Let A be the complex square matrix of size $2018 $ whose diagonal entries are all $−2018$ and off-diagonal entries are all $1$. What are the eigenvalues of A and their geometric multiplicities?
My solutions : First i construct the matrix $$A= \begin{pmatrix} -2018 & 1 & \cdots & 1 \\ 1 & -2018 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & -2018 \end{pmatrix}_{2018 \times 2018}$$
After that i find $(\lambda I-A_{2018}) =(\lambda+2019)^{2017}(\lambda +4037)$
as eigenvalue are $\lambda = -2019 $ with geometric multiplicity $2017$
And $\lambda = -4037$ with geometric multiplicity $ 1$
Is my solution is correct or not ? Pliz tell me
If my solution is not correct then any hints/solution will be appreciated
thanks in advance
In general if $B$ is a matrix with all eigenvalues exist, then eigenvalues of $B+\lambda I$ is same as eigenvalues of $B$ + eigenvalues of $\lambda I$.
Actually your matrix can be written as $$\begin{pmatrix} 1 & 1 & \cdots&1\\1& 1 & \cdots &1 \\ \vdots & \cdots & \cdots & \vdots\\1 & 1& \cdots &1 \end{pmatrix}+(-2019)\begin{pmatrix} 1 & 0 & \cdots&0\\0& 1 & \cdots &0 \\ \vdots & \cdots & \cdots & \vdots\\0 & 0& \cdots &1 \end{pmatrix}=B+(-2019)C$$ Now $B$ is a rank $1$ matrix and so eigenvalues of $B$ are $0(2017 \;\text{times})\; \text{and}\;2018(=\text{trace})$
Eigenvalues of $C$ are $-2019( 2018 \;\text{times})$
So eigenvalues of $B+(-2019)C$ are $$0+(-2019),$$ $$0+(-2019),$$ $$\vdots$$ $$0+(-2019)$$ $$2018+(-2019)$$ so $$\sigma\Big( B+(-2019)C\Big)=\{-2019,-1\}$$
The geometric multiplicity of the eigenvalue is the dimension of the eigenspace corresponding to that eigenvalue and it's already explained by user551819 in the comment!