4
$\begingroup$

Avoiding too many steps, which is the characteristic polynomial of this matrix 7x7? And why?

\begin{pmatrix} 5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\end{pmatrix}

$\endgroup$

marked as duplicate by Marc van Leeuwen linear-algebra Jul 1 '17 at 19:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ Hint: zero is an eigenvalue, so such a polynomial will have no constant term. $\endgroup$ – Sean Roberson Jul 1 '17 at 17:36
  • 1
    $\begingroup$ another hint: $35$ is another eigenvalue, the eigenvector for this eigenvalue is easy to guess. In addition, quite obviously, the matrix has rank $1$ $\endgroup$ – Thomas Jul 1 '17 at 17:38
  • 2
    $\begingroup$ Hint: you need no more hints. ;-) $\endgroup$ – José Carlos Santos Jul 1 '17 at 17:40
  • $\begingroup$ p(t)=t^7, is it correct? $\endgroup$ – Prometeo96 Jul 1 '17 at 17:51
  • 1
    $\begingroup$ @Prometeo96 Right, but I'd rather write that as $t(t-10)$. $\endgroup$ – José Carlos Santos Jul 1 '17 at 18:05
3
$\begingroup$

As it was stated in the commentaries, the rank of this matrix is $1$; so it will have $6$ null eigenvalues, which means the characteristic polynomial will be in the form:

$p(\lambda)=\alpha\,\lambda^6(\lambda-\beta) = \gamma_6\,\lambda^6 +\gamma_7\,\lambda^7$

Using Cayley-Hamilton:

$p(A)=\gamma_6\,A^6+\gamma_7\,A^7 =0$

Any power of this matrix will have the same format, a positive value for all elements.

$B=\begin{bmatrix}1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\end{bmatrix}$

$A = 5\,B$

$A^2 = 5^2\,7\,B$

$...$

$A^6 = 5^6\,7^5\,B$

$A^7=5^7\,7^6\,B$

$p(A) = (\gamma_6+35\,\gamma_7)\,B=0\Rightarrow\gamma_6=-35\gamma_7$

So we have: $\alpha=\gamma_7$ and $\beta = 35$

$p(\lambda)=\alpha\,\lambda^6(\lambda-35)$

$\endgroup$
  • 1
    $\begingroup$ Thanks, but why not assign a value to alfa? $\endgroup$ – Prometeo96 Jul 1 '17 at 18:56
  • 1
    $\begingroup$ About $\alpha$, I am not sure, but I think it will always be $(-1)^n$, so in this case it would be $-1$. Anyway, I think it has not much importance, since we are mostly interested in the roots of the polynomial. $\endgroup$ – Daniel Cunha Jul 1 '17 at 19:17
3
$\begingroup$

It is easy to see, that $v=(1,1,1,1,1,1,1)^T$ is an eigenvector of that matrix. By calculation the corresponding eigenvalue is $35$ (just calculate $Av$).

since the rank of the matrix is $1$ and it has the eigenvalues with their multiplicities as zeros it has to be of the form $p(t) = a t^6 (t-35)$ with $a\neq 0$

$\endgroup$
  • $\begingroup$ Thanks, but why you not assign a value to a? $\endgroup$ – Prometeo96 Jul 1 '17 at 18:55
  • 1
    $\begingroup$ @Prometeo96 If you define the characteristic polynomial as a normalized polynomial then this forces $a=1$. Otherwise it's only determined up to a constant factor. $\endgroup$ – Thomas Jul 1 '17 at 18:57
  • $\begingroup$ Another question: which is the minimum polynomial? How can I calculate it? $\endgroup$ – Prometeo96 Jul 1 '17 at 19:23
  • $\begingroup$ @Prometeo96 consider asking a new question. $\endgroup$ – Thomas Jul 1 '17 at 20:17
2
$\begingroup$

Here is a matrix $P,$ the columns are eigenvectors of your matrix. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal. $$ P = \left( \begin{array}{rrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 \end{array} \right). $$

The columns of $P$ are of varying lengths; lengths $ \sqrt{7}, \sqrt{2}, \sqrt{6}, \sqrt{12},..$ All that is necessary to make an orthogonal matrix $Q$ out of this is to divide each column by its length.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.