Is $\det(c I_n - A^T) = \det(c I_n - A)$? How to prove it?

Question

Problem: Are the following assertions true or false? Prove or give a counterexample:

1) If $A$ is an $(n \times n)$-matrix, then for every $c \in \mathbb{R}$ we have $\det(c I_n - A) = c^n - \det(A)$.

2) If $A$ is an $(n \times n)$-matrix, then for every $c \in \mathbb{R}$, we have $\det(c I_n - A^T) = \det(c I_n - A)$.

Attempt: 1) This is false. Let $c I_2 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. Let $A = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$. Then \begin{align*} \det(c I_2 - A) = \det \begin{pmatrix} 1 & 0 \\ -2 & -1 \end{pmatrix} = -1. \end{align*} But \begin{align*} c^2 - \det(A) = 2^2 - 3 = 1. \end{align*}

2) For the second assertion, I think this is true. I tried to find a counter example, but couldn't. So how can I prove it?

Answer 1

For (2), note that $$cI_n-A^T = (cI_n)^T-A^T = (cI_n-A)^T,$$ and then that $\det A = \det A^T$ for any $A$.

Answer 2

Hint: $\det(M) = \det(M^T)$ for any square matrix $M$.

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Solution: (after comment exchange below)

Let $M = cI_n - A$.

Note that $M^T = cI_n^T - A^T = cI_n - A^T$.

Then, from the hint, you get $\det(cI_n - A) = \det(cI_n - A^T)$

Answer 3

Hint: It is useful to note that $$ cI_n - A^T = [cI_n - A]^T $$