det(lambda∗I−A)=0{\displaystyle det(lambda*I-A)=0} In other words, the determinant of lambda times the identity matrix minus the given transformation matrix.
det(lambda[1001]−[2783])=0{\displaystyle det(lambda{\begin{bmatrix}1&0\0&1\end{bmatrix}}-{\begin{bmatrix}2&7\8&3\end{bmatrix}})=0}
det([lambda−2−7−8lambda−3])=0{\displaystyle det({\begin{bmatrix}lambda-2&-7\-8&lambda-3\end{bmatrix}})=0}
(lambda−2)(lambda−3)−(−8)(−7)=0{\displaystyle (lambda-2)(lambda-3)-(-8)(-7)=0} lambda2−5lambda−50=0{\displaystyle lambda^{2}-5lambda-50=0} lambda=−5{\displaystyle lambda=-5} and lambda=10{\displaystyle lambda=10} These two values are the eigenvalues.
AV=lambda∗V{\displaystyle AV=lambdaV} In other words, a given transformation matrix (A) times the eigenvector (V) equals the eigenvalue (lambda) times the eigenvector (V). This can be rewritten as: lambda∗V−AV=0{\displaystyle lambdaV-AV=0} (lambda∗I−A)V=0{\displaystyle (lambda*I-A)V=0} Where I is the identity matrix.
(−5[1001]−[2783])V=0{\displaystyle (-5{\begin{bmatrix}1&0\0&1\end{bmatrix}}-{\begin{bmatrix}2&7\8&3\end{bmatrix}})V=0} ([−500−5]−[2783])V=0{\displaystyle ({\begin{bmatrix}-5&0\0&-5\end{bmatrix}}-{\begin{bmatrix}2&7\8&3\end{bmatrix}})V=0} [−7−7−8−8]V=0{\displaystyle {\begin{bmatrix}-7&-7\-8&-8\end{bmatrix}}V=0}
Our starting matrix: [−7−7−8−8]V=0{\displaystyle {\begin{bmatrix}-7&-7\-8&-8\end{bmatrix}}V=0} The matrix in reduced row echelon form: [1100]V=0{\displaystyle {\begin{bmatrix}1&1\0&0\end{bmatrix}}V=0}
Our starting matrix equation: [1100]∗[V1V2]=[00]{\displaystyle {\begin{bmatrix}1&1\0&0\end{bmatrix}}*{\begin{bmatrix}V_{1}\V_{2}\end{bmatrix}}={\begin{bmatrix}0\0\end{bmatrix}}} Writing the matrix as a linear equation: V1+V2=0{\displaystyle V_{1}+V_{2}=0} V1=−V2{\displaystyle V_{1}=-V_{2}} Write the components as an eigenvector: [−11]{\displaystyle {\begin{bmatrix}-1\1\end{bmatrix}}}
[78]{\displaystyle {\begin{bmatrix}7\8\end{bmatrix}}}
P^-1 * A * P = D Where P is the matrix of eigenvectors, A is the given matrix, and D is the diagonal matrix of A.
[−1718]{\displaystyle {\begin{bmatrix}-1&7\1&8\end{bmatrix}}} Where the first column is the eigenvector of lambda = -5, and the second column is the eigenvector for lambda = 10.
[−8/157/151/151/15]{\displaystyle {\begin{bmatrix}-8/15&7/15\1/15&1/15\end{bmatrix}}} Check out our expert guide to finding the inverse of a 3x3 matrix if you need a refresher!
[−8/157/151/151/15][2783][−1178]=D{\displaystyle {\begin{bmatrix}-8/15&7/15\1/15&1/15\end{bmatrix}}{\begin{bmatrix}2&7\8&3\end{bmatrix}}{\begin{bmatrix}-1&1\7&8\end{bmatrix}}=D} D=[−50010]{\displaystyle D={\begin{bmatrix}-5&0\0&10\end{bmatrix}}} You’re done! You have diagonalized a matrix. For more linear algebra info, check out how to divide matrices and how to transpose a matrix.
