Why is the determinant of a nilpotent matrix 0




















This is always possible. Tags: Cayley-Hamilton theorem eigenvalue eigenvector Jordan canonical form linear algebra nilpotent matrix triangularizable matrix upper triangular matrix.

Review Nilpotent matrix and eigenvalues of the matrix […]. Your email address will not be published. Save my name, email, and website in this browser for the next time I comment.

Notify me of follow-up comments by email. Notify me of new posts by email. This site uses Akismet to reduce spam. Learn how your comment data is processed. The list of linear algebra problems is available here. But why is this the case? However, for real numbers, we cannot have the product of two non-zero numbers be zero.

We are now ready to put this together into a proof. As we worked through this, we got to work more with properties of determinants. Such theorems are can be extremely helpful in that we now have another way to determine if a matrix is singular. Hopefully this helped you better understand determinants and get a little extra work with proofs. If this did help, make sure to share this post on Social Media and subscribe to our YouTube channel so that you see the videos that accompany these posts.

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