Matrix Diagonalization Tricks

Cheatsheet Content

1. Definition of Diagonalizability A square matrix $A$ is diagonalizable if there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^{-1}$. This means $A$ is similar to a diagonal matrix. Equivalently, $A$ is diagonalizable if and only if it has a basis of eigenvectors. 2. Key Condition for Diagonalizability A matrix $A$ is diagonalizable if and only if for every eigenvalue $\lambda$ of $A$, its algebraic multiplicity (AM) equals its geometric multiplicity (GM) . Algebraic Multiplicity (AM): The number of times $\lambda$ appears as a root of the characteristic polynomial $p(\lambda) = \det(A - \lambda I) = 0$. Geometric Multiplicity (GM): The dimension of the eigenspace corresponding to $\lambda$, i.e., $GM(\lambda) = \dim(Nul(A - \lambda I))$. Important Property: For any eigenvalue $\lambda$, $1 \le GM(\lambda) \le AM(\lambda)$. 3. Quick Checks for Diagonalizability a. Distinct Eigenvalues If an $n \times n$ matrix $A$ has $n$ distinct eigenvalues, then $A$ is diagonalizable. Reason: If all eigenvalues are distinct, then $AM(\lambda) = 1$ for all $\lambda$. Since $1 \le GM(\lambda) \le AM(\lambda)$, it implies $GM(\lambda) = 1$ for all $\lambda$. Thus, $AM(\lambda) = GM(\lambda)$ for all eigenvalues. Caveat: The converse is NOT true. A matrix can be diagonalizable even if it has repeated eigenvalues. b. Symmetric Matrices If a matrix $A$ is symmetric (i.e., $A = A^T$), then $A$ is always diagonalizable. Property: Eigenvectors corresponding to distinct eigenvalues of a symmetric matrix are orthogonal. Benefit: For symmetric matrices, we can even find an orthogonal matrix $P$ such that $A = PDP^T = PDP^{-1}$ (orthogonal diagonalization). c. Triangular Matrices (Upper/Lower) For a triangular matrix, the eigenvalues are its diagonal entries. If a triangular matrix has distinct diagonal entries, it's diagonalizable. If it has repeated diagonal entries, you must check $AM$ vs $GM$. Example: $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$. Eigenvalue $\lambda = 1$ with $AM(1)=2$. To find $GM(1)$, consider $A - 1I = \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}$. $Nul(A-I)$ has dimension 1 (one free variable). So $GM(1)=1$. Since $AM(1) \ne GM(1)$, $A$ is not diagonalizable. d. Identity Matrix and Scalar Multiples of Identity The identity matrix $I$ is diagonal, hence diagonalizable. Any scalar multiple $cI$ is also diagonalizable. Example: $A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2I$. This is already diagonal. $\lambda = 2$ has $AM(2)=2$. $A - 2I = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. $Nul(A-2I)$ has dimension 2. So $GM(2)=2$. $AM=GM$. e. Projection Matrices A projection matrix $P$ (i.e., $P^2=P$) is always diagonalizable. Its eigenvalues are always 0 or 1. f. Nilpotent Matrices (If not zero matrix) If a matrix $A$ is nilpotent ($A^k = 0$ for some $k \ge 1$) and not the zero matrix, it is generally NOT diagonalizable. Reason: The only eigenvalue of a nilpotent matrix is 0. If it were diagonalizable, it would be similar to the zero matrix, meaning $A$ itself would have to be the zero matrix. Example: $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. $A^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. $\lambda = 0$ is the only eigenvalue with $AM(0)=2$. $A - 0I = A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. $Nul(A)$ has dimension 1. $GM(0)=1$. Since $AM(0) \ne GM(0)$, $A$ is not diagonalizable. 4. General Strategy for Non-Obvious Cases Calculate Eigenvalues: Find the roots of the characteristic polynomial $p(\lambda) = \det(A - \lambda I) = 0$. Determine the Algebraic Multiplicity ($AM$) for each eigenvalue. Calculate Geometric Multiplicity: For each eigenvalue $\lambda$, find the dimension of the null space of $(A - \lambda I)$. This is $GM(\lambda) = n - \text{rank}(A - \lambda I)$. Compare: If $AM(\lambda) = GM(\lambda)$ for ALL eigenvalues $\lambda$, then the matrix $A$ is diagonalizable. Otherwise, it is not.