Question

Let each matrix in Exercises 1–6 act on \({\mathbb{C}^2}\). . Find the eigenvalues and a basis for each eigenspace in \({\mathbb{C}^2}\).

6. \(\left( {\begin{aligned}{}4&{}&3\\{ - 3}&{}&{\,4}\end{aligned}} \right)\)

Short answer

Eigenvalues are \(\lambda = 4 + 3i\) and \(\lambda = 4 - 3i\). The basis for each Eigenspace in \({\mathbb{C}^2}\) is \({{\bf{v}}_1} = \left( \begin{aligned}{} - i\\1\end{aligned} \right)\) and \({{\bf{v}}_2} = \left( \begin{aligned}{}i\\1\end{aligned} \right)\).

Step-by-step solution

Find the characteristic equation 

If \(A\) is a\(n \times n\) matrix, then \(det\left( {A - \lambda I} \right) = 0\), is called the characteristic equation of matrix \(A\).

It is given that\(A = \left( {\begin{aligned}{}4&{}&3\\{ - 3}&{}&{\,\,4}\end{aligned}} \right)\)and\(I = \left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)\)is the identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{aligned}{}A - \lambda I = A &= \left( {\begin{aligned}{}4&{}&3\\{ - 3}&{}&{\,\,4}\end{aligned}} \right) - \lambda \left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{4 - \lambda }&{}&3\\{ - 3}&{}&{4 - \lambda }\end{aligned}} \right)\end{aligned}\)

Now calculate the determinant of the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{}{4 - \lambda }&{}&3\\{ - 3}&{}&{4 - \lambda }\end{aligned}} \right)\\ &= \left( {4 - \lambda } \right)\left( {4 - \lambda } \right) + 9\\ &= {\lambda ^2} - 8\lambda + 25\end{aligned}\)

So, the characteristic equation of the matrix \(A\) is \({\lambda ^2} - 8\lambda + 25 = 0\).

Find the Eigenvalues 

Thus, the eigenvalues of\(A\)are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, solve the characteristic equation\({\lambda ^2} - 8\lambda + 25 = 0\), as follows:

For the quadratic equation, \(a{x^2} + bx + c = 0\) , the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\).

Thus, the solution of the characteristic equation \({\lambda ^2} - 8\lambda + 25 = 0\) is obtained as follows:

\(\begin{aligned}{}{\lambda ^2} - 8\lambda + 25 &= 0\\\lambda &= \frac{{ - \left( { - 8} \right) \pm \sqrt {{{\left( { - 8} \right)}^2} - 4\left( {25} \right)} }}{2}\\ &= \frac{{8 \pm \sqrt { - 36} }}{2}\\ &= 4 \pm 3i\end{aligned}\)

The eigenvalues of \(A\) are \(\lambda = 4 + 3i\) and \(\lambda = 4 - 3i\) .

Find the Eigenvectors 

For the Eigenvalue,\({\lambda _i}\), matrix A satisfies the system of equations\(\left( {A - {\lambda _i}} \right){\rm{x}} = {\rm{0}}\).

For the Eigenvector, \(\lambda = 4 + 3i\), the system \(\left( {A - \lambda I} \right)\)is solved as follows:

\(\begin{aligned}{}A - \left( {4 + 3i} \right)I &= \left( {\begin{aligned}{}4&{}&3\\{ - 3}&{}&{\,\,4}\end{aligned}} \right) - \left( {4 + 3i} \right)\left( {\begin{aligned}{}1&{}&0\\{\,0}&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - 3i}&{}&3\\{ - 3}&{}7{ - 3i}\end{aligned}} \right)\end{aligned}\)

The system of equations \(\left( {A - \lambda I} \right)x = 0\)gives the following equations:

\(\begin{aligned}{}\left( {\begin{aligned}{}{ - 3i}&{}&3\\{ - 3}&{}&{ - 3i}\end{aligned}} \right) &= 0\\\left( { - 3i} \right){x_1} + 3{x_2} &= 0\\ - 3{x_1} + \left( { - 3i} \right){x_2} &= 0\end{aligned}\)

Both the equations are equivalent to the equation, \({x_1} = - i{x_2}\), in which \({x_2}\) the variable is free. So, the general solution is \({x_1}\left( \begin{aligned}{} - i\\\,\,1\end{aligned} \right)\) and the corresponding Eigenvector is \({{\bf{v}}_1} = \left( \begin{aligned}{} - i\\\,\,1\end{aligned} \right)\).

For the Eigenvector, \(\lambda = 4 - 3i\), the system \(\left( {A - \lambda I} \right)\) is solved as follows:

\(\begin{aligned}{}A - \left( {4 - 3i} \right)I &= \left( {\begin{aligned}{}4&{}&3\\{ - 3}&{}&{\,\,4}\end{aligned}} \right) - \left( {4 - 3i} \right)\left( {\begin{aligned}{}1&{}&0\\{\,0}&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{3i}&{}&3\\{ - 3}&{}&{3i}\end{aligned}} \right)\end{aligned}\)

The system of equations \(\left( {A - \lambda I} \right)x = 0\)gives the following equations:

\(\begin{aligned}{}\left( {\begin{aligned}{}{3i}&{}&3\\{ - 3}&{}&{3i}\end{aligned}} \right) &= 0\\3i{x_1} + 3{x_2} &= 0\\ - 3{x_1} + 3i{x_2} &= 0\end{aligned}\)

Both the equations are equivalent to the equation, \({x_1} = i{x_2}\), in which \({x_2}\) the variable is free. So, the general solution is \({x_2}\left( \begin{aligned}{}i\\1\end{aligned} \right)\) and the corresponding Eigenvector is \({{\bf{v}}_2} = \left( \begin{aligned}{}i\\1\end{aligned} \right)\).