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

AI Thread Summary
Effie has accurately determined the eigenvalues of the matrix A, which are λ1 = -3 and λ2 = 2. The corresponding eigenvectors are found by solving the equation A * x = λ * x for each eigenvalue, resulting in eigenvectors of the forms [1, -3] and [-2, 1]. A modal matrix M is constructed from these eigenvectors, yielding M = [1, -2; -3, 1]. The diagonal matrix D, which contains the eigenvalues on its diagonal, is confirmed by the relation D = M^(-1) * A * M. The discussion emphasizes the importance of the characteristic equation in deriving the eigenvalues.
Prove It
Gold Member
MHB
Messages
1,434
Reaction score
20
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 $\displaystyle \begin{align*} D = M^{-1} \, A \, M \end{align*}$...

$\displaystyle \begin{align*} M^{-1} &= \frac{1}{1 \cdot 1 - \left( -2 \right) \cdot \left( -3 \right) } \, \left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \\ &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \\ \\ M^{-1} \, A \, M &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \left[ \begin{matrix} \phantom{-}3 & \phantom{-}2 \\ -3 & -4 \end{matrix} \right] \left[ \begin{matrix} \phantom{-}1 & -2 \\ -3 & \phantom{-}1 \end{matrix} \right] \\ &= -\frac{1}{5}\,\left[ \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right] \left[ \begin{matrix} -3 & -4 \\ \phantom{-}9 & \phantom{-}2 \end{matrix} \right] \\ &= -\frac{1}{5} \, \left[ \begin{matrix} 15 & \phantom{-}0 \\ 0 & -10 \end{matrix} \right] \\ &= \left[ \begin{matrix} -3 & 0 \\ \phantom{-}0 & 2 \end{matrix} \right] \\ &= D \end{align*}$
 
Mathematics news on Phys.org
Prove It said:
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...
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top