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?


closed as off-topic by Brian Borchers, José Carlos Santos, Adrian Keister, Theoretical Economist, Namaste Sep 7 '18 at 0:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, José Carlos Santos, Adrian Keister, Theoretical Economist, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ @Thinking See the title of the question. $\endgroup$ – José Carlos Santos Sep 6 '18 at 23:49
  • 1
    $\begingroup$ Even if you show all eigenvalues are zero, that will not show that $A=0$. $\endgroup$ – GEdgar Sep 7 '18 at 0:01
  • 2
    $\begingroup$ When $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, then it is easy to check $\det(I+tA)=1$ for all $t$, so $\implies$ implication is wrong. $\endgroup$ – Mike Earnest Sep 7 '18 at 0:09
  • $\begingroup$ Perhaps you missed some restriction on $A$ ? $\endgroup$ – leonbloy Sep 7 '18 at 0:12
  • $\begingroup$ I'd like to close it because the statement is wrong. See Guido A.'s answer. $\endgroup$ – Zhenduo Cao Sep 7 '18 at 1:15

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.

  • $\begingroup$ Thanks! That helps a lot. $\endgroup$ – Zhenduo Cao Sep 7 '18 at 0:09
  • $\begingroup$ Glad I could help :) $\endgroup$ – Guido A. Sep 7 '18 at 0:10

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 $t\in\mathbb{R}$.

  • $\begingroup$ Thanks. That's a useful expansion of |I+tA|. $\endgroup$ – Zhenduo Cao Sep 7 '18 at 2:17

Not the answer you're looking for? Browse other questions tagged or ask your own question.