How to determine the characteristic polynomial of the $4\times4$ real matrix of ones?

Question

$$\left[\begin{array}{l}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{array}\right]$$

I am having difficulties with calculating this. Currently I'm stuck at finding the determinant of the following matrix:

$$\left[\begin{array}{l}\lambda-1&-1&-1&-1\\-1&\lambda-1&-1&-1\\-1&-1&\lambda-1&-1\\-1&-1&-1&\lambda-1\end{array}\right]$$

Is there a "smart" way/trick to get the result or should I just crunch the numbers through Gaussian elimination? Normally it is not that difficult to do this, but if all the elements of the matrix are the same, it somehow got me confused.

Answer 1

Consider any $n \times n$ rank-$1$ matrix $A$ as follows: one of its eigenvalues is its trace and the remaining eigenvalues are zero. Hence, the characteristic polynomial is $$x^{n-1}(x-\mbox{Tr}(A))$$ and the spectrum is $\{0,4\}$.

Answer 2

Using the Weinstein-Aronszajn determinant identity,

$$\begin{array}{rl} \det (s \mathrm I_n - 1_n 1_n^\top) &= \det \left( s \cdot \left( \mathrm I_n - s^{-1}1_n 1_n^\top \right) \right)\\ &= s^n \cdot \det \left( \mathrm I_n - s^{-1}1_n 1_n^\top \right)\\ &= s^n \cdot \left( 1 - n \, s^{-1} \right)\\ &= s^{n-1} \left( s-n \right)\end{array}$$

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linear-algebra matrices determinant characteristic-polynomial