I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?

The linear transformation $T:M\_{2x2}(R) \to R\_3[x]$ is defined by:
$$T\begin{pmatrix}a&b\\c&d \end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$ 

for every $$\begin{pmatrix}a&b\\c&d \end{pmatrix}$$$\in M\_{2x2}(R)$ 

I reasoned in the following way:

For KerT:

a=d

b=-c

kerT:
$$\begin{pmatrix}a&b\\-b&a \end{pmatrix}$$=$$a\begin{pmatrix}1&0\\0&1 \end{pmatrix}$$+$$b\begin{pmatrix}0&1\\-1&0 \end{pmatrix}$$

So KerT = Sp{$$\begin{pmatrix}1&0\\0&1 \end{pmatrix}$$,$$\begin{pmatrix}0&1\\-1&0 \end{pmatrix}$$}
and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.

For ImT:
$(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp\{x^2,x,1\}$. This set is per definition linearly independent and therefore $Sp\{x^2,x,1\}$ is a basis to ImT and its dimension is 3.

The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M\_{2x2}(R)=4$