# $\det(I+A) = 1 + tr(A) + \det(A)$ for $n=2$ and for $n>2$?

Let $I$ the identity matrix and $A$ another general square matrix. In the case $n=2$ one can easily verifies that $$\det(I+A) = 1 + tr(A) + \det(A)$$ or $$\det(I+tA) = 1 + t\ tr(A) + t^2\det(A)$$ for some scalar $t \in \mathbb{R}$.

I have tried to see if there exists a similar formula for $n>3$. This is a natural question. But the calculations are very big and difficulty to see. Then I do the answer. Is there a similar formula for $n>2$?

• Are you looking for this formula : $\det(I+tA) = \sum_{k=0}^n t^k \operatorname{Tr}(\Lambda^{k} A)$ ? (cf here). – user10676 Aug 1 '13 at 13:40
• $\frac{d}{dt}|_{t=0} \det(I+tA)=Tr(A)$ so perhaps we can think of this as the integral form. – Squirtle Aug 1 '13 at 13:41

$$\left| \begin{array}{ccc} 1+a_{11} & a_{12} & a_{13} \\ a_{21} & 1+a_{22} & a_{23} \\ a_{31} & a_{32} & 1+a_{33} \\ \end{array} \right| = \left| \begin{array}{ccc} 1 & a_{12} & a_{13} \\ 0 & 1+a_{22} & a_{23} \\ 0 & a_{32} & 1+a_{33} \\ \end{array} \right| + \left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & 1+a_{22} & a_{23} \\ a_{31} & a_{32} & 1+a_{33} \\ \end{array} \right|=$$ $$= \left| \begin{array}{cc} 1+a_{22} & a_{23} \\ a_{32} & 1+a_{33} \\ \end{array} \right| + \left| \begin{array}{ccc} a_{11} & 0 & a_{13} \\ a_{21} & 1 & a_{23} \\ a_{31} & 0 & 1+a_{33} \\ \end{array} \right| + \left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & 1+a_{33} \\ \end{array} \right| =\ldots$$ $$= 1+{\rm tr}A + M_{11} + M_{22} + M_{33}+\det A$$ and further by induction.

If $a_1,a_2,\ldots,a_n$ are the eigenvalues of $A$ then the characteristic polynomial of $A$ is $$\det(tI-A)=(t-a_1)(t-a_2)\cdots(t-a_n).$$ Therefore

• $\det(I+A)=(1+a_1)(1+a_2)\cdots(1+a_n) = 1 + \operatorname{tr}A+ \ldots + \det A$

and

• $\det(I+tA)=(1+ta_1)(1+ta_2)\cdots(1+ta_n)= 1 + t\operatorname{tr}A+ \ldots + t^n\det A$.

One can formally expand $\det(I+tA)$ as a power series of $t$ and get:

\begin{align} & \det(I + tA)\\ = & \exp\left(\text{Tr}\log(I+tA)\right)\\ = & \exp\left(t\,\text{Tr}A - \frac{t^2}{2}\text{Tr}A^2 + \frac{t^3}{3}\text{Tr}A^3 + \cdots \right)\\ = & 1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right) + \frac{t^3}{3!}\left((\text{Tr}A)^3 - 3 (\text{Tr}A)(\text{Tr}A^2) + 2 \text{Tr}A^3 \right) + \cdots \end{align}

When $A$ is a $n \times n$ matrix, the above expansion terminate at the $t^n$ term with coefficient equal to $\det A$. With this, you can obtain formula similar to what you have for $n = 2$:

$$\det(I+tA) = \begin{cases} 1 + t\,\text{Tr}A + t^2 \det(A) & n = 2\\ \\ 1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right) + t^3 \det(A) & n = 3\\ \\ 1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right) \\\;\;\;+ \frac{t^3}{3!}\left((\text{Tr}A)^3 - 3 (\text{Tr}A)(\text{Tr}A^2) + 2 \text{Tr}A^3 \right) + t^4 \det(A) & n = 4\\ \end{cases}$$

• how do you obtain $\text{det(I+tA)}=\text{exp(Tr log(I+tA))}?$ – vidyarthi May 3 '17 at 9:33
• @vidyarthi any matrix $A$ with complex coefficients is similar to one in Jordan normal form. It is easy to verify the expression when $A$ is in jordon normal form. Since both $\exp$ and $\log$ is compatible with similarity transform ( i.e $\exp(SAS^{-1}) = S\exp(A)S^{-1}$, $\log(SAS^{-1}) = S\log(A)S^{-1}$ ), the expression works generally. – achille hui May 3 '17 at 11:10

it is like the development of the characteristic polynomial (each term of the polynomial is a combination of symmetric sub-determinants of the initial matrix), I think you can easily find a proof of this fact on the internet. Or you can even prove it by yourself, using the multi-linearity of the determinant, and the fact that it is alternate.

• This is not a answer, this is a comment. – user29999 Aug 1 '13 at 15:57