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)


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

$$\det(A - n{\rm I}_n) = 0.$$

  • $\begingroup$ My professor has not taught us eigenvalue and eigenvector. Is there any other way of solving this problem? Thanks a lot for your reply! Really appreciate it! :D $\endgroup$ – antotony Dec 1 '13 at 2:34
  • $\begingroup$ @antotony You don't really need eigenvalues and eigenvectors. Notice that $(A-n{\rm I}_n)e = 0$. Since $e \ne 0$, this means that $A-n{\rm I}_n$ is singular, and all singular matrices have zero determinant. $\endgroup$ – Vedran Šego Dec 1 '13 at 11:30
  • $\begingroup$ Excellent! Thank you for your help! Appreciate it a lot! :D $\endgroup$ – antotony Dec 1 '13 at 19:40

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