Verify that each of the following is an isomorphism:
1) $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $T(x,y,z)=(x+y,y+z,z+x)$.
2) $T: M_{2,2} \to \mathbb{R}^4$ defined by $T\left(\begin{bmatrix}a & b\\ c & d \end{bmatrix}\right) = (a+b, d, c, a-b)$.
For the 1), should we find $S: \mathbb{R}^3 \to \mathbb{R}^3$ such that $S(x,y,z)=(x, y, z)$ first? And in general do we have to show that each of them is linear.
I will leave you to check that these are actually linear transformations. However, the fact that I am about to express them as matrices implies that they are linear transformations.
1) We can represent $T$ by the matrix $$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{pmatrix}_.$$
By this I mean that $T(v) = Av$.
Then $T$ is invertible if and only if $\det(A) \neq 0$.
Since $\det(A) = 2$, we have that $T$ is invertible.
2) Let $$e_{1,1} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \ e_{1,2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \ e_{2,1} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \ e_{2,2} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}_.$$
These form a basis of $M_{2,2}$. If we use the standard basis for $\mathbb{R}^4$, the matrix of this transformation becomes:
$$A = \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & -1 & 0 & 0\end{pmatrix}_.$$
Since $\det(A) = 2 \neq 0$, we have that $T$ is invertible.