Prove $\det(1+tA)=1+t\cdot tr(A)+O(t^2)$

Question

I need help proving

$\det(1+tA)=1+t\cdot \operatorname{tr}(A)+O(t^2)$

I'm not really sure where to start due to the $(1+tA)$, the $1$ is throwing me off.

Answer 1

Although the most common definition of the determinant is a sum of products of matrix elements, for the computation the definition as $(Tv_1)\wedge\dotsb\wedge(Tv_n)=\det(T)v_1\wedge\dotsb\wedge v_n$ is the most useful.

Putting $T=1+tA$, we use the distributive law and gather like terms to get

$$ (1+tA)v_1\wedge\dotsb\wedge(1+tA)v_n = v_1\wedge\dotsb\wedge v_n\\+t(Av_1\wedge v_2\wedge\dotsb\wedge v_n + v_1\wedge Av_2\wedge\dotsb\wedge v_n +\dotsb+v_1\wedge\dotsb\wedge Av_n)+\\ \dotsb + t^n(Av_1\wedge Av_2\wedge\dotsb\wedge Av_n). $$

If the $v_i$ are a basis, then we have $Av_i=A_i^jv_j,$ and so $$v_1\wedge\dotsb\wedge Av_i\wedge\dotsb\wedge v_n=A_i^i(v_1\wedge\dotsb\wedge v_i\wedge\dotsb\wedge v_n),$$

so our sum becomes

$$ (1+tA)v_1\wedge\dotsb\wedge(1+tA)v_n = v_1\wedge\dotsb\wedge v_n\\+t(\text{tr}(A))+ \dotsb + t^n(Av_1\wedge Av_2\wedge\dotsb\wedge Av_n)\\=(1+\text{tr}(A)t+O(t^2))v_1\wedge\dotsb\wedge v_n. $$

Thus $\det(1+tA)=1+\text{tr}(A)t+O(t^2)$ as was to be proved.

Answer 2

Assume $A$ is $M_{n\times n}(\mathbb{C}).$ Suppose that $A$ is diagonalizable, with eigenvalues $\lambda_1,\dotsc,\lambda_n.$ Then $1+tA$ is also diagonalizable, with eigenvalues $1+t\lambda_i.$ The determinant of a diagonalizable matrix is the product of its eigenvalues, so we have

$$ \det(1+tA)=(1+t\lambda_1)\cdot\dotsb\cdot(1+t\lambda_n)\\=1+t(\lambda_1+\dotsb+\lambda_n)+\dotsb + t^n(\lambda_1\cdot\dotsb\cdot\lambda_n)\\=1+\text{tr}(A)t+\dotsb+t^n\det(A)=1+\text{tr}(A)t + O(t^2). $$

If $A$ is not diagonalizable, observe that the diagonalizable matrices are dense in $M_{n\times n}(\mathbb{C}).$ By continuity, the above equation also holds for $A$.