Rank of $ A^2 +A + I$ Let $A$ be $6 \times 6$ real symmetric matrix of rank $5$. Find the rank of $A^2 +A +I$.
Well i dont know any tool that can solve this question. The book says answer is $5$ but it could be wrong also.
 A: Then rank of  $A^2+A+I$ is $6$. To see this, note that since $A$ is symmetric, it can be diagonalized, i.e. there exists a nonsingular matrix $P$ such that $P^{-1}AP=D$ where $D$ is diagonal.
Suppose $D=\mbox{diag}(\lambda_1,\lambda_2,\cdots,\lambda_6)$. Since rank of $A$ is $5$, 
rank of $D$ is also $5$. This implies that exactly one of the $\lambda_i$ is zero. 
Without loss of generality, assume that $\lambda_i\neq 0$ for $1\leq i\leq 5$ and $\lambda_6=0$. 
Now consider $A^2+A+I$. Note that $P^{-1}(A^2+A+I)P$ has the same rank as 
$A^2+A+I$ since $P$ is nonsingular, and 
$$P^{-1}(A^2+A+I)P=D^2+D+I=\mbox{diag}(\lambda_i^2+\lambda_i+1)$$
which has rank $6$, since  $\lambda_i^2+\lambda_i+1\neq 0$
for $1\leq i\leq 5$ and $\lambda_6^2+\lambda_6+1=1$. 
A: An equivalent, but slightly different, viewpoint is as follows: as $A$ is real symmetric, all eigenvalues of $A$ are real. Hence if $A-cI$ is not invertible for some complex number $c$, then $c \in \mathbb{R}$. Now $A^{2}+A+I = (A-\omega I)(A-\omega^{2}I),$ where $\omega$ is a primitive cube root of unity ( so $\omega = e^{\frac{2 \pi i}{3}}$). Now both $(A - \omega I)$ and $(A-\omega^{2}I)$ are invertible matrices, so their product is invertible, and $A^{2}+A+I$ is invertible, so of rank $6$.
A: It is not possible to tell, the rank of $A^2+A+I$  it can be $5$ or $6$.
If $A = \begin{bmatrix}1&&&&&\\&1&&&&\\&&1&&&\\&&&1&&\\&&&&1&\\&&&&&0\\\end{bmatrix}$ Then $A^2+A+I$ has rank 6. 
Following example works only for complex matrices (I did not see your comment that we look fo real matrices first, sorry):
If $A = e^{2\pi i/3}\begin{bmatrix}1&&&&&\\&1&&&&\\&&1&&&\\&&&1&&\\&&&&1&\\&&&&&0\\\end{bmatrix}$ then the rank of $A^2+A+I$ is $1$. 
And similarly you can create all examples in between.
