Gershgorin theorem when a matrix row is $a_{ij} = \delta_{ij}b$

Question

Given a matrix $A \in \mathbb{R}^{N \times N}$ where there is a row (say the $i$-th row) such that

$$a_{ij} = \delta_{ij}b ~~~~\forall j \in \{1, \ldots, N\}$$

where $$\delta_{ij} = \left\{ \begin{array}{ll}0 & i \neq j \\ 1 & i = j\end{array}\right.$$

From Gershgorin theorem, we know that there is a degenerate circle which coincides with the point $b$ in $\mathbb{C}$. Then, can we assert that this matrix has at least one eigenvalue equal to $b$?

CORRECTION

$A$ is not necessarily diagonal. $A$ has one row, say the $i$-th, such that

$$a_{ij} = \delta_{ij}b ~~~~\forall j \in \{1, \ldots, N\}$$

Answer 1

The way you described it, $A=b I$ where $b\in\mathbb R$ and $I$ is the identity matrix... This means all eigenvalues of $A$ equal $b$.

Edit: after you edited the question, you say you have one ($j$-th) row which has $a_{jj}=b$ and $a_{ij}=0$ for $i\neq j$. In that case, you can still claim that one eigenvalue will equal to $b$, since the determinant of $A-\lambda I$ will be $(b-\lambda)\cdot p(\lambda)$ for some polynomial $p$.

Answer 2

I perfectly agree with what @5xum said. In general if you have $k\le N$ rows with only one non zero element, then those are going to be the eigenvalues of the matrix. This can be seen either by considering the determinant $\det{(\lambda I-A)}$ or by the fact that those eigenvalues will correspond to Gershgorin circles with radii $0$ centered at those values.