Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 14 / a

Expand the determinant along the first row:Thus $\boxed{\det A=45}$. The signs in the middle cofactor account for its positive contribution.

The characteristic polynomial, using the convention $\det (\lambda I-A)$, isIts zeros give the eigenvalues, including algebraic multiplicity:Their product is $45$ and their sum is $-1$, agreeing with the determinant and trace of the matrix.

For $\lambda =5$, the eigenvalue equation reduces to a one-dimensional eigenspace spanned by $(1,2,-1)$. For $\lambda =-3$, the rows of $A+3I$ are multiples of $(1,2,-3)$, so its eigenspace is the plane $x+2y-3z=0$. One possible set of three independent eigenvectors isEvery eigenvector for $-3$ is a nonzero linear combination of the last two. Their two-dimensional eigenspace shows that the repeated root is not defective: the matrix is diagonalizable.

For the particular eigenvectors chosen in (iii), the dot products areThus only ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ are orthogonal for this choice. The matrix is not symmetric, so distinct eigenvalues alone do not imply orthogonality. Another basis within the repeated eigenspace may change which chosen vectors are perpendicular; the dot products above specify the convention completely.