# Are the following transformations linear?

I'm preparing for my exam and I am stuck at these two exercises in which I must prove that the given transformatios are linear. I know that a transformation is linear, if it's closed under adition and scalar multiplication.

1) A vector $\mathbf{a}$ is given in $\mathbb{R}^3$. Let the transformation be defined in $A: \mathbb{R}^3\to\mathbb{R}^3$ as $Ax = \langle x,\mathbf{a}\rangle x$

Is this transformation linear?

2) A transformation is defined in $T: \mathbb{R}_2[x]\to\mathbb{R}_2[x]$ as: $(T(p))(x) = x^2p(1/x)$. Prove, that it's linear and find it's transformation matrix in standard basis.

I know that, to find the transformation matrix, I have to find out where the transformation transforms vectors of the standard basis. I saw a similar example, where the values $\{1,x,x^2\}$ were simply inserted into the equation of the transformation, so this must be somewhat similar.

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Do you know what it means for a transformation to be linear? Also, I am unfamiliar with the notation R2[x] – Moss Feb 5 '13 at 12:52
ALL linear transformations are linear. Perhaps you want to ask "are the following transformations linear?" The context seems to say that. But I want to make sure you realize that the word linear does not always precede the word transformation. – rschwieb Feb 5 '13 at 12:54
I know how to prove, that it's linear (by showing that it's closed under adition and some scalar multiplication). The notation R2[x] denotes a space of polynomials of degree 2. – Trom Feb 5 '13 at 12:55
@MarcvanLeeuwen No, what I wrote is exactly what I meant, and is correct. It looks like you are looking at the current edit and not what used to be there. – rschwieb Feb 5 '13 at 13:04
@MarcvanLeeuwen OK, I finally see the interpretation you were using. While "is always being preceded" would be an ungrammatical substitute here, I think you're right that "Y isn't always preceded by X" is what I really meant when I said "X doesn't always precede Y." To native speakers in everyday conversation, though, they will be considered equivalent. Sorry to have not seen it earlier. I overlooked the subtle point. – rschwieb Feb 5 '13 at 14:56

You must be using the notation $\,(t,r)\,$ to denote some inner product in $\,\Bbb R^3\,$ , most probably the usual euclidean one , so using the bilinearity of the inner product:
$$A(x+y):=(x+y,a)a=\left((x,a)+(y,a)\right)a=(x,a)a+(y,a)a=\ldots\\{}\\\forall\,\,k\in\Bbb R\;\;,\;\;A(kx):=(kx,a)a=k(x,a)a=\ldots$$
$$T(p(x)+q(x)):=x^2\left(p\left(\frac{1}{x}\right)+q\left(\frac{1}{x}\right)\right)=\ldots\\\forall\,k\in\Bbb R\;\;,\;\;T(kp(x)):=x^2\left(kp\left(\frac{1}{x}\right)\right)=kx^2p\left(\frac{1}{x}\right)\ldots$$