# Determinant of a matrix of the form $M = I + xx^t$

$$M=\begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \\ x_2x_1 & 1+x_2^2 &...&x_2x_n \\...&...& &...& \\x_nx_1 & x_nx_2& ...&1+x_n^2&\end{pmatrix}.$$

So I noticed that $$M$$ is a symmetric matrix and $$M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$$. That's all I can get.

• Can you determine the characteristic polynomial of $M-I$? Jan 22, 2019 at 20:24
• Could you give me some more clue about the polynomial? Jan 22, 2019 at 20:30

You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).

For a column vector $$x = (x_1,...,x_n)$$, and an $$n\times n$$ identity matrix $$I$$, we have

$$M = I + x x^T$$

hence, in view of the aforementioned lemma,

$$\det M = (1 + x^T I x)\det I = 1 + \|x\|^2 = 1 + x_1^2 + \cdots +x_n^2.$$

Consider the matrix $$N=M-I.$$ It has rank $$\le1$$, so its characteristic polynomial $$P(\lambda)=\det(\lambda I-N)$$ has the form $$\lambda^n-a\lambda^{n-1}$$. But $$a$$ is the trace of $$N$$, so $$P(\lambda)=\lambda^n-(x_1^2+\cdots+x_n^2)\lambda^{n-1}.$$ Now $$\det M=\det(I+N)=(-1)^nP(-1)=1+x_1^2+\cdots+x_n^2.$$

• Sorry,but why rank $\leq 1$ means the ch. polynomial would be of the above form? Can you please elaborate a little? Jan 29, 2020 at 13:39

Denote by $$M_n$$ the matrix we want to compute the determinant. Letting $$\mathbf{x}=(x_1,\dots,x_{n-1})$$, we express $$M_n$$ as a block matrix: $$M_n=\pmatrix{M_{n-1}& x_n\mathbf x^t\\ x_n\mathbf{x}&1+x_n^2}.$$ Since $$\pmatrix{ x_n\mathbf x^t\\1+x_n^2}=\pmatrix{ 0\\1}+\pmatrix{ \mathbf x_nx^t\\x_n^2}$$, it follows by multi-linearity of the determinant that $$\det\left(M_n\right)=\det\pmatrix{M_{n-1}& 0\\ x_n\mathbf{x}&1}+\det\pmatrix{M_{n-1}& x_n\mathbf x^t\\ x_n\mathbf{x}&x_n^2}.$$ The first term of the right hand side is $$\det\left(M_{n-1}\right)$$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that $$\det\left(M_{n}\right)=\det\left(M_{n-1}\right)+x_n^2\det\pmatrix{M_{n-1}& \mathbf x^t\\ \mathbf{x}&1}.$$ The last determinant is one: one can make the linear combinations $$C_i\leftarrow x_iC_n$$ to see this.