Advanced Linear Algebra: Foundations to Frontiers

A set \({\cal S} \subset \Cm \) is a subspace if and only if for all \(\alpha \in \C \) and \(x,y \in \Cm \) two conditions hold:

• \(x \in {\cal S} \) implies that \(\alpha x \in {\cal S} \text{.}\)

• \(x, y \in {\cal S} \) implies that \(x + y \in {\cal S} \text{.}\)

• \(x \in {\cal E}_{\lambda}( A )\) implies \(\alpha x \in {\cal E}_{\lambda}( A )\text{:}\)

  \(x \in {\cal E}_{\lambda}(A) \) means that \(A x = \lambda x \text{.}\) If \(\alpha \in \C \) then \(\alpha A x = \alpha \lambda x \) which, by commutivity and associativity means that \(A ( \alpha x ) = \lambda ( \alpha x ) \text{.}\) Hence \((\alpha x) \in {\cal E}_{\lambda}(A) \text{.}\)

• \(x,y \in {\cal E}_{\lambda}( A )\) implies \(x+y \in {\cal E}_{\lambda}( A )\text{:}\)

  \begin{equation*} A( x + y ) = A x + A y = \lambda x + \lambda y = \lambda( x + y ) . \end{equation*}

Let

Then

is singular if and only if \(\lambda = \delta_i \) for some \(i \in \{ 0, \ldots , m-1 \} \text{.}\) Hence \(\Lambda( D ) = \{ \delta_0, \delta_1, \ldots, \delta_{m-1} \} \text{.}\)

Now,

and hence \(e_j \) is an eigenvector associated with \(\delta_j \text{.}\)

The eigenvalues can be found on the diagonal: \(\{ -2, 1, 2 \} \text{.}\)

• To find an eigenvector associated with \(-2\text{,}\) form

  \begin{equation*} (-2) I - A = \left( \begin{array}{rrr} 0 \amp -3 \amp 7 \\ 0 \amp -3 \amp -1 \\ 0 \amp 0 \amp -4 \\ \end{array}\right) \end{equation*}

  and look for a vector in the null space of this matrix. By examination,

  \begin{equation*} \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) \end{equation*}

  is in the null space of this matrix and hence an eigenvector of \(A \text{.}\)

• To find an eigenvector associated with \(1\text{,}\) form

  \begin{equation*} (1) I - A = \left(\begin{array}{rrr} 3 \amp -3 \amp 7 \\ 0 \amp 0 \amp -1 \\ 0 \amp 0 \amp -1 \\ \end{array}\right) \end{equation*}

  and look for a vector in the null space of this matrix. Given where the zero appears on the diagonal, we notice that a vector of the form

  \begin{equation*} \left( \begin{array}{c} \chi_0 \\ 1 \\ 0 \end{array} \right) \end{equation*}

  is in the null space if \(\chi_0 \) is choosen appropriately. This means that

  \begin{equation*} 3 \chi_0 - 3 (1) = 0 \end{equation*}

  and hence \(\chi_0 = 1 \) so that

  \begin{equation*} \left( \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right) \end{equation*}

  in the null space of this matrix and hence an eigenvector of \(A \text{.}\)

• To find an eigenvector associated with \(2\text{,}\) form

  \begin{equation*} (2) I - A = \left(\begin{array}{rrr} 4 \amp -3 \amp 7 \\ 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \\ \end{array}\right) \end{equation*}

  and look for a vector in the null space of this matrix. Given where the zero appears on the diagonal, we notice that a vector of the form

  \begin{equation*} \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ 1 \end{array} \right) \end{equation*}

  is in the null space if \(\chi_0 \) and \(\chi_1\) are choosen appropriately. This means that

  \begin{equation*} \chi_1 - 1(1) = 0 \end{equation*}

  and hence \(\chi_1 = 1 \text{.}\) Also,

  \begin{equation*} 4 \chi_0 - 3 (1) + 7 ( 1) = 0 \end{equation*}

  so that \(\chi_0 = -1 \text{,}\)

  \begin{equation*} \left( \begin{array}{r} -1 \\ 1 \\ 1 \end{array} \right) \end{equation*}

  is in the null space of this matrix and hence an eigenvector of \(A \text{.}\)

Let

Then

is singular if and only if \(\lambda = \upsilon_{i,i} \) for some \(i \in \{ 0, \ldots , m-1 \} \text{.}\) Hence \(\Lambda( U ) = \{ \upsilon_{0,0}, \upsilon_{1,1}, \ldots, \upsilon_{m-1,m-1} \} \text{.}\)

Let \(\lambda \) be an eigenvalue of \(U \text{.}\) Things get a little tricky if \(\lambda \) has multiplicity greater than one. Partition

where \(\upsilon_{11} = \lambda \text{.}\) We are looking for \(x \neq 0 \) such that \(( \lambda I - U ) x = 0 \) or, partitioning \(x \text{,}\)

If we choose \(x_2 = 0 \) and \(\chi_1 =1 \text{,}\) then

and hence \(x_0 \) must satisfy

If \(\upsilon_{11} I - U_{00} \) is nonsingular, then there is a unique solution to this equation, and

is the desired eigenvector. HOWEVER, this means that the partitioning

must be such that \(\upsilon_{11} \) is the FIRST diagonal element that equals \(\lambda \text{.}\)