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?
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.