When is the minimum polynomial of a scalar matrix kI equal to t-k?

[Robb]

Homework Statement

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

Homework Equations

The Attempt at a Solution

f(A) is monic
f(A) = 0 since A = kI

Next we must show that deg(g) < deg(f)

I guess I'm not sure where g comes from. Is it merely an assumption? I assume g is a polynomial with root A.
And, if deg(g) < deg(f) then deg(g) must be 0 because deg(f) is 1 so g must be a constant. I think I'm just having trouble wrapping my mind around g. Please advise.

 

[fresh_42]

I guess I'm not sure where g comes from

Neither am I. You throw in polynomials $f(t)$ and $g(t)$ without having expalined / defined them.
What does it mean for $m(t)=t-k$ being the minimal polynomial of $A\,$?

We have two directions to prove, which you should clearly separate:
a) Given $A=k\cdot I \,\Longrightarrow\, (t-k)(A) = 0\text{ minimal }$
b) Given $m(t)=t-k \text{ minimal polynomial of }A \,\Longrightarrow \, A= k \cdot I$

 

[nuuskur]

On the one hand, if you observe $m(A) = A - k\cdot I$, this yields the zero matrix because of the assumption. So the minimal polynomial of $A$ must be $t-k$, there are no lower degree polynomials that could possibly work. Constants are not suitable.
Conversely, if $t-k$ is the minimal polynomial of $A$, then by definition $A - k\cdot I = 0$.

Word of advice. Always start from the definitions. Make sure you understand key concepts in this problem: matrices, (minimal) polynomials, root(s) of a polynomial. That way there will be no mysterious symbols $f$ and $g$ appearing out of nowhere.