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Given an $n\times n$ matrix $C= [c_{ij}]$ which is symmetric (i.e. $c_{ij}=c_{ji}\ \forall i,j$) calculate the determinant of the following matrix (assume $c_{ij} \neq 0\ \forall i,j$):

$$\left(\begin{array}{cccccc} \left(\sum_{i=1}^n\sum_{j=1}^n c_{ij}\right)^2 & \left(\sum_{j=1}^nc_{1j}\right)^2 & \left(\sum_{j=1}^nc_{2j}\right)^2 &\dots & \left(\sum_{j=1}^nc_{nj}\right)^2\\\left(\sum_{j=1}^nc_{1j}\right)^2 & c_{11}^2 & c_{12}^2 & \dots &c_{1n}^2\\ \left(\sum_{j=1}^nc_{2j}\right)^2 & c_{21}^2 & c_{22}^2 & \dots &c_{2n}^2\\ \vdots & \vdots &\vdots & \dots &\vdots \\ \left(\sum_{j=1}^nc_{nj}\right)^2 & c_{n1}^2 & c_{n2}^2 & \dots &c_{nn}^2 \end{array}\right)$$

Remark: For instance when $n=2$ the determinant is equal to $2(c_{11}c_{22}-c_{12}^2)^3$ (note $c_{12}=c_{21}$) which is of a nice form, but it seems hard to generalise this to the $(n+1) \times (n+1) $ case.

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Where is the problem from? –  Davide Giraudo Sep 11 '12 at 9:41
    
It appeared in my research on tight frames of orthogonal polynomials. –  user31899 Sep 11 '12 at 10:31
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up vote 1 down vote accepted

I belive that a approach by block matrix decompositon give a satisfactory answer. Seting

$$ A=\left(\sum_{i=1}^{n}\sum_{i=1}^{n} c_{ij}\right)^2, \quad v^T= \left(\left(\sum_{j=1}^{n}c_{1j}\right)^2, \dots,\left(\sum_{j=1} c_{nj}\right)^2\right) $$ $$ v=\begin{pmatrix} \left(\sum_{j=1}^{n}c_{1j}\right)^2\\ \vdots \\ \left(\sum_{j=1}^{n}c_{nj}\right)^2 \end{pmatrix}, \quad B = \begin{pmatrix} c_{11}^2&\dots&c_{1j}^2&\dots&c_{1n}^2\\ \vdots& &\vdots& &\vdots \\ c_{i1}^2&\dots&c_{ij}^2&\dots&c_{in}^2\\ \vdots& &\vdots& &\vdots \\ c_{n1}^2 &\dots& c_{nj }^2&\dots&c_{nn}^2 \\ \end{pmatrix} $$ We have $(v^TB)=(Bv)^T$ and $v^TBv\in\mathbb{R}$.

Suppose $A$, $D$, $C$, and $B$ are $n\times n$, $n\times m$, $m\times n$-, and $m\times m$ matrices, respectively. Then

$$\det\begin{pmatrix}A& 0\\ C& B\end{pmatrix} = \det\begin{pmatrix}A& D\\ 0& B\end{pmatrix} = \det(A) \det(B) . $$

This can be seen from the Leibniz formula for determinants or by induction on ''n''. When ''A'' is invertible matrix, employing the following identity

$$ \begin{pmatrix}A& D\\ C& B\end{pmatrix} = \begin{pmatrix}A& 0\\ C& I\end{pmatrix} \begin{pmatrix}I& A^{-1} D\\ 0& B - C A^{-1} D\end{pmatrix}$$

leads to

$$\det\begin{pmatrix}A& D\\ C& B\end{pmatrix} = \det(A) \det(B - C A^{-1} D) .$$

When ''B'' is invertible, a similar identity with $\det(B)$ factored out can be derived analogously,These identities were taken from http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/proof003.html that is,

$$\det\begin{pmatrix}A& D\\ C& B\end{pmatrix} = \det(B) \det(A - D B^{-1} C) .$$

When the blocks are square matrices of the same order further formulas hold. For example, if $C$ and $B$ commute (i.e., $CB = BC$), then the following formula comparable to the determinant of a 2-by-2 matrix holds:Proofs are given in J.R. Silvester, Math. Gazette, 10 (2000), pp. 460-467, available at http://www.mth.kcl.ac.uk/~jrs/gazette/blocks.pdf $$ \det\begin{pmatrix}A& D\\ C& B\end{pmatrix} = \det(AB - DC).$$

Them $\det(C)= \det(B)\cdot( A-v^TB^{-1}v)$, whit $v^TB^{-1}v, A\in\mathbb{R}$ or $$ \det(C)= \det(B)\left[ \left(\sum_{i=1}^{n}\sum_{i=1}^{n} c_{ij}\right)^2 - \left(\left(\sum_{j=1}^{n}c_{1j}\right)^2, \dots,\left(\sum_{j=1} c_{nj}\right)^2\right) B^{-1} \begin{pmatrix} \left(\sum_{j=1}^{n}c_{1j}\right)^2\\ \vdots \\ \left(\sum_{j=1}^{n}c_{nj}\right)^2 \end{pmatrix} \right] $$ whit $ \left(\left(\sum_{j=1}^{n}c_{1j}\right)^2, \dots,\left(\sum_{j=1} c_{nj}\right)^2\right) B^{-1} \begin{pmatrix} \left(\sum_{j=1}^{n}c_{1j}\right)^2\\ \vdots \\ \left(\sum_{j=1}^{n}c_{nj}\right)^2 \end{pmatrix} \in\mathbb{R}$.

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