I have to say whether exists two bases, respectively, for $\mathbb{R}^5$ and $\mathbb{R}^4$ for which the matrix $$ A = \begin{bmatrix} 1 & 3 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & -1 \\ -1 & -3 & -1 & -1 & -1 \\ 1 & 5 & -1 & -1 & 5 \end{bmatrix} $$ represents the same homomorphism $f \colon \mathbb{R}^5 \to \mathbb{R}^4$ of the matrix $$ \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{bmatrix} $$
My attempt was: I know that $\operatorname{Null}{A} = 1$ (the fourth column is equal to the third), so the first element of the base $\mathcal{B}_5$ for $\mathbb{R}^5$ is a vector from $\ker{A}$. Then I consider the matrix $A_{4}$ without the fourth column $$ A_4 = \begin{bmatrix} 1 & 3 & 0 & 1 \\ 1 & 1 & 1 & -1 \\ -1 & -3 & -1 & -1 \\ 1 & 5 & -1 & 5 \end{bmatrix} $$ and I compute its characteristic polynomial $\operatorname{char}{A_4}$ that should be $(x - 1)^4$. Then if $\dim{V_{1}} = 4$, the answer is yes, they represents the same homomorphism and they are similar.
The problem is that calculating $\operatorname{char}{A_4}$ is really hard and time-consuming. I calculated with MATLAB the eigenvalues of $A_4$ and $1$ has an algebraic multiplicity of $1$, so the answer should be "no". But during the text I cannot use any calculator, so was I really supposed to find the eigenvalues of $A_4$, or is there another method?
Let $\{e_1,e_2,e_3,e_4,e_5\}$ be the standard basis of $\Bbb R^5$. Since the third and the fourth columns of $A$ are equal, $A.(e_3-e_4)=0$. Let $f_1=e_3-e_4$. Now, consider the subspace $V$ of $\Bbb R^5$ spanned by $\{e_1,e_2,e_3,e_5\}$. The intersection of the space $V$ with $\Bbb Rf_1$ is $\{0\}$ and the map from $V$ into $\Bbb R^4$ defined by $v\mapsto A.v$ is injective. Let $\{f_2,f_3,f_4,f_5\}$ be a basis of $V$ and defined $g_4=A.f_2$, $g_3=A.f_3$, $g_2=A.f_4$, and $g_1=A.f_5$. Then