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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 $\mathbf{R^4}$.

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  • $\begingroup$ 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, 2017 at 8:22

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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.

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  • $\begingroup$ @John Hughes Thanks for your comments. You are perfectly right (+1). $\endgroup$
    – Robert Z
    Aug 26, 2017 at 8:31
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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.

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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 $(\forall k\in\{1,\ldots,n\}):T(e_k)=w_k$.

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  • $\begingroup$ 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. $\endgroup$ Aug 26, 2017 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. $\endgroup$ Aug 26, 2017 at 8:15

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