Show that $A$ is a scalar matrix $kI$ if and only if the minimum polynomial of $A$ is $m(t) = t - k$.

Question

Can someone help me solve this or hint in the right direction?

Some helpful notes may be:

• The characteristic polynomial of an $n$-square matrix $A$ is $\det(tI - A)$ where $I$ is the identity matrix.
• $\det(tI - A) = 0$ is the characteristic equation of $A$.
• Cayley-Hamilton Theorem: Every matrix $A$ is a zero of its characteristic polynomial.
• The minimum polynomial of a matrix (linear operator) $A$ divides every polynomial that has $A$ as a zero. In particular, the minimum polynomial divides the characteristic polynomial of $A$.
• The characteristic polynomial and the minimum polynomial of a matrix $A$ have the same irreducible factors.
• A scalar is an eigenvalue of the matrix $A$ if and only if the scalar is a root of the minimum polynomial of $A$.

Answer 1

It is clear that $k\operatorname{Id}$ is a root of the polynomial $x-k$ and thatefore that its minimal polynomial is $x-k$.

On the other hand, Is $x-k$ is the minimal polynomial of $M$, then $M-k\operatorname{Id}=0$, which means the $M=k\operatorname{Id}$.

Answer 2

Hints:

From your post: A scalar is an eigenvalue of the matrix A if and only if the scalar is a root of the minimum polynomial of A.

Can you think of an eigenvalue of the transformation kI? Can you think of a polynomial that's a factor of a polynomial if and only if that polynomial has k as a zero?