Are determinants additive?

Question

I was studying eigen values and suddenly they try to prove a property that states the only way eigenvalue of a matrix will be $0$ if the matrix itself was singular.

So, they went on to prove their statement by taking the determinant of $ A-\lambda I $

So, $ | A- \lambda I | = 0 $

Then they did this

$$ | A | - \lambda | I | =0 $$

Is this step valid? and if so what is the reason?

Another question: They tried to prove the characteristic equation for $ A-\lambda I $ and $ A^T -\lambda I $ is the same.

They took the determinant of the two matrices $ | A- \lambda I | $ and $ | A^T -\lambda I | $

Then they concluded that since $ | A | = | A^T | $ then the two aforementioned determinants should be the same. Were they using the same law?

Answer 1

According to your explanation for the first question, it seems that the equation $$|A-\lambda I|=|A|-\lambda |I|$$ appaers in their attempt to ensure the following statement:

Statement 1: If $A$ is not singular, then $0$ cannot be an eigenvalue of $A$.

Statement 1 is equivalent to the following statement:

Statement 2: If $\lambda=0$ is an eigenvalue of $A$, then $A$ is singular.

Perhaps their proof is for Statement 2? If this is the case, then the equality holds trivially from the assumption $\lambda=0$.

As a matter of fact, the converse of Statement 2 is also correct. See Show that a matrix $A$ is singular if and only if $0$ is an eigenvalue.