Consider $J$ to be an $n\times n$ matrix whose entries are all $1s$ .

If $P$ is an $n\times n$ matrix such that

$P=$ \begin{bmatrix} v_1| v_2|,\ldots |v_{n-1} |v_n\end{bmatrix}

where the columns $v_i=e_i-e_n$ for $1\le i\le n-1$ and $v_n=\sum e_i$ where $e_i$ is the ith column of the Identity Matrix

Note that $v_i,1\le i\le i-1$ are the eigen vectors corresponding to $0$ of $J$ and $v_n$ is an eigen vector corresponding to $n$ of $J$

Find $\det P$.

It is very difficult to expand by Laplace Expansion

Is there any efficient way to to find the determinant?

  • 1
    $\begingroup$ How about doing it by induction and expanding with respect to the first line? $\endgroup$
    – Mindlack
    Dec 17 '18 at 8:56
  • 1
    $\begingroup$ What does the matrix $J$ have to do with anything? $\endgroup$
    – BigbearZzz
    Dec 17 '18 at 9:04
  • $\begingroup$ @BigbearZzz;Sorry forgot to write that $\endgroup$
    – user596656
    Dec 17 '18 at 9:07


Add each row above to the last row, then expand by the last row.


Use row operations to simplify the matrix in an upper triangular form. The form of $P$ is, $$P=\begin{pmatrix} 1 &0&0& ...&1 \\ 0&1&0&...&1\\ \vdots\\ -1&-1&-1&...&1 \end{pmatrix},$$ So carry out the row operation (which doesn't change the determinant) $R_n=R_n+R_k$ for $1 \leq k \leq n-1$. The resulting matrix is of the form $$P'=\begin{pmatrix} 1 &0&0& ...&1 \\ 0&1&0&...&1\\ \vdots\\ 0&0&0&...&1+(n-1) \end{pmatrix},$$ so $\det P=\det P'=1 \cdot 1 \cdot...\cdot1 \cdot n=n$.

  • $\begingroup$ It will be $n$ in the last entry of the matrix ,not $n+1$ $\endgroup$
    – user596656
    Dec 17 '18 at 9:23
  • $\begingroup$ That's what I wrote $\endgroup$
    – clm
    Dec 17 '18 at 9:29

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