Let \(T:\mathbb{R}^2\rightarrow \mathbb{R}\) be given
by \(T\left(\begin{bmatrix} x\\y\end{bmatrix}\right) = 1\)
for all \(\begin{bmatrix} x\\y \end{bmatrix}\in \mathbb{R}^2\).
Is \(T\) a linear transformation?
The answer is “No”.
There are two ways to see this.
Let \(u = \begin{bmatrix} 0\\0\end{bmatrix}\).
Then \(u+u = u\). Hence,
\(T(u+u) = T(u) = 1\). But \(T(u) + T(u) = 2\).
So the first property for being a linear transformation
is not satisfied.
Another way to to see this is that \(T\) does not map the zero
vector to the zero vector.