I wish to show that $$ \det \begin{pmatrix} x & a & a & a\\ a & x & a & a\\ a & a & x & a\\ a & a & a & x \end{pmatrix}=(x-a)^3(x+3a).$$ Obviously, I could expand it out and try to notice some factors as I go, but that seems rather tedious. Is there a particularly fast way of demonstrating the above identity?
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2Observe that $x-a$ is an eigenvalue of multiplicity $3$ (or higher), and use the trace to show that the remaining eigenvalue is $x+3a$. – Erick Wong Jun 20 '15 at 05:26
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@ErickWong, awesome! It works for a general dimension $n$, right? As in the eigenvalues are $x-a$ of multiplicity of $n-1$, and $x+(n-1)a$. – Indominus Jun 20 '15 at 05:31
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See Determinant of a specially structured matrix and other posts linked there. – Martin Sleziak Jun 20 '15 at 07:38
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$$ \det \begin{pmatrix} x & a & a & a\\ a & x & a & a\\ a & a & x & a\\ a & a & a & x \end{pmatrix}~\quad R_1\to R_1-R_2~\&~ R_2\to R_2-R_3 ~\&~ R_3\to R_3-R_4\\$$ $$==- \det \begin{pmatrix} x-a & a-x & 0 & 0\\ 0 & x-a & a-x & 0\\ 0 & 0 & x-a & a-x\\ a & a & a & x \end{pmatrix}$$ $$=-(x-a)^3\det \begin{pmatrix} 1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a & a & a & x \end{pmatrix}$$
Now if you expand this then you'll get answer