# Evaluating determinant $\det (\delta_{ij}+a_ia_j)_{ij}$

Given $$a_1, \cdots, a_n \in \mathbb R$$, how can we evaluate this? $$\det \left(\delta_{ij}+a_ia_j\right)_{ij}$$

I think the answer should be $$1+a_1^2+\cdots a_n^2$$ and this question is motivated by computing the volume form of a Riemannian submanifold whose parametrization can be represented by a graph. (See Riemannian volume form on surface of a smooth function. In the last answer of the link, $$\partial f/\partial x_i$$ is just replaced to $$a_i$$ in my question.)

Aside: I am also curious whether there is an another way to find the volume form not by evaluating the above determinant.

• – achille hui Sep 9 at 6:33

The determinant of a matrix is the product of its eigenvalues, and adding $$\delta^i_j$$ shifts all eigenvalues by $$1$$. Therefore, if $$\lambda_1,\dots, \lambda_n$$ are the eigenvalues of the matrix $$(a^ia_j)$$, we obtain $$\det(\delta^i_j + a^ia_j) = \Pi_{k=1}^n (\lambda_k+1)$$. Let $$(v_k)_{k=1}^n$$ be an orthogonal base, and $$v_1 = a$$. Then $$(a^ia_j) v_k = a^ia_jv^j_k e_i= a^ie_i = \delta^1_k |a|^2 a$$, where $$(e_k)_{k=1}^n$$ denotes the standard base. Thus the eigenvalues are $$|a|^2,0,\dots,0$$, which yields $$\det(\delta_{ij} + a_ia_j) = \det(\delta^i_j + a^ia_j) = 1 + |a|^2$$.
The above computation can be interpreted in the context of the volume element of a graph as follows: Let $$\nabla f(x), v_2, \dots, v_n$$ be an orthogonal base of $$T_x\mathbb{R}^n$$. Now, $$v_2,\dots,v_n$$ are orthogonal to $$\nabla f$$, hence their pushforwards on the graph are just parallel transports of themselves, with the same length. The first vector, $$\nabla f(x)$$, transforms to $$(\nabla f(x), |\nabla f(x)|^2)^T$$, thus its stretched by a factor of $$1+|\nabla f|^2$$. Therefore, the volume element is stretched by a total factor of $$1+|\nabla f|^2$$.