- #1

SetepenSeth

- 16

- 0

## Homework Statement

Let ##T:M_2 \to M_2## a linear transformation defined by

##T \begin{bmatrix}

a&b\\

c&d

\end{bmatrix} =

\begin{bmatrix}

a&0\\

0&d

\end{bmatrix}##

Describe ##ker(T)## and ##range(T)##, and find their basis.

## Homework Equations

For a linear transformation ##T:V\to W##

##range(T)={T(x) \epsilon W : x \epsilon V}##

##ker(T)= {x \epsilon V : T(x)= 0 \epsilon W}##

## The Attempt at a Solution

Skipping the first part of the proof, I get to the part where I describe the range of the transformation and express it as a linear combination of two ##M_2## matrix##T \begin{bmatrix}

a&b\\

c&d

\end{bmatrix} =

\begin{bmatrix}

a&0\\

0&d

\end{bmatrix}##

##T \begin{bmatrix}

a&b\\

c&d

\end{bmatrix} =

a\begin{bmatrix}

1&0\\

0&0

\end{bmatrix}+

d\begin{bmatrix}

0&0\\

0&1

\end{bmatrix}

##

So $$\begin{bmatrix}

1&0\\

0&0

\end{bmatrix} and \begin{bmatrix}

0&0\\

0&1

\end{bmatrix}$$ span the range for ##T##, also they are linearly independent, thus forming a basis for the range.

The kernel can be expressed as

##ker(T)={A \epsilon M_2 : Ax=0 \epsilon M_2}##

##ker(T)=A \epsilon M_2 :## ##A =

\begin{bmatrix}

a&b\\

c&d

\end{bmatrix} \forall a,b,c,d \epsilon ℝ, a=d=0## If I got it right up to this point then

##dim[range(T)]=2##

But then according to the theorem that says

##dim[range(T)]+dim[ker(T)]=dim(V)##

Being ##V## the space of ##2_x####2## square matrix, then ##dim(V)=2## but that would make ##dim[ker(T)]=0## which doesn't make sense to me, so I believe I got a concept wrong somewhere on my analysis.

Any advise would be appreciated.