I am preparing to take an exam on Linear Algebra.

I found this question on previous exams.

Here is what I think is needed to be done.

1)First find the eigenvalues, I get:

$\displaystyle \det(kI-A)=0$

$\displaystyle \left| \begin{array}{ccc}k+1&7&-1\\0&k-1&0\\0&15&k+2 \end{array} \right|=0$

$\displaystyle (-1)^{1+1}(k+1)\left| \begin{array}{cc}k-1&0\\15&k+2 \end{array}\right|=(k+1)(k-1)(k+2)=0$

$\displaystyle k=-2,-1,1$

2)There are exactly three eigenvalues, thus there are exactly three linearly independant eigenvectors. Since the eigenvectors are a subspace of the nullspace of $\displaystyle (kI-A)$ which is of dimension three it tells us that there is basis for the eigenspace.

3)To find the basis for the eigenspace I need to substitute the eigenvalues and solve the homogenous system $\displaystyle (kI-A)\bold{x}=\bold{0}$. Thus I need to,

$\displaystyle \left[ \begin{array}{cccc}k+1&7&-1&0\\0&k-1&0&0\\0&15&k+2&0 \end{array} \right]$

Solve it by Gaussian elimination for each eigenvalue $\displaystyle k$.

4)Then the basis I get for the eigenspace in #3 will be the diagnolizable matrix for $\displaystyle A$ by making the vectors as coloum vectors.

Is this approach correct?

I have a second question, how is "Eigenvalue" pronounced.

(At least it is not as bad, Lebesque. I never would have guess it, unless I heard someone pronouce it).