# what is the Transformation matrix for the linear transformation $T:M_{2times 2}(mathbb{R})to M_{2times 2}(mathbb{R})$

Mathematics Asked by Jneven on October 30, 2020

Given $$A_{1}=frac{1}{3}left(begin{array}{cc} 2 & 2\ 1 & 0 end{array}right),A_{2}=frac{1}{3}left(begin{array}{cc} 2 & -2\ 0 & 1 end{array}right)$$ , as $$W=sp{A_{1},A_{2}}$$

Let $$T:M_{2times 2}(mathbb{R})to M_{2times 2}(mathbb{R})$$ be a linear transformation defines as:
$$T(A_1)=A_2, T(A_2)=-A_1$$. $$ker T=W^{perp}$$.

Find an Orthonormal base for $$W^{perp}$$ and determine whether $$T$$ is a normal transformation.

My Attempt:

what I did what to treat $$A_1, A_2$$ as vectors in $$mathbb{R}^4$$, hence $$A_1 = frac{1}{3}(2, 2,1,0), A_2 = frac{1}{3} (2, -2,0,1)$$.

$$A_1$$ and $$A_2$$ are linear independent, therefore $$dim W =2$$ and $$dim (KerT)= dim (W^{perp}) = 2$$, which also means that $$dim(ImT) =2$$.

I solved the system $$left(begin{array}{c} A_{1}\ A_{2} end{array}right)x=0, x=(x,y,z,w)$$ to find $$W^{perp}$$, hence $$left(begin{array}{cccc} 2 & 2 & 1 & 0\ 2 & -2 & 0 & 1\ 0 & 0 & 0 & 0\ 0 & 0 & 0 & 0 end{array}right)$$ , and $$W^{perp}= sp {(-1,-1,4,0),(-1,1,0,4)}$$.

Now, I’d would like to use the transformation matrix in order to determine whether $$T$$ is normal.

Let the transformation matrix be called $$[T]$$. it needs to be a $$4times 4$$ matrix, and $$rank([T])=2$$.

is that the correct matrix: $$[T]=left(begin{array}{cccc} frac{2}{3} & -frac{2}{3} & 1 & 0\ -frac{2}{3} & -frac{2}{3} & 0 & 1\ 0 & frac{1}{3} & 0 & 0\ frac{1}{3} & 0 & 0 & 0 end{array}right)$$, I think this is the transformation matrix because if we choose a base to $$mathbb{R}^4$$ constructed by $$B={A_1, A_2, A_3, A_4) as {A_3,A_4}$$ are the vectorsmatrix which composes the base for $$KerT$$, then this each column vector describes the image of T over the vectors belongs to the base $$B$$.

From that, I conclude that $$T$$ is a normal transformation, because $$[T][T]^t=[T]^{t}[T]$$.