Let $A=(a_{ij})$ be a $10 \times 10$ matrix such that $a_{ij}=1$ for $i \neq j$ and $a_{ii}=\alpha +1$, Where $\alpha >0$

Question

Let $A=(a_{ij})$ be a $10 \times 10$ matrix such that $a_{ij}=1$ for $i \neq j$ and $a_{ii}=\alpha +1$, Where $\alpha >0$ . let $\lambda$ and $\mu$ be the largest and smallest eigenvalues of $A,$ respectively. If $\lambda+ \mu =24,$ then $\alpha$ equals

My Idea:

By the given information the matrix is of the form all non-diagonal elements are $1$ and diagonal elements are $\alpha +1$ but i don't how to processed further

What is $Av$? If $v$ is an eigenvector what can you then say about $\lambda$ and $v$. No need to use the determinant here to calculate $\lambda$'s — Paul, Jan 08 '18 at 12:24

score 2 · Accepted Answer · answered Jan 08 '18 at 12:23

2

Hint: Write $A=U+\alpha I$, where $U$ is the matrix having $1$ in all entries. Then the eigenvalues of $A$ are of the form $\lambda+\alpha$, where $\lambda$ is an eigenvalue of $U$.

answered Jan 08 '18 at 12:23

lhf

216,483

See also https://math.stackexchange.com/questions/217521/what-are-the-eigenvalues-of-matrix-that-have-all-elements-equal-1 and https://math.stackexchange.com/a/2541168/589. – lhf Jan 08 '18 at 12:26
thank you i understood now – Inverse Problem Jan 08 '18 at 12:38

Let $A=(a_{ij})$ be a $10 \times 10$ matrix such that $a_{ij}=1$ for $i \neq j$ and $a_{ii}=\alpha +1$, Where $\alpha >0$

1 Answers1