Can someone help me check if this matrix is diagonalisable?

$\left[\begin{matrix}4 & - 5 \\ 2 & 3\end{matrix}\right]$

May 11, 2018

See below

Explanation:

$\left[\begin{matrix}4 & - 5 \\ 2 & 3\end{matrix}\right]$

From the characteristic equation:

${\lambda}^{2} - T r \left(M\right) \lambda + \det \left(M\right) = 0$

$\implies {\lambda}^{2} - 7 \lambda + 22 = 0$

• ${\lambda}_{1 , 2} = \frac{7 \pm i \sqrt{39}}{2}$

These will deliver 2 distinct eigenvectors so you can diagonalise .

For eigenvectors ${\boldsymbol{\alpha}}_{1 , 2}$ , use $M {\boldsymbol{\alpha}}_{i} = {\lambda}_{i} {\boldsymbol{\alpha}}_{i}$:

$\left[\begin{matrix}4 & - 5 \\ 2 & 3\end{matrix}\right] {\boldsymbol{\alpha}}_{1} = \frac{7 + i \sqrt{39}}{2} {\boldsymbol{\alpha}}_{1}$

• $\implies 4 x - 5 y = \frac{7 + i \sqrt{39}}{2} x$

• $5 y = \frac{8 - 7 - i \sqrt{39}}{2} x$

$\implies y = \frac{1 - i \sqrt{39}}{10} x$

A first eigenvector is:

${\boldsymbol{\alpha}}_{1} = \left[\begin{matrix}1 \\ \frac{1 - i \sqrt{39}}{10}\end{matrix}\right]$

For the second:

$\left[\begin{matrix}4 & - 5 \\ 2 & 3\end{matrix}\right] {\boldsymbol{\alpha}}_{2} = \frac{7 - i \sqrt{39}}{2} {\boldsymbol{\alpha}}_{2}$

• $4 x - 5 y = \frac{7 - i \sqrt{39}}{2} x$

$\implies y = \frac{1 + i \sqrt{39}}{10} x$

A second eigenvector is:

${\boldsymbol{\alpha}}_{2} = \left[\begin{matrix}1 \\ \frac{1 + i \sqrt{39}}{10}\end{matrix}\right]$