how to determine a matrix has a single eigenvalue Find the jordan form of the matrix
$$A = \begin{pmatrix}
 1 &amp; 1 &amp; 2 &amp; 2\\
 1 &amp; -2 &amp; -1 &amp; -1\\
 -2 &amp; 1 &amp; -1 &amp; -1\\
1 &amp; 1 &amp; 2 &amp; 2 \\
\end{pmatrix}$$

Hint: check that the matrix A has a single eigenvalue, and $trace(A) = 0$.

How can I check that the matrix has a single eigenvalue without using the determinant?

That's what I managed to do so far:

Let $G$ the Jordan form of $A$, using the hint that there is a single eigenvalue:
$$trace(A) = 0\implies 0 = trace(A) = trace(G) = 4a \implies a = 0$$

meaning that theres a single eigenvalue which is zero.

Then the characteristic polynomial is $A^4$  and the minimal polynomial is $A^3$.

Then:
$G = diag\{J_3(0), J_1(0)\}$

But how do I prove that there is a single eigenvalue?
The rank is 2 (the first two columns span the image) and trace is zero. If there is one non-zero eigenvalue $a$, there is another non-zero eigenvalue $-a$. Hence the image would be spanned by the two eigenvectors of the non-zero eigenvalues and the same would automatically hold for all powers of $A$. But the first column is contained in the kernel and in the image of the matrix, hence the rank of $A^2$ will be at most $1$ (since $e_1$ is in the kernel of $A^2$, but not in the kernel of $A$), contradiction to the conclusion above!
Since the characteristic polynomial is $\chi_A (x) = x^4 $, this tells you that all eigenvalues are zero.

A matrix satisfies its characteristic polynomial by the Cayley Hamilton theorem, hence $A^4 = 0$. If $v$ is a non zero eigenvector with eigenvalue $\lambda$ then $0 = A^4 v = \lambda^4 v$ and so $\lambda = 0$.