# Linear transformation proof

• Oct 23rd 2012, 09:45 PM
M.R
Linear transformation proof
Hi,

I am trying to prove the following are not a linear transformation:

1. $\displaystyle T: R^3 -> R$ defined by $\displaystyle T(x) = x_{1}x_{2}x_{3}$

2. $\displaystyle P -> P, T(p)=p+2p'+3p''$, where P is the vector space of polynomials (of any degree)

3. $\displaystyle det: M_{22} -> R, det \left(\begin{array}{cc}a&b\\c&d\end{array}\right) = ad-bc$

My attempt:

I have tried the 0 vector & -1 scaler multiplication and it staisfies all the 3 eqations. Furthermore they satisfy addition as well. So why are they not linear transformation?
• Oct 24th 2012, 04:33 AM
TheEmptySet
Re: Linear transformation proof
Quote:

Originally Posted by M.R
Hi,

I am trying to prove the following are not a linear transformation:

1. $\displaystyle T: R^3 -> R$ defined by $\displaystyle T(x) = x_{1}x_{2}x_{3}$

2. $\displaystyle P -> P, T(p)=p+2p'+3p''$, where P is the vector space of polynomials (of any degree)

3. $\displaystyle det: M_{22} -> R, det \left(\begin{array}{cc}a&b\\c&d\end{array}\right) = ad-bc$

My attempt:

I have tried the 0 vector & -1 scaler multiplication and it staisfies all the 3 eqations. Furthermore they satisfy addition as well. So why are they not linear transformation?

The first one is not linear because it does not satisfy

$\displaystyle T(\mathbf{v}_1+\mathbf{v_2})=T(\mathbf{v}_2)+T (\mathbf{v}_2)$

Let $\displaystyle v_1=e_1+e_2+e_3$ and $\displaystyle v_2=-e_1$

This gives

$\displaystyle T(v_1)=1 \quad T(v_2)=0$ but

$\displaystyle T(v_1+v_2)=T(e_2+e_3)=0 \ne T(v_1)+T(v_2)=1+0=1$

For number 2. It is most definitely a linear operator. Differentiation is one of the first linear operators people learn.

For number 3. Try addition again

Note that

$\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}$

use the same idea as in number 1