# Definition: Let $A$ be an $n\times n$ matrix. A scalar $\lambda$ is an eigenvalue of $A$ if a nonzero vector $\bar{x}$ exists such that $A\bar{x}=\lambda\bar{x}$. Such a vector $\bar{x}$ is called an eigenvector of $A$ corresponding to $\lambda$. # Intuition An eigenvector is a vector that remains on its original span after a linear transformation. Eigenvalues represent the scaling factor experienced by the linear transformation. $\overbrace{A\bar{x}}^{\text{matrix-vector multiplication}}=\underbrace{\lambda\bar{x}}_\text{scalar multiplication}$ # Computation To work out difference between matrix-vector multiplication and scalar multiplication, rewrite right hand side as matrix-vector multiplication: $ \lambda = \begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}=\lambda\begin{bmatrix}1&0\\0&1\end{bmatrix}=\lambda I $ $ A\bar{x}=(\lambda I)\bar{x} $ $ A\bar{x}-(\lambda I)\bar{x}=\bar{0} $ $ (A-\lambda I)\bar{x}=\bar{0} $ Because we want a non-zero $\bar{x}$, the only way for matrix-vector multiplication to result in zero is for the [[Determinant|determinant]] to be equal to zero: $ \det(A-\lambda I)= 0$ $ \begin{vmatrix}a_{11}-\lambda&0\\0&a_{22}-\lambda\end{vmatrix}=[(a_{11}-\lambda)(a_{22}-\lambda)]-[0(0)]=0 $ # Examples: Show that $x=\begin{bmatrix}2 \\ 1 \end{bmatrix}$ is an eigenvector of $A=\begin{bmatrix}3&2\\3&-2\end{bmatrix}$ corresponding to $\lambda = 4$ $ \begin{gather} A\bar{x}=\lambda\bar{x}\\ \begin{bmatrix}3&2\\3&-2\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=4\begin{bmatrix}2\\1\end{bmatrix}\\ [3(2)+2(1)]+[3(2)+(-2)(1)]=4(2)+4(1)\\ 8+4=8+4\\ 12=12 \end{gather} $