What could be the matrix of Transformation that has no effect on vectors? The vectors are $T(x)=(x)$.

Question

We call this matrix Identity and denote it by $I$. But I am not clear with how to assume one matrix to proceed with its solution. The Form of Transformation here is $\mathbb R^2 \to \mathbb R^2$ subspaces. I am quoting the exact question here:

All of the transformations here have the form $T: \mathbb R^2 \to \mathbb R^2$. Find the matrix of the transformation that has no effect on vectors, that is $T(x)= x$. We call this matrix the identity matrix and denote it by $I$.

Answer 1

Any transformation is represented by a matrix. Now, for a $\mathbb{R}^2 \to \mathbb{R}^2$ transformation, this would be a $2 \times 2$ matrix

$$T = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$$

Now, we are given that

$$T\vec{x} = \vec{x}$$

$$\implies \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}ax+by \\ cx + dy \end{bmatrix} = \begin{bmatrix}x \\ y\end{bmatrix}$$

Since this is true for all $x,y \in R$, we have

$$a = 1, b = 0, c = 0, d =1$$

Answer 2

$T(x)=x\implies (T-I)x=0\forall x\in\mathbb R^2$

The minimal polynomial must be $m(t)=t-1$ which suggests $T$ is diagonizable. $m(t)$ also divides the characteristic polynomial so that is $c(t)=(t-1)^2$. Looking for the possible Jordan canonical forms for the matrix representation of $T$, you have $T=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ ($\color{red}{Why?}$)