Wednesday, September 5, 2018

linear algebra - Determinant of a special $ntimes n$ matrix




Compute the determinant of the nun matrix:
$$
\begin{pmatrix}
2 & 1 & \ldots & 1 \\

1 & 2 & \ldots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\ldots & 2
\end{pmatrix}
$$



For $n=2$, I have$$
\begin{pmatrix}
2 & 1 \\
1 & 2

\end{pmatrix}
$$



Then $det = 3$.



For $n=3$, we have
$$
\begin{pmatrix}
2 & 1 & 1\\
1 & 2 & 1\\

1 & 1 & 2 \\
\end{pmatrix}
$$



Then $det = 4$.



For $n=4$ again we have



$$
\begin{pmatrix}

2 & 1 & 1 & 1 \\
1 & 2 & 1 & 1\\
1 & 1 & 2 & 1\\
1 & 1 & 1 & 2
\end{pmatrix}
$$
Then $det = 5$



How can I prove that the determinant of nun matrix is $n+1$.


Answer




A standard result (http://en.wikipedia.org/wiki/Matrix_determinant_lemma) is $\det(I+AB) = \det(I+BA)$.



Since the matrix above can be written as $I+ e e^T$, where $e$ is a vector of ones, we have $\det(I+ e e^T) = \det(1+ e^T e) = 1+e^Te = n+1$.


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