Question

Let \(A\) be a real \(2 \times 2\) matrix with a complex eigenvalue \(\lambda = a - bi\)\(\left( {b \ne 0} \right)\) and an associated eigenvector \({\bf{v}}\) in \({\mathbb{}^2}\).

1. Show that \(A({\mathop{\rm Re}\nolimits} {\bf{v}}) = a{\mathop{\rm Re}\nolimits} {\bf{v}} + b{\mathop{\rm Im}\nolimits} {\bf{v}}\) and \(A({\mathop{\rm Im}\nolimits} {\bf{v}}) = - b{\mathop{\rm Re}\nolimits} {\bf{v}} + a{\mathop{\rm Im}\nolimits} {\bf{v}}\). (Hint: Write \({\bf{v}} = {\mathop{\rm Re}\nolimits} {\bf{v}} + i{\mathop{\rm Im}\nolimits} {\bf{v}}\), and compute \(A{\bf{v}}\).)
2. Verify that if \(P\) and \(C\) are given as in Theorem 9, then \(AP = PC\)

Short answer

1. \(A\left( {{\mathop{\rm Re}\nolimits} \left( {\bf{v}} \right)} \right) = {\mathop{\rm Re}\nolimits} \left( {A{\bf{v}}} \right) = a{\mathop{\rm Re}\nolimits} \left( {\bf{v}} \right) + b{\mathop{\rm Im}\nolimits} \left( {\bf{v}} \right)\)\(A\left( {{\mathop{\rm Im}\nolimits} \left( {\bf{v}} \right)} \right) = {\mathop{\rm Im}\nolimits} \left( {A{\bf{v}}} \right) = - b{\mathop{\rm Re}\nolimits} \left( {\bf{v}} \right) + aIm\left( {\bf{v}} \right)\)
2. \(AP = \left( {A\left( {{\mathop{\rm Re}\nolimits} \left( {\bf{v}} \right)} \right)\,\,\,\,\,A\left( {{\mathop{\rm Im}\nolimits} \left( {\bf{v}} \right)} \right)} \right) = \left( {P\left( {\begin{aligned}{}a\\b\end{aligned}} \right)\;\;\;P\left( {\begin{aligned}{}{ - b}\\a\end{aligned}} \right)} \right) = P\left( {\begin{aligned}{}a&{}&{ - b}\\b&{}&a\end{aligned}} \right) = PC\)

Step-by-step solution

Formula of eigenvector 

Remember that an eigenvalue\(\lambda \)and an eigenvector\(x\)for a square matrix A satisfy the equation\(Ax = \lambda x\).

Proof

(a)

Given an eigenvector \(v\) and eigenvalue \(\lambda = a - bi\). Consider \(Ax = \lambda x\) and solve it as follows:

\(\begin{aligned}{}Av &= \lambda v\\ &= \left( {a - bi} \right)\left( {{\mathop{\rm Re}\nolimits} \left( v \right) + i{\mathop{\rm Im}\nolimits} \left( v \right)} \right)\\ &= \left( {a{\mathop{\rm Re}\nolimits} \left( v \right) + b{\mathop{\rm Im}\nolimits} \left( v \right)} \right) + i\left( {a{\mathop{\rm Im}\nolimits} \left( v \right) - b{\mathop{\rm Re}\nolimits} \left( v \right)} \right)\end{aligned}\)

So, from the above equation, it can be concluded that

\(\begin{aligned}{}A\left( {{\mathop{\rm Re}\nolimits} \left( v \right)} \right) = {\mathop{\rm Re}\nolimits} \left( {Av} \right) = a{\mathop{\rm Re}\nolimits} \left( v \right) + b{\mathop{\rm Im}\nolimits} \left( v \right)\\A\left( {{\mathop{\rm Re}\nolimits} \left( v \right)} \right) = {\mathop{\rm Im}\nolimits} \left( {Av} \right) = - b{\mathop{\rm Re}\nolimits} \left( v \right) + a{\mathop{\rm Im}\nolimits} \left( v \right)\end{aligned}\)

Verify that if P and C are given as in Theorem 9, then AP = PC. 

(b)

By theorem 9, let \(P = \left( {{\mathop{\rm Re}\nolimits} \left( v \right)\,\,\,\,\,\,\,\,{\mathop{\rm Im}\nolimits} \left( v \right)} \right)\)

From part (a),

\(\begin{aligned}{}A\left( {{\mathop{\rm Re}\nolimits} \left( v \right)} \right) &= P\left( {\begin{aligned}{}a\\b\end{aligned}} \right)\\A\left( {{\mathop{\rm Im}\nolimits} \left( v \right)} \right) &= P\left( {\begin{aligned}{}{ - b}\\a\end{aligned}} \right)\end{aligned}\)

Now, Solve AP as follows:

\(\begin{aligned}{}AP &= \left( {A\left( {{\mathop{\rm Re}\nolimits} \left( v \right)} \right)\,\,\,\,\,A\left( {{\mathop{\rm Im}\nolimits} \left( v \right)} \right)} \right)\\ &= \left( {P\left( {\begin{aligned}{}a\\b\end{aligned}} \right)\;\;\;P\left( {\begin{aligned}{}{ - b}\\a\end{aligned}} \right)} \right)\\ &= P\left( {\begin{aligned}{}a&{}&{ - b}\\b&{}&a\end{aligned}} \right)\\ &= PC\end{aligned}\)

Hence, it proved that \(AP = PC\).