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$M_{n\times k}$ is defined as a matrix whose all elements are '-1'.

The following block matrix is as such:

$A =\begin{bmatrix}m\cdot I_{n-1} & M_{n-1\times m}\\M_{m\times n-1} & n\cdot I_{m}\end{bmatrix}$

prove the following: $det A = n^{m-1}\cdot m^{n-1}$

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  • $\begingroup$ You could perhaps start by multiplying on the left by the matrix $[I, 0 \\ M/m, I]$ $\endgroup$ – Gribouillis Dec 24 '16 at 13:10
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Let more generally let $C(\gamma) = \gamma 1_{n-1}1_{m}^T $ denote the ${n-1}\times m$ matrix with each element equal to $\gamma$ (here $1_k$ denotes the k-dimensional column vector of all ones) and let $A(\gamma) = \begin{pmatrix} mI_{n-1} & C(\gamma) \\ C(\gamma)^T & nI_m\end{pmatrix}$. Your problem is to compute the determinant of $A(-1)$.

Using the following identity $$ \begin{pmatrix} I_{n-1} & 0 \\ - \dfrac{C(\gamma)^T}{m} & I_m \end{pmatrix} \begin{pmatrix} mI_{n-1} & C(\gamma) \\ C(\gamma)^T & nI_m\end{pmatrix} = \begin{pmatrix} mI_{n-1} & C(\gamma)\\ 0 & nI_m - \dfrac{C(\gamma)^TC(\gamma)}{m} \end{pmatrix}.$$

we have on taking determinants $$ \begin{align} \operatorname{det}(A(\gamma)) &=\operatorname{det}(mI_{n-1})\operatorname{det}(nI_m - \dfrac{C(\gamma)^TC(\gamma)}{m})\\ &= m^{n-1} \times n^{m} \det(I_m - \dfrac{C(\gamma)^TC(\gamma)}{mn})\\ &= m^{n-1}n^{m} \det(I_{m} - \gamma^2 \dfrac{(n-1)}{nm} 1_{m}1_{m}^{T}) \tag{*} \\ &= m^{n-1} n^m \det(I_m - \delta 1_m 1_m^T) \end{align} $$ where $\delta = \gamma^2\dfrac{n-1}{nm}.$

In (*) we have used $C(\gamma) = \gamma 1_{n-1}1_{m}^T$.

Consider the following identities: $$ \begin{pmatrix} I_m & 0 \\ -1_m^T & 1 \end{pmatrix} \begin{pmatrix} I_m & \delta 1_m \\ 1_m^T & 1 \end{pmatrix} = \begin{pmatrix} I_m & \delta 1_m \\ 0 & 1 - m \delta \end{pmatrix},$$ $$\begin{pmatrix} I_m & -\delta 1_m \\ 0 & 1 \end{pmatrix} \begin{pmatrix} I_m & \delta 1_m \\ 1_m^T & 1 \end{pmatrix} = \begin{pmatrix} I_m - \delta 1_m 1_m^T & 0 \\ 1_m^T & 1 \end{pmatrix}. $$ On taking determinants we have $\det\begin{pmatrix} I_m - \delta 1_m 1_m^T \end{pmatrix} = 1 - m \delta.$

This gives $\det(A(\gamma))= m^{n-1}n^m(1 - m \delta) = m^{n-1}n^m (1 - \gamma^2\dfrac{n-1}{n})$ so $ \det(A(-1))=m^{n-1}n^{m-1}.$

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  • $\begingroup$ I will fix it. Thanks. $\endgroup$ – Arin Chaudhuri Dec 24 '16 at 16:17
  • $\begingroup$ @ArinChaudhuri Can you elaborate on this transition? \begin{align} &m^{n-1} \times n^{m} \det(I_m - \dfrac{C(\gamma)^TC(\gamma)}{mn})\\ &= m^{n-1}n^{m} \det(I_{m} - \gamma^2 \dfrac{(n-1)}{nm} 1_{m}1_{m}^{T}) \\ \end{align} $\endgroup$ – Konstantin Sartinin Dec 24 '16 at 16:31
  • $\begingroup$ Just plug in $C(\gamma) = \gamma 1_{n-1}1_m^T$ so then we have $C(\gamma)^TC(\gamma) = \gamma^2 1_{m}1_{n-1}^T1_{n-1}1_m^T = \gamma^2 (n-1) 1_m 1_m^T$, note $1_k^T 1_k = k$. $\endgroup$ – Arin Chaudhuri Dec 24 '16 at 16:35

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