This is a short one but: Consider the real matrix $\alpha = \begin{pmatrix} 7 &3 &-4 \\ -2&-1 &2 \\ 6&2 &-3 \end{pmatrix}$ and let $\beta \in M_n (\mathbb{C})$ be the same matrix but considered as a complex matrix.
Problem
with use of a computer program calculate the eigenvalues of the two matrices. i.e $\alpha$ over the real numbers and $\beta$ over the complex numbers.
Question
What does this actually mean. They are the same but over different numbers? Is it just the same matrix and thus the same result in this problem but got new rules in different problems where the fact that it is over complex numbers makes a difference? Does it make a difference here?
What does the matrix actually look like - I am thinking it is the same but the fact it is over complex numbers do not matter here.
The characteristic polynomial of your matrix is $$-x^3+3x^2-x+3 = -(x-3)(x^2+1)$$ The real root of this polynomial is $3$ which is an eigenvalue for $\alpha$ and $\beta$: $$\begin{pmatrix} 7 &3 &-4 \\ -2&-1 &2 \\ 6&2 &-3 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 3\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$$ Non-real roots of the polynomial are $\pm i$, which are eigenvalues only for $\beta$: $$\begin{pmatrix} 7 &3 &-4 \\ -2&-1 &2 \\ 6&2 &-3 \end{pmatrix} \begin{pmatrix} 1+i \\ -2i \\ 2 \end{pmatrix} = i\begin{pmatrix} 1+i \\ -2i \\ 2 \end{pmatrix}, \quad \begin{pmatrix} 7 &3 &-4 \\ -2&-1 &2 \\ 6&2 &-3 \end{pmatrix} \begin{pmatrix} 1-i \\ 2i \\ 2 \end{pmatrix} = -i\begin{pmatrix} 1-i \\ 2i \\ 2 \end{pmatrix}$$ Note that there are no real nonzero vectors $v$ such that $\alpha v = \pm i v$ so they are not eigenvalues of $\alpha$ in this sense.