Question

Let \(A\) be a real \(n \times n\) matrix, and let \(x\) be a vector in \({\mathbb{C}^n}\). Show that \({\mathop{\rm Re}\nolimits} \left( {Ax} \right) = A\left( {{\mathop{\rm Re}\nolimits} x} \right)\) and \({\mathop{\rm Im}\nolimits} \left( {Ax} \right) = A\left( {{\mathop{\rm Im}\nolimits} x} \right)\).

Short answer

Since \({\bf{x}}\) is some nonzero vector in \({\bf{C}}\), so we can write it as \({\bf{x}} = {\mathop{\rm Re}\nolimits} {\bf{x}} + i \cdot {\mathop{\rm lm}\nolimits} {\bf{x}}\).

Spiting it in real and imaginary parts, although both \(\{ Re\} {\bf{x}}\) and \(\{ Im\} {\bf{x}}\) are real.

Step-by-step solution

Define Non zero vector

A vector with a magnitude not equal to zero is nonzero vector.

Prove the condition

Since \({\bf{x}}\) is some nonzero vector in \({\bf{C}}\) , so we can write it as \({\bf{x}} = {\mathop{\rm Re}\nolimits} {\bf{x}} + i \cdot {\mathop{\rm lm}\nolimits} {\bf{x}}\).

Splitting it in real and imaginary parts, although both \(\{ Re\} {\bf{x}}\) and \(\{ Im\} {\bf{x}}\) are real.

Now we have:

\(\begin{aligned}{}A{\bf{x}} &= A({\mathop{\rm Re}\nolimits} {\bf{x}} + i \cdot {\mathop{\rm lm}\nolimits} {\bf{x}})\\ &= A({\mathop{\rm Re}\nolimits} {\bf{x}}) + i \cdot A({\mathop{\rm lm}\nolimits} {\bf{x}})\end{aligned}\)

Since \(A\) is real. Then we have \(A({\mathop{\rm Re}\nolimits} {\bf{x}})\) and \(A({\mathop{\rm Im}\nolimits} {\bf{x}})\) both real, and hence

\({\mathop{\rm Re}\nolimits} (A{\bf{x}}) = A({\mathop{\rm Re}\nolimits} {\bf{x}}){\rm{ and }}{\mathop{\rm Im}\nolimits} (A{\bf{x}}) = A({\mathop{\rm lm}\nolimits} {\bf{x}})\).

We could think of \({\mathop{\rm Re}\nolimits} (A{\bf{x}})\) as taking the real part of \(A{\bf{x}}\).

Since,

\(\begin{aligned}{}Ax &= A\left( {{\mathop{\rm Re}\nolimits} {\bf{x}} + i \cdot {\mathop{\rm lm}\nolimits} {\bf{x}}} \right)\\ &= A\left( {{\mathop{\rm Re}\nolimits} {\bf{x}}} \right) + i \cdot A\left( {{\mathop{\rm Im}\nolimits} {\bf{x}}} \right)\end{aligned}\)

So, from the above, it is obvious that \({\mathop{\rm Re}\nolimits} \left( {A{\bf{x}}} \right) = A({\mathop{\rm Re}\nolimits} {\bf{x}})\).

The same goes for the imaginary part of \(A \to {\mathop{\rm lm}\nolimits} \left( {A{\bf{x}}} \right) = A\left( {{\mathop{\rm lm}\nolimits} {\bf{x}}} \right)\)

Since \({\bf{x}}\) is some nonzero vector in \({\bf{C}}\) we can write it as \({\bf{x}} = {\mathop{\rm Re}\nolimits} {\bf{x}} + i \cdot {\mathop{\rm lm}\nolimits} {\bf{x}}\).

Splitting it in real and imaginary part, although both \(\{ Re\} {\bf{x}}\) and \(\{ Im\} {\bf{x}}\) are real.