# Thread: Linear Transformation with matrixes

1. ## Linear Transformation with matrixes

Greetings , got stuck on the following queston
Let M(3x3) be vector space of 3x3 matrices , let F : M(3x3) ---> M(3x3) be a linear transformation defined the following way :

for every A in M(3x3) be F(A) = B is a symmetrice matrice

F(A) = 1/2(A+A(T)) -1/3*Tr(A)*I
Tr(B) = 0

Find a basis for Ker F and Im F
anyways while playing around with Ker F i found the way the matrix must look like for the Ker F

also i think i got a basis for the ker F , but not sure how to go about with finding the Im F , also is my basis right ? here it is

2. Yes, that's correct so far. Since the dimension of M(3x3) is 9 and the dimension of the kernel is 4, the rank-nullity theorem tells you that you're looking for a 5-dimensional range. But a symmetric 3x3 matrix with trace 0 must look like $\begin{bmatrix}a&c&d\\c&b&e\\d&e&-a-b\end{bmatrix}$.

Now you do the rest!

3. I see , i mistakengly tried to find a matrix with the trace parts all being zero , so i got a 3 dimension each time , wich didnt go well with the dimV = dimkerf dim imf , but now i got it and solved it , thx for the help.