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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

Expert verified

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

01

Define Non zero vector

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

02

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.

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Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

20. Let \(p\left( t \right){\bf{ = }}\left( {t{\bf{ - 2}}} \right)\left( {t{\bf{ - 3}}} \right)\left( {t{\bf{ - 4}}} \right){\bf{ = - 24 + 26}}t{\bf{ - 9}}{t^{\bf{2}}}{\bf{ + }}{t^{\bf{3}}}\). Write the companion matrix for \(p\left( t \right)\), and use techniques from chapter \({\bf{3}}\) to find the characteristic polynomial.

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