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The question says:

Prove that the function $f(x)=\sqrt{x}$ is not a linear transformation
(particularly $\sqrt{1+x^2}≠1+x$)

I think that this is because the exponent of $\sqrt{x}$ is $1/2$, and with that exponent, $T(cu)=c^{1/2}T(u)$, which does not follow the rules of a linear transformation...

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    $\begingroup$ It's pretty straight forward. They are try to guide you in noting if it were linear then $\sqrt{1 + x^2} = \sqrt{1} + \sqrt{x^2} = 1 + x$. This is obviously not true so it isn't linear. You noted (perhaps more intelligently than the original question that $\sqrt{cT} = \sqrt{c}\sqrt{T} \ne c\sqrt{T}$ for $c \ne 1; T \ne 0$, so it can't be linear. That's all. Both arguments are valid and unequivicably that sqrt is not linear. $\endgroup$
    – fleablood
    Commented May 23, 2016 at 17:58
  • $\begingroup$ Oh, I guess the question is asking you to prove $\sqr{1 + x^2} \ne 1 + x$.... Well, that's easy to do. $\endgroup$
    – fleablood
    Commented May 23, 2016 at 18:02
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    $\begingroup$ This is exactly because it don't satisfies the definition of linear, in particular because $\sqrt{\lambda x}\ne \lambda \sqrt x$ $\endgroup$
    – Piquito
    Commented May 24, 2016 at 0:28

7 Answers 7

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I suppose you could start with the definition of a linear transformation.

I think you have it right that $T(cu)\ne c T(u)$ for scalar $c$.

You might look at the other property of a linear transformation:

Is $T(x+y) = T(x)+T(y)$?

(i.e. does $\sqrt{x+y}=\sqrt{x}+\sqrt{y}$?)

Edit: If we want to be more abstract. (I doubt this is what the question is aiming for, but others have pointed this out.) We can define addition in our vector space as multiplication and scalar multiplication as exponentiation. This would work for $\mathbb{R}_{>0}$. We can then show that the $f$ does work. It is likely the problem you are working on intends the reals in which case you can ignore everything past "edit".

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    $\begingroup$ Yeah, you can clearly show that $f(x) = \sqrt{x}$ violates both properties of a linear transformation (i.e. $T(cx) = cT(x)$ and $T(x+y) = T(x)+T(y)$, but the question does seem to be asking for a violation of the latter. $\endgroup$ Commented May 22, 2016 at 23:51
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    $\begingroup$ You overlooked that the domain of $f$ cannot be the vector space $\mathbb{R}$ with usual addition as vector addition and usual multiplication as scalar multiplication. The requirements of the definition of a linear map are, however, met when the domain is the vector space of positive-real numbers, $\mathbb{R}_{>0}$ (over the field $\mathbb{R}$). It is an easy exercise to show that $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ is a linear map. Just think about what the vector space operations of $\mathbb{R}_{>0}$ are. $\endgroup$ Commented May 23, 2016 at 21:52
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Others have gone the symbolic route... However, it is sufficient to show one (valid) assignment of the variables that fails to achieve equality. For instance, if $x = 1$, we would need to show that $\sqrt{1+1^2} = \sqrt{2}$ does not equal $\sqrt{1} + \sqrt{1^2} = 1 + 1 = 2$. And this should be evident.

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  • $\begingroup$ @LaurentDuval : using $\sqrt{9 + 16}$ is simpler. No irrational numbers or approximations are necessary. $\endgroup$
    – zyx
    Commented May 23, 2016 at 18:45
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    $\begingroup$ @zyx : Fails to address the "Hint". $\endgroup$ Commented May 23, 2016 at 18:48
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    $\begingroup$ $\sqrt{1+\left(\frac{4}{3}\right)^2}$ would do the job. Only fractions, no need to factorize $3^2$ under a root $\endgroup$ Commented May 23, 2016 at 18:51
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    $\begingroup$ @LaurentDuval : An infinite number of choices would "do the job". What criterion shall we use to discriminate among them? I choose simplicity. $\endgroup$ Commented May 23, 2016 at 18:52
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    $\begingroup$ @Eric Towers I was just answering your "Fails to address the "Hint"" $\endgroup$ Commented May 23, 2016 at 18:54
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As Matthew Gunn pointed out, the question definitely seems to be asking you to prove $\sqrt{1+x^2}≠1+x$ in order to disprove the statement: "$\sqrt{X+Y}=\sqrt{X}+\sqrt{Y}$ for each $(X,Y)$ in $R^2$".
You may start by supposing $\sqrt{X+Y}=\sqrt{X}+\sqrt{Y}$ for each $(X,Y)$, applying that to $X=1$ and $Y=x^2$, and quickly reaching an absurdity (for example, if $x=y$ then $x^2=y^2$).

You then end up in contradiction with the very definition of a linear transformation and can conclude on that.

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Suppose that: $\sqrt{1+x^2}=1+x.$

Then: $1+x^2=1+x^2+2x.$

Then: $x=0$.

So, since in general $x \in \mathbb{R}$, the hypothesis cannot be true.

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I'll go for something different.

A one variable linear transformation must have a vector space as it's domain, yet the square root has a restricted domain, only the non-negative real numbers. The non-negative real numbers (under addition) is not a vector space (can you see why?).

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  • $\begingroup$ Would the down voter please explain? $\endgroup$ Commented May 23, 2016 at 15:47
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    $\begingroup$ Not a downvoter, I think this answer is a bit less satisfactory than others, because the “vector space” definition is not the only possible definition of linearity. Most elementarily, one can define linearity for any real-valued function on a subset of $\newcommand{\R}{\mathbb{R}}\R$, by asking it to satisfy the usual equations whenever all required inputs are in the domain of the function (and some courses do introduce linearity this way). More abstractly, one can generalise the vector space definition to e.g. modules over rigs; and $\R^{>0}$ is a perfectly good rig/module. $\endgroup$ Commented May 23, 2016 at 17:49
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    $\begingroup$ Thanks @PeterLeFanuLumsdaine. I think your points are quite valid. I saw the OP used the term "Linear Transformation", which in my experience implies the domain of discussion is vector spaces. On the other point, I think the other answers covered the non-linearity of the operation well, so I wanted to give an alternate flavor. $\endgroup$ Commented May 23, 2016 at 19:12
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The statement

Statement 1: The square root is not a linear transformation.

is not generally true, at least as it stands.

According to the most often applied definition (see, e.g., Wikipedia), a linear transformation $f : V \to W$ is a mapping between two vector spaces $V$ and $W$ over the same field. (In a weaker form of the definition $V$ and $W$ can be modules over the same ring.)

The difficulty is to acknowledge that the operations on $V$ and $W$ may well be different from usual addition and usual multiplication, even if the symbols $+$ and $\cdot$ are often used to denote vector addition and and scalar multiplication when we talk about abstract vector spaces. The terms vector addition and scalar multiplication themselves are suggestive, but misleading. Indeed, when the vector space under consideration is the concrete vector space $\mathbb{R}$, then vector addition and scalar multiplication agree with usual addition and usual multiplication; but there are numerous examples in which this is not the case. An important and simple example is the concrete vector space of the positive-real numbers, $\mathbb{R}_{>0}$, in which multiplication and exponentiation take on the role of vector addition and scalar multiplication.

Now consider the OP's square-root function $f$. The domain of $f$ cannot be $\mathbb{R}$, because the square root is not defined for negative values (let us leave aside the complex-valued case). Therefore, it is questionable if the operations of $\mathbb{R}$ are, after all, validly used to proof that Statement 1 is true. It may be argued that such proofs are flawed, because the requirements of the definition of a liear map are not met. The appendage in the OP's question, $\sqrt{1 + x^2} \neq 1 + x$, is actually irrelevant for this matter.

In fact, one can easily prove, meeting all requirements of the definition of a linear map, the converse of Statement 1:

Proposition 1: The map $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}, \, x \mapsto \sqrt{x}$ is a linear transformation.

Proof: Let $x,y \in \mathbb{R}_{>0}$, and let $\lambda \in \mathbb{R}$. Then $$ \begin{array}{rclr} f(x \cdot y) =& \sqrt{x \cdot y} = \sqrt{x} \cdot \sqrt{y} &= f(x) \cdot f(y) & \text{(preservation of vector addition)} \\ f(x^\lambda) =& \sqrt{x^\lambda} = \left( \sqrt{x} \right)^\lambda &= f^\lambda(x) & \text{(preservation of scalar multiplication)} \end{array} $$ $$\tag*{$\square$}$$

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If linearity is understood over the field of reals, then since $f(-x)$ does not exist for $x> 0$, case closed. Even if we restrict to a weak form of linearity, $\sqrt{cx} = c\sqrt{x}$ for $x> 0$ only when $c=1$.

Using the given hint, $(\sqrt{1+x^2})^2 - (1+x)^2 = 2x$, hence both quantities on the LHS are equal only when $x=0$. So almost always: $(\sqrt{1+x^2})\neq (1+x)$. A single and simple counter-example using Pythagorean triples $3^2+4^2=5^2$ (from a comment):

$$ \sqrt{1+\left( \frac{4}{3}\right)^2} = \sqrt{\left( \frac{3^2+4}{3^2}\right)} =\sqrt{\left( \frac{5^2}{3^2}\right)} = \frac{5}{3} $$ and $$1+\frac{4}{3}=\frac{7}{3}$$ and of course $$ \frac{5}{3}\neq \frac{7}{3}\,.$$

These are two reasons to conclude that $f$ is not linear. Each one is sufficient, since a counter-example suffices.

In passing, one sometimes linearize this function around $1$. A standard first order development exists when $y\sim 0$: $\sqrt{1+y} \sim 1+\frac{y}{2} $.

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    $\begingroup$ If someone has trouble understanding that $\sqrt x$ is not linear, I doubt that talking about asymptotic expansions is going to help $\endgroup$
    – Ant
    Commented May 23, 2016 at 13:26
  • $\begingroup$ @Ant Possibly, but the hint on $\sqrt{1+x^2}$, and not on the product, made me wonder about an other question. $\endgroup$ Commented May 23, 2016 at 13:32

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