Find $\det(A – nI_n)$, where $A$ is an $n \times n$ matrix whose entries are all 1, and $I_n$ is the $n \times n$.
identity matrix.
I have no clue how to approach this. If $A$ is an $n \times n$ matrix whose entries are all $1$, then the determinant is $0$?
What does $nI_n$ mean? The identity matrix multiplied by the number of rows/columns? (I realize that they are equal because it is a square matrix)
Best Answer
I assume you mean
$$\det(A - n{\rm I}_n) = ?,$$
where ${\rm I}_n$ is an identity matrix, so $n{\rm I}_n = \operatorname{diag}(n,n,\dots,n)$.
In this case, note that $n$ is an eigenvalue of $A$, with the associated eigenvector $e = \begin{bmatrix} 1 & 1 & \dots & 1 \end{bmatrix}^T$. So, zero is an eigenvalue of $A - n{\rm I}_n$, which means that
$$\det(A - n{\rm I}_n) = 0.$$