Find eigenvalues and eigenvectors of this matrix Problem: Let \begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*} Compute all the eigenvalues and eigenvectors of $A$.
Attempt at solution: I found the eigenvalues by computing the characteristic polynomial. This gives me \begin{align*} \det(A - x \mathbb{I}_4) = \det \begin{pmatrix} 1-x & 1 & 1 & 1 \\ 1 & 1-x & 1 & 1 \\ 1 & 1 & 1-x & 1 \\ 1 & 1 & 1 & 1-x \end{pmatrix} = -x^3 (x-4) = 0 \end{align*} after many steps. So the eigenvalues are $\lambda_1 = 0$ with multiplicity $3$ and $ \lambda_2 = 4$ with multiplicity $1$. 
Now I was trying to figure out what the eigenvectors are corresponding to these eigenvalues. Per definition we have $Av = \lambda v$, where $v$ is an eigenvector with the corresponding eigenvalue. So I did for $\lambda_2 = 4$: \begin{align*} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = 4 \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \end{align*} I think this is only possible when $x_1 = x_2 = x_3 = x_4 = 1$. So am I right in stating that all the eigenvectors corresponding to $\lambda_2$ are of the form $t \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$ with $t$ some number $\neq 0$? For $\lambda_1 = 0$, I'm not sure how to find the eigenvectors. The zero vector is never an eigenvector. This means $x_1, x_2, x_3$ and $x_4$ can be anything aslong as they add to zero?
 A: You are exactly right!  $x_1,x_2,x_3$ and $x_4$ can be anything as long as they sum to zero.  This is already a complete solution in some sense (you have "found" all of the eigenvectors).  If you would like, you could find 3 linearly independent such vectors, and then you would be sure that these three span the whole space.
A: the matrix you have has rank one and can be written as $$A = uu^\top  \text{ where } u = \pmatrix{1,1,1,1}^\top.$$  now, $$Av = uu^\top v=(u^\top v)u.$$  note that  $u^\top v$ is a scalar, therefore $A$ has eigenvalue $0$ of multiplicity $3$ corresponding eigenvector any vector orthogonal to $u$ which has dimension $3.$  $A$ has also the nonzero eigenvalue $u^\top u = 4$ and the  corresponding eigenvector $u.$
the same ideas can be used to find the eigenvalues and eigenvectors of any rank one matrix $ab^\top.$ 
A: You are right, and to show how this applies more generally, look at the eigenspace corresponding to the other eigenvalue, $\lambda = 0$. This has multiplicity 3, so we expect exactly 3 linearly independent vectors in this space. The main equation looks like $A \vec{x} = 0 \vec{x} = \vec{0}$, in other words, 
$$
\begin{pmatrix}
1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1
\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}
 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
$$
It's not hard to see that we have 4 identical equations, so the only constraint is $x + y + z + w = 0$, so we define $w = -x-y-z$ and any vector in the eigenspace now looks like
$$
\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}
= \begin{pmatrix} x \\ y \\ z \\ -x-y-z \end{pmatrix}
= x \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix}
+ y \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}
+ z \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix}
$$
and the three vectors which form the basis of the eigenspace are now specified.
