Algebra - Matrices

Chapter 3. Eigen vectors of $2 \times 2$ matrices

Prerequisites: In order to understand Eigen vectors the reader must be able to
  • Find the Eigen values of the matrix.


  • Definition 3.1: Let $A$ be a square matrix of order $n$. If there exists a non zero matrix $X$ of order $n \times 1$ and a scalar $\lambda$ such that $AX=\lambda X$ then $X$ is called the Eigen vector (column wise) corresponding to $\lambda$


  • Definition 3.2: Let $A$ be a square matrix of order $n$. If there exists a non zero matrix $X$ of order $1\times n $ and a scalar $\lambda$ such that $XA=X\lambda $ then $X$ is called the Eigen vector (row wise) corresponding to $\lambda$


  • The choice of choosing the Eigen vector row wise or column wise is left to the reader.

Problems

  • Problem 3.1 Find the Eigen values and corresponding Eigen vectors of the matrix $\begin{bmatrix} 1&2\\3&2 \end{bmatrix}$
    Solution: Let $A=\begin{bmatrix} 1&2\\3&2 \end{bmatrix}$
    Consider $|A-\lambda I|=0$ $\Rightarrow \begin{vmatrix} 1-\lambda&2\\3&2-\lambda \end{vmatrix}=0$
    $\Rightarrow (1-\lambda)(2-\lambda)-6=0$
    $\Rightarrow \lambda^2-3\lambda-4=0$ is the Eigen equation of $A$
    $\Rightarrow \lambda=-1,4$ are the Eigen values of $A$
    Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 1-\lambda&2\\3&2-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
    Put $\lambda =-1$ in (1) then
    $\Rightarrow \begin{bmatrix} 2&2\\3&3 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $2x+2y=0$
    $3x+3y=0$
    Consider $2x+2y=0$ $\Rightarrow$ $x+y=0$ $\Rightarrow$ $x=-y$
    Let $y=k$ then $x=-k$
    $\therefore$ $X=\begin{bmatrix}-k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =-1$
    Put $\lambda =4$ in (1) then
    $\Rightarrow \begin{bmatrix} -3&2\\3&-2 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $-3x+2y=0$
    $3x-2y=0$
    Consider $3x-2y=0$ $\Rightarrow$ $x=\frac{2y}{3}$
    Let $y=k$ then $x=\frac{2k}{3}$
    $\therefore$ $X=\begin{bmatrix}\frac{2k}{3}\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =4$ $\spadesuit$


  • Problem 3.2 Find the Eigen values and corresponding Eigen vectors of the matrix $\begin{bmatrix} 1&2\\5&4 \end{bmatrix}$
    Solution: Let $A=\begin{bmatrix} 1&2\\5&4 \end{bmatrix}$
    Consider $|A-\lambda I|=0$ $\Rightarrow \begin{vmatrix} 1-\lambda&2\\5&4-\lambda \end{vmatrix}=0$
    $\Rightarrow (1-\lambda)(4-\lambda)-10=0$
    $\Rightarrow \lambda^2-5\lambda-6=0$ is the Eigen equation of $A$
    $\Rightarrow \lambda=-1,6$ are the Eigen values of $A$
    Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 1-\lambda&2\\5&4-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
    Put $\lambda =-1$ in (1) then
    $\Rightarrow \begin{bmatrix} 2&2\\5&5 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $2x+2y=0$
    $5x+5y=0$
    Consider $2x+2y=0$ $\Rightarrow$ $x+y=0$ $\Rightarrow$ $x=-y$
    Let $y=k$ then $x=-k$
    $\therefore$ $X=\begin{bmatrix}-k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =-1$
    Put $\lambda =6$ in (1) then
    $\Rightarrow \begin{bmatrix} -5&2\\5&-2 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $-5x+2y=0$
    $5x-2y=0$
    Consider $5x-2y=0$ $\Rightarrow$ $x=\frac{2y}{5}$
    Let $y=k$ then $x=\frac{2k}{5}$
    $\therefore$ $X=\begin{bmatrix}\frac{2k}{5}\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =6$ $\spadesuit$


  • Problem 3.3 Find the Eigen values and corresponding Eigen vectors of the matrix $\begin{bmatrix} 1&2\\2&1 \end{bmatrix}$
    Solution: Let $A=\begin{bmatrix} 1&2\\2&1 \end{bmatrix}$
    Consider $|A-\lambda I|=0$ $\Rightarrow \begin{vmatrix} 1-\lambda&2\\2&1-\lambda \end{vmatrix}=0$
    $\Rightarrow (1-\lambda)(1-\lambda)-4=0$
    $\Rightarrow \lambda^2-2\lambda-3=0$ is the Eigen equation of $A$
    $\Rightarrow \lambda=-1,3$ are the Eigen values of $A$
    Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 1-\lambda&2\\2&1-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
    Put $\lambda =-1$ in (1) then
    $\Rightarrow \begin{bmatrix} 2&2\\2&2 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $2x+2y=0$
    $2x+2y=0$
    Consider $2x+2y=0$ $\Rightarrow$ $x+y=0$ $\Rightarrow$ $x=-y$
    Let $y=k$ then $x=-k$
    $\therefore$ $X=\begin{bmatrix}-k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =-1$
    Put $\lambda =3$ in (1) then
    $\Rightarrow \begin{bmatrix} -2&2\\2&-2 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $-2x+2y=0$
    $2x-2y=0$
    Consider $2x-2y=0$ $\Rightarrow$ $x=y$
    Let $y=k$ then $x=k$
    $\therefore$ $X=\begin{bmatrix}k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =3$ $\spadesuit$


  • Problem 3.4 Find the Eigen values and corresponding Eigen vectors of the matrix $\begin{bmatrix} 2&\sqrt{2}\\\sqrt{2}&1 \end{bmatrix}$
    Solution: Let $A=\begin{bmatrix} 2&\sqrt{2}\\\sqrt{2}&1 \end{bmatrix}$
    Consider $|A-\lambda I|=0$ $\Rightarrow \begin{vmatrix} 2-\lambda&\sqrt{2}\\\sqrt{2}&1-\lambda \end{vmatrix}=0$
    $\Rightarrow (2-\lambda)(1-\lambda)-2=0$
    $\Rightarrow \lambda^2-3\lambda=0$ is the Eigen equation of $A$
    $\Rightarrow \lambda=0,3$ are the Eigen values of $A$
    Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 2-\lambda&\sqrt{2}\\\sqrt{2}&1-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
    Put $\lambda =0$ in (1) then
    $\Rightarrow \begin{bmatrix} 2&\sqrt{2}\\\sqrt{2}&1 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $2x+\sqrt{2}y=0$
    $\sqrt{2}x+y=0$
    Consider $2x+\sqrt{2}y=0$ $\Rightarrow$ $x=\frac{-\sqrt{2}y}{2}$
    Let $y=k$ then $x=\frac{-\sqrt{2}k}{2}$
    $\therefore$ $X=\begin{bmatrix}-\frac{\sqrt{2}k}{2}\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =0$
    Put $\lambda =3$ in (1) then
    $\Rightarrow \begin{bmatrix} -1&\sqrt{2}\\\sqrt{2}&-2 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $-x+\sqrt{2}y=0$
    $\sqrt{2}x-2y=0$
    Consider $-x+\sqrt{2}y=0$ $\Rightarrow$ $x=\sqrt{2}y$
    Let $y=k$ then $x=\sqrt{2}k$
    $\therefore$ $X=\begin{bmatrix}\sqrt{2}k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =3$ $\spadesuit$


  • Problem 3.5 Find the Eigen values and corresponding Eigen vectors of the matrix $\begin{bmatrix} 4&1\\-1&2 \end{bmatrix}$
    Solution: Let $A=\begin{bmatrix} 4&1\\-1&2 \end{bmatrix}$
    Consider $|A-\lambda I|=0$ $\Rightarrow \begin{vmatrix} 4-\lambda&1\\-1&2-\lambda \end{vmatrix}=0$
    $\Rightarrow (4-\lambda)(2-\lambda)+1=0$
    $\Rightarrow (\lambda-3)^2=0$ is the Eigen equation of $A$
    $\Rightarrow \lambda=3,3$ are the Eigen values of $A$
    Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 4-\lambda&1\\-1&2-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
    Put $\lambda =3$ in (1) then
    $\Rightarrow \begin{bmatrix} 1&1\\-1&-1 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $x+y=0$
    $-x-y=0$
    Consider $x+y=0$ $\Rightarrow$ $x=-y$
    Let $y=k$ then $x=-k$
    $\therefore$ $X=\begin{bmatrix}-k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =3$ $\spadesuit$


  • Problem 3.6 Find the Eigen values and corresponding Eigen vectors of the matrix $\begin{bmatrix} 3&4\\-2&-3 \end{bmatrix}$
    Solution: Let $A=\begin{bmatrix} 3&4\\-2&-3 \end{bmatrix}$
    Consider $|A-\lambda I|=0$ $\Rightarrow \begin{vmatrix} 3-\lambda&4\\-2&-3-\lambda \end{vmatrix}=0$
    $\Rightarrow (3-\lambda)(-3-\lambda)+8=0$
    $\Rightarrow \lambda^2-1=0$ is the Eigen equation of $A$
    $\Rightarrow \lambda=-1,1$ are the Eigen values of $A$
    Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 3-\lambda&4\\-2&-3-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
    Put $\lambda =1$ in (1) then
    $\Rightarrow \begin{bmatrix} 2&4\\-2&-4 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $2x+4y=0$
    $-2x-4y=0$
    Consider $2x+4y=0$ $\Rightarrow$ $x=-2y$
    Let $y=k$ then $x=-2k$
    $\therefore$ $X=\begin{bmatrix}-2k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =1$
    Put $\lambda =-1$ in (1) then
    $\Rightarrow \begin{bmatrix} 4&4\\-2&-2 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
    $4x+4y=0$
    $-2x-2y=0$
    Consider $4x+4y=0$ $\Rightarrow$ $x=-y$
    Let $y=k$ then $x=-k$
    $\therefore$ $X=\begin{bmatrix}-k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =-1$ $\spadesuit$

Exercise

  • Ecercise 3.1 Find the Eigen values and corresponding vectors of the matrix $\begin{bmatrix} 1&4\\2&3 \end{bmatrix}$


  • Ecercise 3.2 Find the Eigen values and corresponding vectors of the matrix $\begin{bmatrix} 1&2\\2&-1 \end{bmatrix}$


  • Ecercise 3.3 Find the Eigen values and corresponding vectors of the matrix $\begin{bmatrix} 1&-1\\0&2 \end{bmatrix}$


  • Ecercise 3.4 Find the Eigen values and corresponding vectors of the matrix $\begin{bmatrix} 1&2\\2&-2 \end{bmatrix}$


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