The transformation T(A)=AT-A doesn't makes sense. Since T is a transformation between two 4-dimensional vector spaces so the multiplication AT cannot occur.
Paul T.
tutor
Ah good. There's really no good way of formatting on this unfortunately, so let me put trans(A) to denote the matrix transpose of A. Depending on what major you are, you may see later that people use for matrix A^i_j, the transpose people denote A^j_i. It's a a convention that people use the upper index to indicate columns and bottom to indicated rows. So for example A^2_3 denotes the entry in the 2nd row and 3rd column. So if we apply transpose you see that trans(A^2_3)=A^3_2. Anyways, let's start discussing some of these problems then.
1. I remember when I first started linear algebra, I was also confused when my professor asked me to show a certain transformation was linear. Often times, I would think what exactly did I need to prove? Wasn't it obvious? So let me walk you through additivity and I'll leave homogeneity for you to practice. So we want to show for vectors v,w, T(v+w)=T(v)+T(w), i.e. (trans(A)-A)(v+w)=(trans(A)-A)v+(trans(A)-A)w. To get this, we first have
(trans(A)-A)(v+w)=(trans(A))(v+w)-A(v+w)
this comes from the fact that trans(A) is again a matrix, and matrices correspond to linear functions. For functions, we define (f+g)(x):=f(x)+g(x), i.e. we add in the codomain. Next,
(trans(A))(v+w)-A(v+w)=(trans(A)v+trans(A)w)-(Av+Aw)
This is since A, trans(A) are matrices, they correspond to linear transformations. Thus, I am using additivity of A, trans(A) to get this equality.
(trans(A)v+trans(A)w)-(Av+Aw)=(trans(A)v-Av)+(trans(A)w-Aw)
Here, notice that trans(A)v, trans(A)w, Av, Aw are vectors in the codomain. Thus all the operations you see - addition and subtraction - are happening in the codomain vector space. We know that addition is associative and commutative in a vector space, so we have this equality.
(trans(A)v-Av)+(trans(A)w-Aw)=(trans(A)-A)v+(trans(A)-A)w
Again I am just using the definition of addition of functions: (f+g)(x):=f(x)+g(x).
For 2, the kerT:= matrices A such that trans(A)-A=0 i.e. trans(A)=A. This is where our previous discussion may come in handy. We want matrices such that A^i_j=A^j_i, for i,j=1,2. I'll leave it at that; if you need more help comment below (but try finding some explicit examples of matrices which satisfy that condition)
Part 3 follows 2 quite closely, so I won't say anything unless you specifically need a bit more push.
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12/25/20
Paul T.
tutor
Sorry for the formatting. I originally meant for this answer to be spaced much smoother, but this I guess wyzant's short on good programmers
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12/25/20
Ashley P.
Actually AT was referring to the transpose of matrix A. Sorry for any inconvenience caused. Does the question makes sense now?12/25/20