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 $[T][T]^t=[T]^{t}[T]$.

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