Finding properties of a linear transformation

[papaross]

Homework Statement

Find the domain, target space, image, rank and nullity of the linear transformation T(A)=Av, where v= (1, 2) and A is any 2×2matrix.

Homework Equations

The Attempt at a Solution

I believe I know the domain (R2x2 vector space) and target space (R2), but I am not sure how to solve for the image, rank, and nullity. I have tried making the matrix A an arbitrary 2x2 matrix with the variables a, b, c, d but this doesn't seem to be making it easy to find the image or rank. And I think if I am able to find the rank, I can find the nullity based on the rank-nullity theorem.

 

[Mark44]

Homework Statement

Find the domain, target space, image, rank and nullity of the linear transformation T(A)=Av, where v= (1, 2) and A is any 2×2matrix.

Homework Equations

The Attempt at a Solution

I believe I know the domain (R2x2 vector space)

Not sure what you mean by "R2x2 vector space".

and target space (R2), but I am not sure how to solve for the image, rank, and nullity. I have tried making the matrix A an arbitrary 2x2 matrix with the variables a, b, c, d but this doesn't seem to be making it easy to find the image or rank. And I think if I am able to find the rank, I can find the nullity based on the rank-nullity theorem.

What does the vector Av look like, using the matrix you described?

The nullspace is easy: Solve the equation
$$A\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}$$

Finding the image is similar:
Solve the equation
$$A\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2\end{bmatrix}$$

 

[papaross]

Not sure what you mean by "R2x2 vector space".

What does the vector Av look like, using the matrix you described?

The nullspace is easy: Solve the equation
$$A\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}$$

Finding the image is similar:
Solve the equation
$$A\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2\end{bmatrix}$$

Ok, so if I let
A= a b
c d
then, solving for the nullspace, I would get something like $$t\begin{bmatrix} -2 \\ 0 \end{bmatrix} + s\begin{bmatrix} 0 \\ -0.5\end{bmatrix}$$, meaning $$\begin{bmatrix} -2 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ -0.5\end{bmatrix}$$ spans the nullspace?

and solving for the image would be similar of course.