact that two particular vectors are a basis of the range is completely irrelevant. That doesn't make this sentence wrong (after all, if $p$ implies $q$, then $p$ and $r$ also implies $q$, no matter what $r$ is!), but it certainly makes it misleading.

And that's where your proof ends, without ever mentioning the nullspace of $T$. And since you were supposed to be proving that $null~T = range~T$, that's a pretty serious omission.

$\endgroup$
4
$\begingroup$

Yes, your idea is correct. In order to obtain a linear map $T:\mathbb{R}^{2n}\to\mathbb{R}^{2n}$ such that $\operatorname{null}(T)= \operatorname{range}(T)$ consider the matrix $$ \left[ \begin{array}{c|c} 0_n & I_n \\ \hline 0_n & 0_n \end{array} \right] $$ where $0_n$ is the $n\times n$ zero matrix and $I_n$ is the $n\times n$ identity matrix .

However your proof is a too long and incomplete. I recommend reading carefully the detailed answer given by John Hughes.

$\endgroup$
  • $\begingroup$ @John Hughes Thanks for your comments. You are perfectly right (+1). $\endgroup$ – Robert Z Aug 26 '17 at 8:31
2
$\begingroup$

It is far too long. There is no need to check that $T$ is linear. Given any basis $(e_1,\ldots,e_n)$ of a vector space $V$, given a vector space $W$ and given $w_1,\ldots,w_n\in W$, there is always one and only one lineat map $T\colon V\longrightarrow W$ such that $(\forall k\in\{1,\ldots,n\}):T(e_k)=w_k$.

$\endgroup$
  • $\begingroup$ Actually, there is a need to check that $T$ is linear, or at least a need to fully define $T$, at which point it's obvious. At present, OP has defined $T$ only on four items in the domain, and then implicitly assumes it's linear in proving that it's linear. $\endgroup$ – John Hughes Aug 26 '17 at 8:01
  • $\begingroup$ You might want to look at my detailed answer to see that the proof is not only too long, but incomplete and wrong in multiple places. $\endgroup$ – John Hughes Aug 26 '17 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.