I am new to linear algebra, and was asked to determine if the following is a linear transformation. What is throwing me off is that for the first time we are dealing with matrices instead of vectors, and I am not sure if what I and doing makes any sense. Any feedback would be very much appreciated!
$T:M_{n\times n}^{\Bbb R} \to M_{n\times n}^{\Bbb R}$ defined by $T(X) = X^2-X$
$M,N\in M_{n\times n}^{\Bbb R}$ will be two matrices subjected to the transformation, while $λ_1,λ_2$ are two scalars $ \in\Bbb R$.
$T(λ_1M+λ_2N)=λ_1M^2-λ_1M+λ_2N^2-λ_2N\\=λ_1(M^2-M)+λ_2(N^2-N)\\=λ_1Τ(M)+λ_2Τ(N)^2$
Thank you!
In short, the answer is no. A quick way to see this is using $1\times 1$ matrices with coefficients in $\mathbb{R}$. That is, arrays $[x]$. I make the identification $[x]=x$. Then, our map is $T(x)=x^2-x$ which is not linear. For instance, $$ T(1+1)=T(2)=2^2-2=2\ne 0+0=T(1)+T(1).$$ This mapping also fails to be linear in higher dimensional cases for the same reasons. For instance, try diagonal matrices in $M_{n\times n}(\mathbb{R})$, like $$ \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}\in M_{2\times 2}(\mathbb{R})$$