As you noticed, since $A$ is a real symmetric matrix, it is diagonalizable, and the properties satisfied (or not) by $A$ will be satisfied also by $D$, where $D$ is a diagonal matrix which consists of the eigenvalues of $A$.
Indeed, $\dim\ker(A-I)$ and characteristic polynomial are invariant up a change of basis. To see that, write $A=P^{-1}DP$, and $\ker(A-I)=\ker(D-I)$, as it can be seen by a double inclusion. For the characteristic polynomial, use determinant for example.
- The first is true if we switch $m$ and $n$ in the power of $\lambda$, namely $p_A(\lambda)=(\lambda-1)^m\lambda^{n-m}$, otherwise the degree would be $2m-n$, which is not $m$, except in the case $m=n$.
- The second is false if $k=0$ unless $A=I$, but true for $k\geq 1$.
- The third is also true. Indeed, we look at the rank of $A$ is the same as the rank of $D$, and the rank of a diagonal matrix is easy to determine.
That is a false assumption since a (nXn matrix) a square matrix needs to have at least n different eigenvalues (to make eigenvectors from)
No, they need not be different, as you yourself have noticed (the identity matrix example) and others have also pointed it out.
You can write each matrix in the Jordan normal form:
$$A = S J S^{-1}, \quad J = \operatorname{diag}(J_1, J_2, \dots, J_m),$$
where
$$J_k = \begin{bmatrix} \lambda_k & 1 \\ & \ddots & \ddots \\ & & \lambda_k & 1 \\ & & & \lambda_k \end{bmatrix}.$$
A matrix $A$ is diagonalizable if and only if all $J_k$ are of order $1$, i.e., $J_k = \begin{bmatrix} \lambda_k \end{bmatrix}$. In other words, $m = n$ and
$$J = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n).$$
Now, if $A$ has only one eigenvalue, that means that $\lambda := \lambda_1 = \lambda_2 = \cdots = \lambda_n$, so
$$J = \operatorname{diag}(\lambda, \lambda, \dots, \lambda) = \lambda \operatorname{diag}(1, 1, \dots, 1) = \lambda {\rm I}.$$
Now, let us get back to $A$:
$$A = S J S^{-1} = S (\lambda {\rm I}) S^{-1} = \lambda S S^{-1} = \lambda {\rm I}.$$
So, $A$ is diagonalizable with only one eigenvalue if and only if it is a scalar matrix.
However, not every matrix with only one eigenvalue is diagonalizable. Just put $\lambda := \lambda_1 = \lambda_2 = \cdots = \lambda_m$, but let some of $J_k$ be of order strictly bigger than $1$. For example,
$$B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$
is not diagonalizable.
Best Answer
You have two missing eigenvalues, other than the known eigenvalue $2$. The first equation tells you that their sum is $-4$. The last equation gives you another equation: since $\chi_A(t) = (t-2)(t-\lambda)(t-\mu)$, plugging in $t=1$ allows you to solve for $\lambda$ and $\mu$.