Largest and smallest eigenvalues of a given symmetric matrix Matrix A is given to be 10 by 10 matrix such that it's diagonal entries are all same and equal to a+1 where a>0. All other entries are one. Sum of largest eigenvalue of A and smallest eigenvalue of A is 24 and we are to find value of a.
One eigenvalue is sum of each row that is a+10. But I don't know how to proceed to find largest and smallest eigenvalues. May be these eigenvalues will be in terms of a and the given equation of sum of these eigenvalues will give the value of a but how to find largest and smallest eigenvalues of this matrix. Thanks 
 A: This matrix is the sum of an all-1s matrix and $aI$. Call that all-ones matrix $U$. Then 
$$
M = U + aI
$$
The characteristic polynomial of $M$ is 
$$
c(x) = \det(U + aI - xI) = \det (U + (a-x)I)
$$
When $x = a$, we have a root of some multiplicity. What multiplicity is it? (You might want to try a $3 \times 3$ example to check.) 
And $x = a+10$ is another root. What's its multiplicity? 
A: We have 
$$\tag{1}M=aI+{\bf 1}{\bf 1}^T$$
where ${\bf 1}$ is the vector (or matrix $n \times 1$) with all coefficients equal to 1. 
Let us define $N:=\dfrac{1}{a}M$ and $V:=\dfrac{1}{\sqrt{a}}{\bf 1}$
Using these notations, we can write (1) under the form 
$$N=I+VV^T$$ 
Matrix-determinant lemma (https://en.wikipedia.org/wiki/Matrix_determinant_lemma) gives 
$$\det(N)=\det \left(I+VV^T\right)=1+V^TV=1+\tfrac{1}{a}$$
Thus
$$\det(N)=\tfrac{1}{a^n}\det(M)$$
finaly giving 
$$\det(M)=a^n \det(N) = a^{n-1}(a-1)$$
Now, if we replace in the previous formula $a$ by $a-\lambda$, we have the characteristic polynomial of $M$:
$$\det(M - \lambda I)=(a-\lambda)^{n-1}((a-1)-\lambda)$$
Thus 2 eigenvalues, one equal to $a-1$ (the smallest) and the other, with order of multiplicity $n-1$, equal to $a$ (the largest).
