Saturday, October 20, 2018

linear algebra - Characteristic polynomial of a matrix 7x7?




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}


Answer



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 $\beta = 35$



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


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