0
$\begingroup$

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)

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.