Linear Algebra - Kernel and range of T

[SetepenSeth]

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.

 

[LCKurtz]

Why do you think the dimension of V is 2?

 

[SetepenSeth]

Why do you think the dimension of V is 2?

I see, I just noticed I followed a wrong assumption

$V= a\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}+ b\begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}+ c\begin{bmatrix} 0&0\\ 1&0 \end{bmatrix}+ d\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}$

Thus $dim(V)=4$, right?

 

[member 587159]

I see, I just noticed I followed a wrong assumption

$V= a\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}+ b\begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}+ c\begin{bmatrix} 0&0\\ 1&0 \end{bmatrix}+ d\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}$

Thus $dim(V)=4$, right?

Yes, in general, if you consider the vector space $M_{m,n}(\mathbb{K})$ (the $m \times n$ matrices), it has dimension $mn$. This can easily be proven by showing that the set $\{E_{ij}|1 \leq i \leq m, 1 \leq j \leq n\}$ defines a basis for this space. Here, $E_{ij}$ is the matrix that is everywhere zero, except on place $(i,j)$ where it is $1$ (or another non zero number).

 

[LCKurtz]

Yes.

 

[SetepenSeth]

Thank you both, now it all makes sense.