Prove that $\det(I+tA)=1$ $\forall t \in\mathbb{R} \iff A=0$

Question

I tried to convert it to $\det(A- \lambda I)$ but cannot proceed.

$I$ and $A$ are both n-by-n matrix.

My idea: Let $\lambda_i$ be eigenvalues of A. Then $tA$ has eigenvalues of $t\lambda_i$. $I+tA$ has eigenvalues of $1+t\lambda_i$. Now $\Pi$$(1+t\lambda_i)=1$ $∀t$.

What's the next step?

Answer 1

Note that if $t \neq 0$,

$$ 1 = \det(I + tA) = \det((-t)(-\frac{1}{t}I-A)) = (-t)^n\chi_A(-\frac{1}{t}) $$

dividing by $(-t)^n$, we get that

$$ \left(-\frac{1}{t}\right)^n = \chi_A(-\frac{1}{t}) $$

Since $t \mapsto -\frac{1}{t}$ is a bijection in $\mathbb{R} \setminus \{0\}$, this implies

$$ t^n = \chi_A(t) \quad (\forall t \neq 0) $$

Since this is an equality of two real valued polynomials in one variable in infinitely many points, they must be the same polynomial:

$$ \chi_A(t) \equiv t^n $$

By Cayley-Hamilton, $A^n = \chi_A(A) = 0$ so $A$ is nilpotent. Every step is reversible, so the original statement is false. What it does hold, is that $A$ is nilpotent, which is equivalent to the original condition. However, there are (plenty of) non zero nilpotent matrices.

Answer 2

If $A=0_{n\times n}$ is immediate that $\det(I_{n\times n}-tA)=1$ for all $t\in \mathbb{R}$. Suppose now that $\det(I_{n\times n}-tA)=1$. There are several ways to deal it. And many of them may be familiar to your background or not. The way I find it easiest is to use the multilinearity of the determinant. It is well known that the determinat function depends linearly on the columns of the matrix. Let $ A_1, A_2, \ldots, A_n $ be the columns of the matrix $$ A=\begin{pmatrix}A_{11}&A_{12}&\cdots & A_{1n}\\A_{21}&A_{22}&\cdots & A_{2n}\\ \vdots& \vdots& \ddots&\vdots \\A_{n1}&A_{n2}&\cdots & A_{nn}\end{pmatrix} $$ that is, $$ A_1=\begin{pmatrix}A_{11}\\\vdots\\ A_{n1}\end{pmatrix}, A_2=\begin{pmatrix}A_{12}\\\vdots\\ A_{n2}\end{pmatrix}, \cdots, A_1=\begin{pmatrix}A_{11}\\\vdots\\ A_{n1}\end{pmatrix} $$ Let $ e_1, e_2, \ldots, e_n $ be the columns of the matrix $$ I_{n\times n}=\begin{pmatrix}1& 0&\cdots &0\\0&1&\cdots & 0\\ \vdots& \vdots& \ddots&\vdots \\0&0&\cdots & 1\end{pmatrix} $$ that is, $$ e_1=\begin{pmatrix}1\\0\\\vdots\\ 0\end{pmatrix}, e_2=\begin{pmatrix}0\\ 1\\\vdots\\ 0\end{pmatrix} , \cdots, e_n=\begin{pmatrix}0\\ 0\\\vdots\\ 1\end{pmatrix} $$ We have \begin{align} \det(I_{n\times n}+tA) =& \det[A_{1}+te_1,A_{2}+te_2,\ldots, A_{n}+te_n] \\ =& \det[A_1,A_2,\ldots,A_j,\ldots, A_n] \\ +& t\sum_{j_1=1}^{n}\det[A_1,\ldots,A_{j_1-1},e_{j_1},A_{j_1+1},\ldots, A_n] \\ +& t^2\sum_{1\leq j_1<j_2\leq n}\det[A_1,\ldots,A_{j_1-1},e_{j_1},A_{j_1+1},\ldots,A_{j_2-1},e_{j_2},A_{j_2+1},\ldots, A_n] \\ +& t^3\sum_{1\leq j_1<j_2<j_2\leq n}\det[A_1,\ldots,A_{j_1-1},e_{j_1},A_{j_1+1},\ldots,A_{j_2-1},e_{j_2},A_{j_2+1},\ldots,A_{j_3-1},e_{j_3},A_{j_3+1}\ldots, A_n] \\ \vdots \\ \\ \vdots \\ +& t^{n}\det[e_1,\ldots,e_n] \end{align} Since the identity holds for any $ t \in \mathbb{R} $, making $ t = 0 $ we have $$ \det(A)=0. $$ On the other hand, all derivatives of $ \det (I + tA)$ are zero for all $t\in\mathbb{R}$. Making $t=0$ in $D_t^{(1)}\det (I + tA)=0$ we have $$ \sum_{j_1=1}^{n}\det[A_1,\ldots,A_{j_1-1},e_{j_1},A_{j_1+1},\ldots, A_n]=0 $$ Making $t=0$ in $D_t^{(2)}\det (I + tA)=0$ we have

$$ \sum_{1\leq j_1<j_2\leq n}\det[A_1,\ldots,A_{j_1-1},e_{j_1},A_{j_1+1},\ldots,A_{j_2-1},e_{j_2},A_{j_2+1},\ldots, A_n]=0 $$ Making this n-1 procedure times we conclude that all coefficients (except the coefficient of $t^n$) of the polynomial $\det(I+tA)=1$ are zero. Now this implies that $\det(I+tA)= t^n $ is the characteristic polynomial of the matrix $ A $.We can conclude that the $ A $ matrix is nilpotent. But as pointed out by Guido A. this is the maximum we can conclude from A. Search for an example of a nonzero nilpotent matrix that satisfies $\det(I+tA)=1$ for all $t\in\mathbb{R}$.