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 (so over the complex numbers)
Find the basis of the corresponding Eigenspace
I found found the eigenvalues to be:
$\alpha$: over reals and then only the value $\lambda_1=3$
$\beta$: over complex and then the values $\lambda_1=3$, $\lambda_2=i$ and $\lambda_3=-i$
How would I proceed to find a basis for the Eigenspaces of the two matrices$
Best Answer
For example, we can find the eigenvector corresponding to $\lambda = i$ as follows: compute $$ \beta - \lambda I = \pmatrix{ 7-i &3 &-4 \\ -2&-1-i &2 \\ 6&2 &-3-i }. $$ We can row-reduce this matrix to get the reduced row-echelon form $$ \pmatrix{1 & 0 & -1/2 - i/2\\0 & 1 & i\\ 0 & 0 & 0}. $$ Conclude that the kernel of $A - \lambda I$ (i.e. the eigenspace) is spanned by the vector (i.e. eigenvector) $v_i = (1/2 + i/2, -i, 1)^\top$. In other words, the set $\{(1/2+i/2,-i,1)^\top\}$ forms a basis of the eigenspace associated with $\lambda = i$.
The other two basis (each a set with one vector) can be computed in a similar fashion.
Actually, because $A$ has real entries, we can use our result for $\lambda = i$ to get the eigenvector for $\lambda = -i$: $$ Av_i = iv_i \implies \overline{Av_i} = \overline{iv_i} \implies A \overline{v_i} = -i \overline{v_i}. $$ So, we can take $v_{-i} = \overline{v_i} =(1/2 - i/2, i,1)^\top$.