Question

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

2). \(\left( {\begin{aligned}{}5&{}&{ - 5}\\1&{}&{\,1}\end{aligned}} \right)\)

Short answer

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

Step-by-step solution

Find the characteristic equation

If 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\(A = \left( {\begin{aligned}{}5&{}&{ - 5}\\1&{}&{\,\,1}\end{aligned}} \right)\)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:

\(\begin{aligned}{}A - \lambda I &= \left( {\begin{aligned}{}5&{}&{ - 5}\\1&{}&{\,\,1}\end{aligned}} \right) - \lambda \left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{5 - \lambda }&{}&{ - 5}\\1&{}&{1 - \lambda }\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{}{5 - \lambda }&{}&{ - 5}\\1&{}&{1 - \lambda }\end{aligned}} \right)\\ &= \left( {5 - \lambda } \right)\left( {1 - \lambda } \right) + 5\\ &= {\lambda ^2} - 6\lambda + 10\end{aligned}\)

So, the characteristic equation of the matrix \(A\) is \({\lambda ^2} - 6\lambda + 10 = 0\).

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\({\lambda ^2} - 6\lambda + 10 = 0\), 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 \({\lambda ^2} - 6\lambda + 10 = 0\) is obtained as follows:

\(\begin{aligned}{}{\lambda ^2} - 6\lambda + 10 &= 0\\\lambda &= \frac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( {10} \right)} }}{2}\\ &= \frac{{6 \pm \sqrt { - 4} }}{2}\\ &= 3 \pm i\end{aligned}\)

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

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

\(\begin{aligned}{}A - \left( {2 + i} \right)I = \left( {\begin{aligned}{}5&{}&{ - 5}\\1&{}&{\,\,1}\end{aligned}} \right) - \left( {3 + i} \right)\left( {\begin{aligned}{}1&{}&0\\{\,0}&{}&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{}{2 - i}&{}&{ - 5}\\1&{}&{ - 2 - i}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}\left( {\begin{aligned}{}{2 - i}&{ - 5}\\1&{ - 2 - i}\end{aligned}} \right) &= 0\\\left( {2 - i} \right){x_1} - 5{x_2} &= 0\\{x_1} + \left( { - 2 - i} \right){x_2} &= 0\end{aligned}\)

Both the equations are equivalent to the equation, \({x_1} = - \left( { - 2 - i} \right){x_2}\), in which \({x_2}\) the variable is free. So, the general solution is \({x_2}\left( \begin{aligned}{}2 + i\\\,\,\,\,\,\,1\end{aligned} \right)\) and the corresponding Eigenvector is \({{\bf{v}}_1} = \left( \begin{aligned}{}2 + i\\\,\,\,\,\,\,1\end{aligned} \right)\).

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

\(\begin{aligned}{}A - \left( {3 - i} \right)I &= \left( {\begin{aligned}{}1&{}&{ - 2}\\1&{}&{\,\,3}\end{aligned}} \right) - \left( {3 - i} \right)\left( {\begin{aligned}{}1&{}&0\\{\,0}&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{2 + i}&{ - 5}\\1&{ - 2 + i}\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{c}\left( {\begin{aligned}{}{2 + i}&{ - 5}\\1&{ - 2 + i}\end{aligned}} \right) = 0\\\left( {2 + i} \right){x_1} - 5{x_2} = 0\\{x_1} + \left( { - 2 + i} \right){x_2} = 0\end{aligned}\)

Both the equations are equivalent to the equation, \({x_1} = - \left( { - 2 + i} \right){x_2}\), in which \({x_2}\) the variable is free. So, the general solution is \({x_2}\left( \begin{aligned}{}\,\,2 - i\\\,\,\,\,\,\,1\end{aligned} \right)\) and the corresponding Eigenvector is \({{\bf{v}}_2} = \left( \begin{aligned}{}\,\,2 - i\\\,\,\,\,\,\,1\end{aligned} \right)\).