# Thread: matrix problems

1. ## matrix problems

Let A= (1/3 1/3 1/3)^T. Think of A as an operator from R^1(a one roll matrix with real number) to R^3(a matrix with 3 rows and one column) via matrix vector multiplication.

My questions are that how can I show that

1) the operator is one to one
2) how can i find 2 linear left- inverses for A
3) how can i find a left inverse for A that is not linear

A= (1/3 1/3 1/3)^T this, T mean transpose

Plz help

2. Originally Posted by oxxiissiixxo
Let A= (1/3 1/3 1/3)^T. Think of A as an operator from R^1(a one roll matrix with real number) to R^3(a matrix with 3 rows and one column) via matrix vector multiplication.

My questions are that how can I show that

1) the operator is one to one
2) how can i find 2 linear left- inverses for A
3) how can i find a left inverse for A that is not linear

A= (1/3 1/3 1/3)^T this, T mean transpose

Plz help
the operator corresponding to $\displaystyle A$ is $\displaystyle T: \mathbb{R} \longrightarrow \mathbb{R}^3$ defined by $\displaystyle T(x)=\begin{bmatrix}\frac{x}{3} & \frac{x}{3} & \frac{x}{3} \end{bmatrix}^T.$ it's obviously 1-1 because $\displaystyle T(x)=\bold{0}$ iff $\displaystyle x=0.$ now define the maps: $\displaystyle S_1,S_2: \mathbb{R}^3 \longrightarrow \mathbb{R}$ by:

$\displaystyle S_1( \begin{bmatrix}x & y & z \end{bmatrix}^T)=3x, \ S_2( \begin{bmatrix}x & y & z \end{bmatrix}^T)=3y.$ clearly $\displaystyle S_1,S_2$ are linear and $\displaystyle S_1T(x)=S_2T(x)=x.$ hence $\displaystyle S_1,S_2$ are left inverse of $\displaystyle T.$ also, for example, the map $\displaystyle S_3: \mathbb{R}^3 \longrightarrow \mathbb{R}$

defined by: $\displaystyle S( \begin{bmatrix}x & y & z \end{bmatrix}^T)=x(x-y+3),$ is a non-linear map which is a left inverse of $\displaystyle T$ because $\displaystyle S_3T(x)=x.$

3. $\displaystyle$
$\displaystyle S_1( \begin{bmatrix}x & y & z \end{bmatrix}^T)=3x, \ S_2( \begin{bmatrix}x & y & z \end{bmatrix}^T)=3y.$

how do you get this. do you just manually make the S1 and S1 = to 3x and 3 y?

4. $\displaystyle S( \begin{bmatrix}x & y & z \end{bmatrix}^T)=x(x-y+3),$ is a non-linear map which is a left inverse of $\displaystyle T$ because $\displaystyle S_3T(x)=x.$[/quote]

how come this S_3of T(x) is x for this?