1

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

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.

Rodrigo de Azevedo
  • 1
BlackAdder
  • 19

2 Answers2

2

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 $\{0,4\}$.

Rodrigo de Azevedo
  • 1
Devendra Singh Rana
  • 1,780
  • 1
  • 15
  • 32
  • How did you find the eigenvalues? – Zanzi Jan 17 '18 at 16:23
  • @Zanzi Actually we have a result which says about any n x n rank 1 matrix I just applied that. – Devendra Singh Rana Jan 17 '18 at 16:26
  • 1
    @Zanzi The eigenvalues can be found by inspection. $A$’s column space is obviously one-dimensional, so that immediately gives you three of the eigenvalues, and you can always find the last eigenvalue via the trace. – amd Jan 17 '18 at 21:01
0

Using the Weinstein-Aronszajn determinant identity,

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

Rodrigo de Azevedo
  • 1