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If the matrix $A$ of size $6\times6$ satisfies equation $A^2 + 12I = 7A$ and the rank of $A-3I$ is $1$, then what is the rank of $A-4I$?

I understand that if $A-3I = B$ and $\ B$ has the rank $1$ then all the rows of $B$ except one has to be $0$. Then, let's assume, first row of $A$, looks like this: $[(a_{11}+3) ...... a_{16}]$. The whole matrix:

$$A = \begin{pmatrix} a_{11}+3 & a_{12} & a_{13} & a_{14} & a_{15} & a_{16}\\ 0 & 3 & 0 & 0 & 0 & 0\\ 0 & 0 & 3 & 0 & 0 & 0\\ 0 & 0 & 0 & 3 & 0 & 0\\ 0 & 0 & 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 0 & 0 & 3\end{pmatrix}$$

I believe that the rank of $A-4I$ has to be at least $5$. But it can be $6$ as well if the $a_{11}-1$ is different than zero.

So I use the equation $A^2 + 12E = 7A$ to get:

  • $A^2 + 12I - 7A = 0$

But we are looking for the first element of the matrix only, so:

  • $(a_{11}+3)^2 + 12 - 7 (a_{11}+3)= 0$

And so I got:

  • $a_{11}(a_{11}-1)=0$

Is my thought process correct? Will I end up with two situations or is something wrong?

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  • $\begingroup$ "and $B$ has the rank $1$ then all the rows of $B$ except one has to be $0$" This is a false assumption. $\begin{bmatrix}1&2&3\\2&4&6\\3&6&9\end{bmatrix}$ has rank $1$ and does not have any rows of all zero. $\endgroup$
    – JMoravitz
    Jan 26 '18 at 7:13
  • $\begingroup$ Can you factorise $x^2-7x+12$? $\endgroup$ Jan 26 '18 at 7:14
  • $\begingroup$ For clarification, what is the matrix $E$? Is this perhaps a notation being used for the identity matrix here? $\endgroup$
    – JMoravitz
    Jan 26 '18 at 7:15
  • $\begingroup$ Yes, its identity matrix. $\endgroup$
    – Patryk
    Jan 26 '18 at 7:29
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Well it looks fine to me, you can also try it by another way. As the given matrix is Upper triangular , so the diagonal entries are actually its Eigenvalues . Clearly 5 of it's Eigenvalues are 3 that means everything depends on the first entry or we can say $a_{11}$, if it is 1 then the rank of $A-4I$ will be 5 and for any other value rank will surely be 6.

It's as simple as that. Hope it works.

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