Eigenvector , eigenvalue and eigenbasis
Then \(\lambda\) is the eigenvalue of \(A\), and \(x\) is the eigenvector of \(A\).
Geometrically, \(x\) stays in the same direction after linear transformation \(A\), with it’s length multiplied by \(\lambda\).
Since \(\lambda x = \lambda I x\), \(Ax = \lambda x \Leftrightarrow (A - \lambda I)x = 0\).
\(A - \lambda I\) is also a linear transformation.
\[\exists x \quad s.t \quad (A - \lambda I) x = 0 \Leftrightarrow det(A - \lambda I) = 0\]In order to squeeze a vector into zero through linear transformation \(A\), \(A\) has to squeeze a vectorspace into lower dimension, which means it’s determinant is zero.
Eigenbasis
If every basis vectors are eigenvectors they are called \(eigenbasis\).
\(A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}\) transforms \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) to \(\begin{bmatrix} -1 \\ 0 \end{bmatrix}\), and \(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\) into \(\begin{bmatrix} 0 \\ 2 \end{bmatrix}\)
Eigenbasis is useful because it’s coresponding matrix is diagonal. Diagonal matrices are useful because it’s easy to multiply.
\[D = \begin{bmatrix} d_1 & 0 \\ 0 & d_2 \end{bmatrix} \rightarrow D^n = \begin{bmatrix} d_1^n & 0 \\ 0 & d_2^n \end{bmatrix}\]Suppose you have to compute \(A^{100}\). It’s a very hard problem. But if you can find eigenvectors of \(A\), you can easily compute \(A^{100}\) by changing basis to \(A\)’s eigenvectors.
If you choose \(A\)’s basis, it becomes eigenbasis of that vectorspace because only their length changes after multiplying \(A\).
In this new vectorspace’s perspective, \(A\) is a diagonal matrix.
So, in order to compute \(A^{100}\), you can move the coordinate system to which has \(A\)’s eigenvectors as basis, compute \(A_{\mathbb{A}}^{100}\) (which is diagonal), and change the basis back to standard basis.
\[A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} \rightarrow \lambda_1 = 3, v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \lambda_2 = 2, v_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\] \[\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}^{-1} A \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} = A_{\mathbb{A}}\]