**Eigenvector:** An eigenvector of a matrix $A$ is a vector that *remains in the same direction* (or exactly opposite direction) after a linear [[Matrix Transformations|transformation]] represented by $A$. Mathematically, an eigenvector $x$ satisfies: $ Ax= \lambda x$ **Eigenvalue:** Each eigenvector has a corresponding eigenvalue $\lambda$, which represents the *scaling factor* applied to the eigenvector by the transformation $A$. This scaling changes the vector's magnitude but not its direction. ## Derivation We can find an eigenvector $x$ when the transformation matrix $A$ is applied on the vector $x$ results in the same vector $x$ just scaled by some factor $\lambda$. $Ax= \lambda x$ where: - $A$: Transformation matrix with shape $(n \times n)$ - $x$: Vector of shape $(1 \times n)$ - $\lambda$: Scalar value *Step 1: Setup the Characteristic Equation* $ \begin{align} Ax &= \lambda x \tag{1}\\[2pt] Ax - \lambda x &= 0 \tag{2}\\[2pt] x*(A- \lambda \mathbf I) &=0 \tag{3} \end{align} $ (3) We need to add the identity matrix to $\lambda$ as subtracting a scalar from a matrix is not defined. The product of $\lambda \mathbf I$ will have $\lambda$ on all diagonal elements and zero elsewhere. *Step 2: Solve for lambda* For the equation to hold, either $x$ or $(A- \lambda \mathbf I)$ has to equal $0$. Since $x=0$ is a trivial solution, where all vectors have zero [[Vector Length|length]] and no direction, we focus on the scenario where $(A-\lambda \mathbf I)=0$. $ \begin{align} A-\lambda \mathbf I &=0 \tag{4}\\ \det(A- \lambda \mathbf I)&=0 \tag{5}\\ \end{align} $ (5) A matrix operation only results in $0$ if its [[Matrix Determinant]] is also $0$. This determinant equation is called the characteristic equation, and solving it gives the eigenvalues $\lambda$. *Step 3: Solve for the eigenvector* Once $\lambda$ is found substitute it back into the following to find the eigenvector $x$. $ (A-\lambda \mathbf I) x=0 $ ## Example Computing the determinant for a high dimensional matrix is complicated. However we can show calculations for the case where $A$ has shape $(2 \times 2)$. $ \begin{align} \det\left( \begin{bmatrix} a&b\\c&d \end{bmatrix} \begin{bmatrix} 1&0\\0&1 \end{bmatrix} *\lambda \right)&=0\\[10pt] \det\left( \begin{bmatrix} a-\lambda &b\\c&d-\lambda \end{bmatrix} \right)&=0\\[12pt] (a-\lambda)*(d-\lambda)-bc&=0\\[10pt] \lambda^2-(a+d)\lambda+(ad-bc)&=0 \end{align} $ For a given matrix $A$ we can solve this equation w.r.t. $\lambda$ with the [[Quadratic Formula]]. Finally, we plug the solution for $\lambda$ into equation (3) to derive vector $x$.