For
$$
A=\left[\begin{array}{lllll}
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1
\end{array}\right]
$$
Using MATLAB, one can easily show that $\lambda_1 = 0\,\,\,\,\,\&\,\,\,\,\lambda_2 = 5$ are the eigenvalues of the $A$.
However, I would like to do it by hand.
Since $det(A) = 0$, we know that $\lambda_1 = 0$ is an eigen value.
However, I am having difficulty finding the other value. I know if we take $det(A-\lambda I) = 0$, then we get the characteristic equation and can find eigenvalues.
But, I think it is too much to do for such a special matrix.
On the other hand, if I can establish that algebraic multiplicity of $\lambda_1 = 0$ is $4$, then I can say that since $A$ must $5$ eigenvalues (not necessarily different), and $4$ of them being equal to $0$, then the other eigenvalue $\lambda_2 = \sum{a_{ii}} – 4 * \lambda_1 = 5 – 4*0 = 5$.
Is it possible to efficiently compute the algebraic multiplicity of $\lambda_1 = 0$ as $4$ by observing the matrix?
Best Answer
It is obvious that $A$ has $\mbox{rank}(A)=1$. Therefore, by the rank-nullity theorem, the Nullspace of $A$ has dimension $4$.
This means that the eigenspace for the eigenvalue $\lambda=0$ is 4-dimensional.
It is a fundamental result of Linear algebra that the multiplicity of an eigenvalue is greater or equal to the dimension of the eigenspace. [This is typically stated as "algebraic multiplicity" $\geq$ "geometric multiplicity"]
This means that $\lambda=0$ has multiplicity $4$.
It follows that $\lambda_1=\lambda_2=\lambda_3=\lambda_4=0$.
Finally, $$\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5= \mbox{tr}(A)=5$$
This gives $\lambda_5=5$.