Eigenvalues

Eigenvalues.utf8

Eigenvalues

Let \(A\) be an \(n\times n\) matrix, \(I_{n}\) the identity \(n\times n\) matrix and \(\lambda I_{n}\) the diagonal with \(\lambda\)-scalar. Then \(A-\lambda I_{n}\) is an \(n\times n\) matrix and the determinant \(|A-\lambda I_{n}|\) is a polynomial which we denote by \(P_{N}(\lambda).\) \(P_{N}(\lambda)\) is called the characteristic polynomial of \(A\). The roots of \(P_{N}(\lambda)\) are called eigenvalues of \(A\).

Find the eigenvalues of the matrix

\[ A=\left(\begin{array}{ccc} 1 & 2 & 1\\ 2 & 0 & -2\\ -1 & 2 & 3\end{array}\right).\]

Solution \[\begin{eqnarray*} P_{n}(\lambda)=\left|A-\lambda I_{3}\right| & = & \left|\begin{array}{ccc} 1-\lambda & 2 & 1\\ 2 & -\lambda & -2\\ -1 & 2 & 3-\lambda\end{array}\right|\\ \, & = & (1-\lambda)\left|\begin{array}{cc} -\lambda & -2\\ 2 & 3-\lambda\end{array}\right|-2\left|\begin{array}{cc} 2 & -2\\ -1 & 3-\lambda\end{array}\right|+\left|\begin{array}{cc} 2 & -\lambda\\ -1 & 2\end{array}\right|\\ \, & = & \lambda^{3}-4\lambda^{2}+4\lambda\\ \, & = & \lambda(\lambda-2)^{2}.\end{eqnarray*}\]

To get the eigenvalues we set \(P_{n}(\lambda)=0).\)

\[\begin{eqnarray*} P_{n}(\lambda)=0\\ \Rightarrow & \lambda(\lambda-2)^{2}= & 0\\ \Rightarrow & \lambda_1=2,\, & \lambda_2=0\:\text{are the eigenvalues.}\end{eqnarray*}\]

To find the eigenvalues using R,first enter the matrix

A<-matrix(c(1,2,1,2,0,-2,-1,2,3),nrow=3,ncol=3,byrow=TRUE)

Then use the command

eigen(A,only.value=TRUE)
## $values
## [1] 2.000000e+00 2.000000e+00 5.055881e-16
##
## $vectors
## NULL

Eigenvectors

Eigenvectors.utf8

Eigenvalues can be equivalently defined as follows:

Let \(A\) be an \(n\times n\) matrix, \(\lambda\) is called an eigenvalue of \(A\) if and only if \(\exists\) a non-zero vector \(\underline{x}\in\mathbb{R}^{n}\) such that \[\begin{equation} A\underline{x}=\lambda\underline{x}\quad (1) \end{equation}\]

Every non-zero vector satisfying equation (1) is called an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\).

Example Find the eigenvectors of \[ A=\left(\begin{array}{ccc} 1 & 2 & 1\\ 2 & 0 & -2\\ -1 & 2 & 3\end{array}\right).\]

Solution

From the previous section we got the eigenvalues of the above matrix as \(\lambda=0\) and \(\lambda=2.\) For the eigenvector corresponding to \(\lambda=0\), we use equation (1)

\[\begin{eqnarray*} \, & A\underline{x}-\lambda I_{n}\underline{x} & =\underline{0}\\ \Rightarrow & (A-\lambda I_{n})\underline{x} & =\underline{0}\\ \Rightarrow & \left(\begin{array}{ccc} 1 & 2 & 1\\ 2 & 0 & -2\\ -1 & 2 & 3\end{array}\right)\left(\begin{array}{c} x\\ y\\ z\end{array}\right) & =\left(\begin{array}{c} 0\\ 0\\ 0\end{array}\right) \end{eqnarray*}\]

\[\begin{eqnarray} \Rightarrow & x+2y+z & =0\quad (2)\\ \, & 2x-2z & =0\quad (3)\\ \, & -x+2y+3z & =0\quad (4) \end{eqnarray}\]

From equation (3) we have \(x=z.\) Substituting this into equation (2) we have \(y=-x\). Set \(x=\alpha\), where \(\alpha\in\mathbb{R}\). The vector is

\[ \left(\begin{array}{c} \alpha\\ -\alpha\\ \alpha\end{array}\right)=\alpha\left(\begin{array}{c} 1\\ -1\\ 1\end{array}\right).\]

Therefore the eigenvector corresponding to \(\lambda=0\), is \[\left(\begin{array}{c} 1\\ -1\\ 1\end{array}\right).\]

For the eigenvector corresponding to \(\lambda=2\), we use equation (1) \[\begin{eqnarray*} \, & (A-\lambda I_{n})\underline{x} & =\underline{0}\\ \Rightarrow & \left(\begin{array}{ccc} -1 & 2 & 1\\ 2 & -2 & -2\\ -1 & 2 & 1\end{array}\right)\left(\begin{array}{c} x\\ y\\ z\end{array}\right) & =\left(\begin{array}{c} 0\\ 0\\ 0\end{array}\right) \end{eqnarray*}\]

\[\begin{eqnarray} \Rightarrow & -x+2y+z & =0\quad (5)\\ \, & 2x-2y-2z & =0\quad (6)\\ \, & -x+2y+z & =0\quad (7) \end{eqnarray}\]

If we add equation (5) and (6) we see that \(x=z.\) If we substitute this into equation (7) we get \(y=0.\) Set \(x=\alpha\), where \(\alpha\in\mathbb{R}\). So the vector is \[ \left(\begin{array}{c} \alpha\\ 0\\ \alpha\end{array}\right)=\alpha\left(\begin{array}{c} 1\\ 0\\ 1\end{array}\right).\]

Therefore the eigenvector corresponding to \(\lambda=2\), is \(\left(\begin{array}{c} 1\\ 0\\ 1\end{array}\right).\)

Trevor. N. Mutusva, February 2020.