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Let $a$ be a real number. What is the number of distinct real roots of the following

$$\left| \begin{array}{ccc} x & a & a & a \\ a & x & a & a \\ a & a & x & a \\ a & a & a & x \end{array} \right|=0.$$

I guess some method is there other than calculating the determinant explicitly.

  • @Amzoti If you read the solution, it has the eigenvalues as well. Also, given the eigenvalues, you can find the determinant trivially. –  May 06 '13 at 02:06
  • @user17762 but wouldn't determining the eigenvalues and eigenvectors of A involve the same time-consuming process of calculating the determinant. – Johnathon Svenkat May 06 '13 at 02:07
  • @JohnathonSvenkat There is no time-consuming process. All the things eigenvalues, determinant, eigenvectors etc are trivial in this case, if you read the duplicate post. –  May 06 '13 at 02:08
  • See also here: http://math.stackexchange.com/questions/86644/determinant-of-a-specially-structured-matrix and http://math.stackexchange.com/questions/84206/how-to-calculate-the-following-determinants-all-ones-minus-i/ – Martin Sleziak Jan 28 '14 at 09:22

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I assume $a\neq 0$, for otherwise the question is trivial.

Note this is the characteristic polynomial $$ p_A(x)=\det(xI_4-A) $$ of the real symmetric matrix $$A=\pmatrix{0&-a&-a&-a\\ -a&0&-a&-a\\-a&-a&0&-a\\-a&-a&-a&0}.$$ We know such a matrix is diagonalizable via an orthogonal matrix. In particular, its spectrum is all real and its characteristic polynomial splits over $\mathbb{R}$.

For $x=a$, we get a rank one matrix $aI_4-A$. This means that $a$ is an eigenvalue and that the corresponding eigenspace has dimension $3$, which is also the multiplicity of $a$ in the characteristic polynomial.

There only remains one eigenvalue $\lambda$ of multiplicity $1$ to find. It suffices to take the trace: $$\mbox{Tr} A=0=3a+\lambda \quad\Rightarrow\quad\lambda=-3a.$$

So $$ p_A(x)=(x-a)^3(x+3a). $$

Julien
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