## Algebra - Matrices

### Chapter 4. Diagonalization of $2 \times 2$ matrix

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

• Definition 4.1: Let $A$ be a square matrix of order $n$. Then matrix $A$ is said to be diagonalizable if there exists a non singular matrix $P$ such that $PAP^{-1}$ or $P^{-1}AP$ is a diagonal matrix. Otherwise $A$ is said to be non diagonalizable.

Problems 4

• Problem 4.1 Diagonalize the matrix $\begin{bmatrix} 2&4\\0&5\end{bmatrix}$
Solution: Let $A=\begin{bmatrix} 2&4\\0&5\end{bmatrix}$
Consider $|A-\lambda I|=0$ $\Rightarrow$ $\begin{vmatrix} 2-\lambda&4\\0&5-\lambda\end{vmatrix}=0$
$\Rightarrow$ $(2-\lambda)(5-\lambda)=0$
$\Rightarrow$ $\lambda=2,5$
Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 2-\lambda&4\\0&5-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
Put $\lambda =2$ in (1) then
$\Rightarrow \begin{bmatrix} 0&4\\0&3 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
$0x+4y=0$
$0x+3y=0$
Consider $0x+4y=0$ $\Rightarrow$ $y=0$
Let $x=k$
$\therefore$ $X=\begin{bmatrix}k\\0\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =2$
Put $\lambda =5$ in (1) then
$\Rightarrow \begin{bmatrix} -3&4\\0&0 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
$-3x+4y=0$
$0x+0y=0$
Consider $-3x+4y=0$ $\Rightarrow$ $x=\frac{4y}{3}$
Let $y=k$ then $x=\frac{4k}{3}$
$\therefore$ $X=\begin{bmatrix}\frac{4k}{3}\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =5$
Let $P=\begin{bmatrix}1&4\\0&3 \end{bmatrix}$
Then $P^{-1}=\frac{1}{3} \begin{bmatrix} 3&-4\\0&1 \end{bmatrix}$
Consider $P^{-1}AP$ $=\frac{1}{3} \begin{bmatrix} 3&-4\\0&1 \end{bmatrix} \begin{bmatrix} 2&4\\0&5 \end{bmatrix} \begin{bmatrix} 1&4\\0&3 \end{bmatrix}$ $=\begin{bmatrix} 2&0\\0&5 \end{bmatrix}$
Hence $A$ is diagonalizable. $\spadesuit$

• Problem 4.2 Diagonalize the matrix $\begin{bmatrix} 3&4\\-2&-3 \end{bmatrix}$
Solution: Let $A=\begin{bmatrix} 3&4\\-2&-3 \end{bmatrix}$
Consider $|A-\lambda I|=0$ $\Rightarrow$ $\begin{vmatrix} 3-\lambda&4\\-2&-3-\lambda\end{vmatrix}=0$
$\Rightarrow$ $\lambda^2-1=0$
$\Rightarrow$ $\lambda=-1,1$
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 $\lambda =-1$
Let $P=\begin{bmatrix}-2&-1\\1&1 \end{bmatrix}$
Then $P^{-1}= \begin{bmatrix} -1&-1\\1&2 \end{bmatrix}$
Consider $P^{-1}AP$ $= \begin{bmatrix} -1&-1\\1&2 \end{bmatrix} \begin{bmatrix} 3&4\\-2&-3 \end{bmatrix} \begin{bmatrix} -2&-1\\1&1 \end{bmatrix}$ $=\begin{bmatrix} 1&0\\0&-1 \end{bmatrix}$
Hence $A$ is diagonalizable. $\spadesuit$

• Problem 4.3 Diagonalize the matrix $\begin{bmatrix} 5&-3\\-6&2\end{bmatrix}$
Solution: Let $A=\begin{bmatrix} 5&-3\\-6&2 \end{bmatrix}$
Consider $|A-\lambda I|=0$ $\Rightarrow$ $\begin{vmatrix} 5-\lambda&-3\\-6&2-\lambda\end{vmatrix}=0$
$\Rightarrow$ $\lambda^2-7\lambda-8=0$
$\Rightarrow$ $\lambda=-1,8$
Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 5-\lambda&-3\\-6&2-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
Put $\lambda =8$ in (1) then
$\Rightarrow \begin{bmatrix} -3&-3\\-6&-6 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
$-3x-3y=0$
$-6x-6y=0$
Consider $-3x-3y=0$ $\Rightarrow$ $x+y=0$ $\Rightarrow$ $x=-y$
Let $y=k$ then $x=-k$
$\therefore$ $X=\begin{bmatrix}-k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =8$
Put $\lambda =-1$ in (1) then
$\Rightarrow \begin{bmatrix} 6&-3\\-6&3 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
$6x-3y=0$
$-6x+3y=0$
Consider $6x-3y=0$ $\Rightarrow$ $x=\frac{y}{2}$
Let $y=k$ then $x=\frac{k}{2}$
$\therefore$ $X=\begin{bmatrix}\frac{k}{2}\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =-1$
Let $P=\begin{bmatrix}-1&1\\1&2 \end{bmatrix}$
Then $P^{-1}=\frac{1}{3} \begin{bmatrix} -2&1\\1&1 \end{bmatrix}$
Consider $P^{-1}AP$ $= \frac{1}{3} \begin{bmatrix} -2&1\\1&1 \end{bmatrix} \begin{bmatrix} 5&-3\\-6&2 \end{bmatrix} \begin{bmatrix} -1&1\\1&2 \end{bmatrix}$ $=\begin{bmatrix} 8&0\\0&-1 \end{bmatrix}$
Hence $A$ is diagonalizable. $\spadesuit$

• Problem 4.4 Diagonalize the matrix $\begin{bmatrix} 2&6\\0&-1\end{bmatrix}$
Solution: Let $A=\begin{bmatrix} 2&6\\0&-1 \end{bmatrix}$
Consider $|A-\lambda I|=0$ $\Rightarrow$ $\begin{vmatrix} 2-\lambda&6\\0&-1-\lambda\end{vmatrix}=0$
$\Rightarrow$ $(2-\lambda)(-1-\lambda)=0$
$\Rightarrow$ $\lambda=-1,2$
Let $X=\begin{bmatrix}x\\y\end{bmatrix}$ be the Eigen vector such that $AX=\lambda X$ $\Rightarrow AX-\lambda X=0$ $\Rightarrow (A-\lambda)X=0$ $\Rightarrow \begin{bmatrix} 2-\lambda&6\\0&-1-\lambda \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$ ......(1)
Put $\lambda =2$ in (1) then
$\Rightarrow \begin{bmatrix} 0&6\\0&-3 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
$0x+6y=0$
$0x-3y=0$
Consider $0x+6y=0$
Let $x=k$ then $y=0$
$\therefore$ $X=\begin{bmatrix}k\\0\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =2$
Put $\lambda =-1$ in (1) then
$\Rightarrow \begin{bmatrix} 3&6\\0&0 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}=0$
$3x+6y=0$
$0x+0y=0$
Consider $3x+6y=0$ $\Rightarrow$ $x=-2y$
Let $y=k$ then $x=-2k$
$\therefore$ $X=\begin{bmatrix}-2k\\k\end{bmatrix}$ is the Eigen vector corresponding to $\lambda =-1$
Let $P=\begin{bmatrix}1&-2\\0&1 \end{bmatrix}$
Then $P^{-1}= \begin{bmatrix} 1&2\\0&1 \end{bmatrix}$
Consider $P^{-1}AP$ $= \begin{bmatrix} 1&2\\0&1 \end{bmatrix} \begin{bmatrix} 2&6\\0&-1 \end{bmatrix} \begin{bmatrix} 1&-2\\0&1 \end{bmatrix}$ $=\begin{bmatrix} 2&0\\0&-1 \end{bmatrix}$
Hence $A$ is diagonalizable. $\spadesuit$

• Problem 4.5 Diagonalize 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 $\lambda =3$
Let $P=\begin{bmatrix} -1&-2\\1&2 \end{bmatrix}$
Then $P^{-1}$ does not exist.
Hence $A$ is not diagonalizable. $\spadesuit$

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