Effie's question via email about Eigenvalues, Eigenvectors and Diagonalisation

[Prove It]

Effie has correctly found that the eigenvalues of $\displaystyle \begin{align*} A = \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \end{align*}$ are $\displaystyle \begin{align*} \lambda_1 = -3 \end{align*}$ and $\displaystyle \begin{align*} \lambda_2 = 2 \end{align*}$. To find the eigenvectors we solve $\displaystyle \begin{align*} A \,\mathbf{x} = \lambda \, \mathbf{x} \end{align*}$ for each $\displaystyle \begin{align*} \lambda \end{align*}$. For $\displaystyle \begin{align*} \lambda_1 \end{align*}$ we have

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= -3\,\left[ \begin{matrix} x \\ y \end{matrix} \right] \\ \left[ \begin{matrix} \phantom{-}6 & \phantom{-}2 \\ -3 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \\ \left[ \begin{matrix} 6 & 2 \\ 0 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \textrm{ after adding half of row 1 to row 2 in row 2...} \end{align*}$

So we can see that $\displaystyle \begin{align*} 6\,x + 2\,y = 0 \implies y = -3\,x \end{align*}$, so by letting $\displaystyle \begin{align*} x = t \end{align*}$ where $\displaystyle \begin{align*} t \in \mathbf{R} \end{align*}$ we find that the eigenvectors are of the family $\displaystyle \begin{align*} t\,\left[ \begin{matrix} \phantom{-}1 \\ -3 \end{matrix} \right] \end{align*}$. We only need one of these eigenvectors to diagonalise the matrix, so $\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 \\ -3 \end{matrix} \right] \end{align*}$ will do.

For $\displaystyle \begin{align*} \lambda_2 \end{align*}$ we have

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= 2\,\left[ \begin{matrix} x \\ y \end{matrix} \right] \\ \left[ \begin{matrix} \phantom{-}1 & \phantom{-}2 \\ -3 & -6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \\ \left[ \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \textrm{ after adding three lots of row 1 to row 2 in row 2...} \end{align*}$

We can see that $\displaystyle \begin{align*} x + 2\,y = 0 \implies x = -2\,y \end{align*}$. If we let $\displaystyle \begin{align*} y = s \end{align*}$ where $\displaystyle \begin{align*} s \in \mathbf{R} \end{align*}$ we find that the eigenvectors are of the family $\displaystyle \begin{align*} s\,\left[ \begin{matrix} -2 \\ \phantom{-}1 \end{matrix} \right] \end{align*}$. We only need one of these eigenvectors to diagonalise the matrix, so $\displaystyle \begin{align*} \left[ \begin{matrix} -2 \\ \phantom{-}1 \end{matrix}\right] \end{align*}$ will do.

So a modal matrix, whose columns are made up of the eigenvectors, is $\displaystyle \begin{align*} M = \left[ \begin{matrix} \phantom{-}1 & -2 \\ -3 & \phantom{-}1 \end{matrix} \right] \end{align*}$. The spectral (diagonal) matrix has the corresponding eigenvalues on the main diagonal and 0 everywhere else, so $\displaystyle \begin{align*} D = \left[ \begin{matrix} -3 & 0 \\ \phantom{-}0 & 2 \end{matrix} \right] \end{align*}$. We can show that $\displaystyle \begin{align*} D = M^{-1} \, A \, M \end{align*}$...

$\displaystyle \begin{align*} M^{-1} &= \frac{1}{1 \cdot 1 - \left( -2 \right) \cdot \left( -3 \right) } \, \left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \\ &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \\ \\ M^{-1} \, A \, M &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \left[ \begin{matrix} \phantom{-}1 & -2 \\ -3 & \phantom{-}1 \end{matrix} \right] \\ &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \left[ \begin{matrix} -3 & -4 \\ \phantom{-}9 & \phantom{-}2 \end{matrix} \right] \\ &= -\frac{1}{5} \, \left[ \begin{matrix} 15 & \phantom{-}0 \\ 0 & -10 \end{matrix} \right] \\ &= \left[ \begin{matrix} -3 & 0 \\ \phantom{-}0 & 2 \end{matrix} \right] \\ &= D \end{align*}$

 

[chwala]

Effie has correctly found that the eigenvalues of $\displaystyle \begin{align*} A = \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \end{align*}$ are $\displaystyle \begin{align*} \lambda_1 = -3 \end{align*}$ and $\displaystyle \begin{align*} \lambda_2 = 2 \end{align*}$. To find the eigenvectors we solve $\displaystyle \begin{align*} A \,\mathbf{x} = \lambda \, \mathbf{x} \end{align*}$ for each $\displaystyle \begin{align*} \lambda \end{align*}$. For $\displaystyle \begin{align*} \lambda_1 \end{align*}$ we have

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= -3\,\left[ \begin{matrix} x \\ y \end{matrix} \right] \\ \left[ \begin{matrix} \phantom{-}6 & \phantom{-}2 \\ -3 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \\ \left[ \begin{matrix} 6 & 2 \\ 0 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \textrm{ after adding half of row 1 to row 2 in row 2...} \end{align*}$

So we can see that $\displaystyle \begin{align*} 6\,x + 2\,y = 0 \implies y = -3\,x \end{align*}$, so by letting $\displaystyle \begin{align*} x = t \end{align*}$ where $\displaystyle \begin{align*} t \in \mathbf{R} \end{align*}$ we find that the eigenvectors are of the family $\displaystyle \begin{align*} t\,\left[ \begin{matrix} \phantom{-}1 \\ -3 \end{matrix} \right] \end{align*}$. We only need one of these eigenvectors to diagonalise the matrix, so $\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}1 \\ -3 \end{matrix} \right] \end{align*}$ will do.

For $\displaystyle \begin{align*} \lambda_2 \end{align*}$ we have

$\displaystyle \begin{align*} \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= 2\,\left[ \begin{matrix} x \\ y \end{matrix} \right] \\ \left[ \begin{matrix} \phantom{-}1 & \phantom{-}2 \\ -3 & -6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \\ \left[ \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] &= \left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \textrm{ after adding three lots of row 1 to row 2 in row 2...} \end{align*}$

We can see that $\displaystyle \begin{align*} x + 2\,y = 0 \implies x = -2\,y \end{align*}$. If we let $\displaystyle \begin{align*} y = s \end{align*}$ where $\displaystyle \begin{align*} s \in \mathbf{R} \end{align*}$ we find that the eigenvectors are of the family $\displaystyle \begin{align*} s\,\left[ \begin{matrix} -2 \\ \phantom{-}1 \end{matrix} \right] \end{align*}$. We only need one of these eigenvectors to diagonalise the matrix, so $\displaystyle \begin{align*} \left[ \begin{matrix} -2 \\ \phantom{-}1 \end{matrix}\right] \end{align*}$ will do.

So a modal matrix, whose columns are made up of the eigenvectors, is $\displaystyle \begin{align*} M = \left[ \begin{matrix} \phantom{-}1 & -2 \\ -3 & \phantom{-}1 \end{matrix} \right] \end{align*}$. The spectral (diagonal) matrix has the corresponding eigenvalues on the main diagonal and 0 everywhere else, so $\displaystyle \begin{align*} D = \left[ \begin{matrix} -3 & 0 \\ \phantom{-}0 & 2 \end{matrix} \right] \end{align*}$. We can show that $\displaystyle \begin{align*} D = M^{-1} \, A \, M \end{align*}$...

$\displaystyle \begin{align*} M^{-1} &= \frac{1}{1 \cdot 1 - \left( -2 \right) \cdot \left( -3 \right) } \, \left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \\ &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \\ \\ M^{-1} \, A \, M &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \left[ \begin{matrix} \phantom{-}1 & -2 \\ -3 & \phantom{-}1 \end{matrix} \right] \\ &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \left[ \begin{matrix} -3 & -4 \\ \phantom{-}9 & \phantom{-}2 \end{matrix} \right] \\ &= -\frac{1}{5} \, \left[ \begin{matrix} 15 & \phantom{-}0 \\ 0 & -10 \end{matrix} \right] \\ &= \left[ \begin{matrix} -3 & 0 \\ \phantom{-}0 & 2 \end{matrix} \right] \\ &= D \end{align*}$

Correct!...the learner has to first come up with what i call characteristic equation; $(3-λ)(-4-λ)+6=0$ in finding the eigenvalues...