Isomorphism of linear map

Question

Suppose we have an equation of linear transformation $\varphi : \mathbb{R}^4 \rightarrow \mathbb{R}^4$. How to show that such transformation is isomorphic and how to find inverse isomorphism ($\varphi^{-1}$)?

EDIT: The transformation is: $\varphi ([x_1,x_2,x_3,x_4])=[x_1+x_2+x_3, x_1+x_3+x_4,x_1+x_4,x_1+x_2+x_3+x_4]$

Answer 1

The matrix of the linear map with respect to the standard basis is $$ A=\begin{bmatrix} 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix} $$ and Gaussian elimination goes as follows \begin{align} \begin{bmatrix} 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & -1 & 0 & 1 \\ 1 & -1 & -1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \\[1ex]&\to \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & -1 & 0 & 1 \\ 0 & -1 & -1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \\[1ex]&\to \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \\[1ex]&\to \begin{bmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{align} which shows the rank of the matrix is $4$. So the map is surjective, hence also injective.

In order to find the inverse, a good method is to proceed with finding the RREF of the matrix $[A\ I_4]$: \begin{align} \left[\begin{array}{@{}cccc|cccc@{}} 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \end{array}\right] &\to \left[\begin{array}{@{}cccc|cccc@{}} 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 1 & -1 & 1 & 0 & 0 \\ 0 & -1 & -1 & 1 & -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right] \\[1ex]&\to \left[\begin{array}{@{}cccc|cccc@{}} 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right] \\[1ex]&\to \left[\begin{array}{@{}cccc|cccc@{}} 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right] \\[1ex]&\to \left[\begin{array}{@{}cccc|cccc@{}} 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right] \\[1ex]&\to \left[\begin{array}{@{}cccc|cccc@{}} 1 & 1 & 0 & 0 & 1 & -1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right] \\[1ex]&\to \left[\begin{array}{@{}cccc|cccc@{}} 1 & 0 & 0 & 0 & 1 & 0 & 1 & -1 \\ 0 & 1 & 0 & 0 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right] \end{align} which shows that $$ A^{-1}= \begin{bmatrix} 1 & 0 & 1 & -1 \\ 0 & -1 & 0 & 1 \\ 0 & 1 & -1 & 0 \\ -1 & 0 & 0 & 1 \end{bmatrix} $$ so that $$ \varphi^{-1}([x_1,x_2,x_3,x_4])= [x_1-x_4,-x_2+x_3,x_1-x_3,-x_1+x_2+x_4] $$

Answer 2

Recall the definition of a linear isomorphism of vector spaces. In this case is sufficient for $\dim(\ker(\varphi)) = 0$ and $\dim(\mathrm{im}(\varphi)) = 4$. To help you further requires knowledge of $\varphi$.

EDIT

Ok now try to show that

(a) if $\varphi(\mathbb{x}) = 0$ then necessarily $\mathbb{x} = 0$. This shows $\dim(\ker(\varphi)) = 0$, thus injectivity.

(b) there are four vectors whose images are linearly independent. This shows that $\dim(\mathrm{im}(\varphi)) = 4$, or surjectivity.

I suggest that trying to find a matrix representation of $\varphi$ would be a very good idea for showing both these points.

For $\varphi^{-1}$, once you have a matrix you can compute its inverse.

Answer 3

$\phi : \mathbb R^4 \to \mathbb R^4$ is given by $\phi ((x_1,x_2,x_3,x_4))=(x_1+x_2+x_3, x_1+x_3+x_4,x_1+x_4,x_1+x_2+x_3+x_4).$ Let $e_1, e_2, e_3, e_4$ be the standard basis of $\mathbb R^4.$ Then $\phi (e_1)= (1, 1, 1, 1), \phi (e_2) = (1, 0, 0, 1), \phi(e_3)= (1, 1, 0,1), \phi(e_4) = (0, 1, 1, 1).$ Now you need to show that $\{\phi(e_1), \phi(e_2), \phi(e_3), \phi(e_4) \}$ forms a basis of $\mathbb R^4.$

Using Rank-Nullity theorem, to show that a given linear transformation is an isomorphism it is sufficient to check that it maps a basis to a basis.