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I am considering block matrices $$\begin{pmatrix} A & v \\ v^T & x \end{pmatrix}$$ with $A \in \mathbb{R}^{(n-1) \times (n-1)}$, $v \in \mathbb{R}^{n-1}$, $x \in \mathbb{R}.$ Is there a rational $(n-1) \times (n-1)$ expression $p(A,v,x)$ I can form in the variables $A,v,x$ such that $$\mathrm{det}\begin{pmatrix} A & v \\ v^T & x \end{pmatrix} = \mathrm{det}(p(A,v,x))\; ?$$

The first thing I tried is the block matrix formula $$\mathrm{det}\begin{pmatrix} A & v \\ v^T & x \end{pmatrix} = x \, \mathrm{det}\Big(A - \frac{1}{x}vv^T \Big).$$ However writing this as just an $(n-1) \times (n-1)$ determinant introduces strange exponents: $$\mathrm{det}\Big( x^{1/(n-1)} A - x^{(2-n)/(n-1)} vv^T \Big)$$ which is obviously not polynomial unless $n = 2$. Since the result does not involve noninteger powers of $x$ I am hoping there is some expression $p$ that also only involves integer powers of $x$.

user399601
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3 Answers3

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In "Algebra" by J.W.Archbold, Pitman, 4e, 1970, s24.16, p.417, it proves,

$\mathrm{det}\begin{pmatrix} A & u^T \\ v & w \end{pmatrix} = -v.adj(A).u^T + w .det(A)$

where $u,v$ are $1$ x $n$ and $A$ is $n$ x $n$ and $w$ is $1$ x $1$.

gjh
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    Your formula can be written $det(A)(w-vA^{-1}u^T)$ (because $adj(A):=det(A).A^{-1}$). – Jean Marie Dec 22 '16 at 09:37
  • Nice. The proof for invertible $A$ follows from @JeanMarie 's comment. And for general $A$ one can use the fact that one can find a sequence of invertible matrices $A_n$ converging to $A$ and both LHS and RHS are continuous functions of $A$. – Arin Chaudhuri Dec 22 '16 at 10:18
  • @JeanMarie: As you point out, the Archbold form can be rewritten by eliminating the adj(A) in favour of det(A) and inv(A). I left it unchanged partly because it was a quote but mainly because it seemed the questioner wanted a rational expression for the determinant of the block matrix and the Archbold form provides this immediately. – gjh Dec 23 '16 at 04:55
  • @gjh You are right. – Jean Marie Dec 23 '16 at 06:54
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For blok matrices $$ \left[ \begin{array}{c|c} A_{11}& A_{12} \\ \hline A_{21} & A_{22} \end{array} \right] $$ if $A_{11}$ is invertible, we have the general formula ( you can see here or here): $$ \det \left[ \begin{array}{c|c} A_{11}& A_{12} \\ \hline A_{21} & A_{22} \end{array} \right]= \det A_{11}\cdot \det\left( A_{22}-A_{21}A_{11}^{-1}A_{12} \right) $$ that, in your case becomes: $$ \det \left[ \begin{array}{c|c} A& v \\ \hline v^T & x \end{array} \right]= \det A\cdot \det\left(x-v^TA^{-1}v \right)=\left(x-v^TA^{-1}v \right)\det A $$

Emilio Novati
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Using multilinearity one can rewrite $\text{det}\begin{pmatrix} A & v \\ v^T &x \end{pmatrix} = \text{det}\begin{pmatrix} A & 0 \\ v^T &x \end{pmatrix} + \text{det}\begin{pmatrix} A & v \\ v^T & 0 \end{pmatrix} = x \text{det}(A)+\text{det}\begin{pmatrix} A & v \\ v^T & 0 \end{pmatrix}.$

Let $v^T = \begin{pmatrix} v_1 & v_2 & \dots & v_n \end{pmatrix}$ so $v = \sum_{i=1}^{n} v_i e_i$ where $e_i$ is the $i^{\text{th}}$ canonical basis vector consisting of $1$ at the $i^{\text{th}}$ coordinate and $0$ elsewhere.

Using multilinearity of the determinant one sees $\text{det}\begin{pmatrix} A & v \\ v^T & 0 \end{pmatrix} = \text{det}\begin{pmatrix} A & \sum_{i=1}^{n} v_i e_i \\ \sum_{i=1}^{n} v_i e_i^T & 0 \end{pmatrix} = \sum_{i=1}^{n}\sum_{j=1}^{n}v_i v_j \operatorname{det}\begin{pmatrix} A & e_i \\ e_j^T & 0 \end{pmatrix}.$

Expanding $\operatorname{det}\begin{pmatrix} A & e_i \\ e_j^T & 0 \end{pmatrix}$ along the last row and then the last column it is easy to see $\operatorname{det}\begin{pmatrix} A & e_i \\ e_j^T & 0 \end{pmatrix} = (-1)^{n+1+i}\operatorname{det}\begin{pmatrix} A_{(-i)} \\ e_j^T \end{pmatrix} = {(-1)}^{n+1+i}(-1)^{n+j}M_{ij}=-C_{ij}$ where $A_{(-i)}$ is the submatrix of $A$ with the $i$th row deleted, $M_{ij}$ is the determinant of the submatrix of $A$ obtained by deleting the $i$th row and $j$th column and $C_{ij}$ is the $i,j$ th cofactor of $A$. Let $C = (C_{ij}) = \operatorname{adj}(A)^T$ then $\operatorname{det}\begin{pmatrix} A & v \\ v^T &x \end{pmatrix} = x \operatorname{det}(A) - v^T C v = x \operatorname{det}(A) - v^T C^T v = x \operatorname{det}(A) - v^T \operatorname{adj}(A) v.$

The above formula is obviously a polynomial in the entries of $A,v$ and $x$.

Arin Chaudhuri
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  • Archbold provides a simple proof that consists basically of (step 1) expand the block determinant by the last row and (step 2) expand the resulting minors, other than det(A), by the last column. – gjh Dec 23 '16 at 04:57