# 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?

• you are absolutely right. – Zhanxiong Jul 13 '15 at 16:41
• the eigenspace which is corresponding to the eigenvalue 0 is also called the kernel of the matrix - and you are right – user190080 Jul 13 '15 at 16:41
• related – Omnomnomnom Jul 13 '15 at 18:01

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.

• Thank you, that was very clear. So all the eigenvectors corresponding to the eigenvalue zero are of the form $x \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} + y \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix} + z \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix}$, with the $x,y$ and $z$ numbers $\neq 0$? – Kamil Jul 13 '15 at 17:09
• @Kamil exactly :) – gt6989b Jul 13 '15 at 17:11
• Worth noting that these are not the only three vectors which work. You can choose any three linearly independent such vectors. – Steven Gubkin Jul 13 '15 at 17:11
• @StevenGubkin of course, once you are in a span, any 3 lin.indep. members will form a basis, but I wanted to show an easy way to construct one, which is untuitive and direct... – gt6989b Jul 13 '15 at 17:12

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.

• So there are infinite eigenvectors corresponding to $\lambda = 0$. How would I express this mathematically, I mean in set notation? The only condition is that they add to zero. – Kamil Jul 13 '15 at 16:47
• @Kamil yes, there are infinite solutions (over an infinite field, of course). They are the solution set of the homogeneous system $\;\det(A-xI)=0\;$ , and in this case this space has dimension three. – Timbuc Jul 13 '15 at 16:53

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 $ab^\top.$