Are all matrices diagonalizable?

\[ A = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \]

\[ A = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= \end{aligned} \]

\[ A = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= \begin{vmatrix} -\lambda & 1\\ -1 & -\lambda \end{vmatrix} \end{aligned} \]

\[ A = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= \begin{vmatrix} -\lambda & 1\\ -1 & -\lambda \end{vmatrix}\\\\ &= \lambda^2 + 1 \end{aligned} \]

\[ A = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= \begin{vmatrix} -\lambda & 1\\ -1 & -\lambda \end{vmatrix}\\\\ &= \lambda^2 + 1 \end{aligned} \]

This polynomial doesn’t have real roots

\[ A = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= \begin{vmatrix} -\lambda & 1\\ -1 & -\lambda \end{vmatrix}\\\\ &= \lambda^2 + 1 \end{aligned} \]

This polynomial doesn’t have real roots

No eigenvalues \(\implies\) no eigenvectors

If the characteristic polynomial of a matrix \(A\) does not have \(n\) eigenvalues, then the matrix is not diagonalizable.

If the characteristic polynomial of a matrix \(A\) does not have \(n\) eigenvalues, then the matrix is not diagonalizable.

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

Are \(v_1\) and \(v_2\) linearly independent?

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

\[ \begin{aligned} \quad \quad \quad \quad \quad \quad \quad \quad \alpha_1 v_1 + \alpha_2 v_2 &= 0 \end{aligned} \]

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

\[ \begin{aligned} \quad \quad \quad \quad \quad \quad \quad \quad \alpha_1 v_1 + \alpha_2 v_2 &= 0\\\\ (A - \lambda_1 I)(\alpha_1 v_1 + \alpha_2 v_2) &= 0 \end{aligned} \]

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

\[ \begin{aligned} \alpha_1 v_1 + \alpha_2 v_2 &= 0\\\\ (A - \lambda_1 I)(\alpha_1 v_1 + \alpha_2 v_2) &= 0\\\\ \alpha_1 (A - \lambda_1 I)v_1 + \alpha_2 (A - \lambda_1 I)v_2 &= 0 \end{aligned} \]

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

\[ \begin{aligned} \alpha_1 v_1 + \alpha_2 v_2 &= 0\\\\ (A - \lambda_1 I)(\alpha_1 v_1 + \alpha_2 v_2) &= 0\\\\ \alpha_1 (A - \lambda_1 I)v_1 + \alpha_2 (A - \lambda_1 I)v_2 &= 0\\\\ \alpha_2 (A - \lambda_1 I)v_2 &= 0 \end{aligned} \]

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

\[ \begin{aligned} \alpha_1 v_1 + \alpha_2 v_2 &= 0\\\\ (A - \lambda_1 I)(\alpha_1 v_1 + \alpha_2 v_2) &= 0\\\\ \alpha_1 (A - \lambda_1 I)v_1 + \alpha_2 (A - \lambda_1 I)v_2 &= 0\\\\ \alpha_2 (A - \lambda_1 I)v_2 &= 0\\\\ \alpha_2 (\lambda_2 v_2 - \lambda_1v_2) &= 0 \end{aligned} \]

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

\[ \begin{aligned} \alpha_1 v_1 + \alpha_2 v_2 &= 0\\\\ (A - \lambda_1 I)(\alpha_1 v_1 + \alpha_2 v_2) &= 0\\\\ \alpha_1 (A - \lambda_1 I)v_1 + \alpha_2 (A - \lambda_1 I)v_2 &= 0\\\\ \alpha_2 (A - \lambda_1 I)v_2 &= 0\\\\ \alpha_2 (\lambda_2 v_2 - \lambda_1v_2) &= 0\\\\ \alpha_2 (\lambda_2 - \lambda_1) v_2 &= 0\\\\ \end{aligned} \]

Let \((\lambda_1, v_1)\) and \((\lambda_2, v_2)\) be two eigenpairs of \(A\)

\(\lambda_1 \neq \lambda_2\)

\[ \begin{aligned} \alpha_1 v_1 + \alpha_2 v_2 &= 0\\\\ (A - \lambda_1 I)(\alpha_1 v_1 + \alpha_2 v_2) &= 0\\\\ \alpha_1 (A - \lambda_1 I)v_1 + \alpha_2 (A - \lambda_1 I)v_2 &= 0\\\\ \alpha_2 (A - \lambda_1 I)v_2 &= 0\\\\ \alpha_2 (\lambda_2 v_2 - \lambda_1v_2) &= 0\\\\ \alpha_2 (\lambda_2 - \lambda_1) v_2 &= 0\\\\ \implies \alpha_2 = \alpha_1 &= 0 \end{aligned} \]

If an \(n \times n\) matrix \(A\) has \(n\) distinct distinct eigenvalues, then it is diagonalizable.

What if a matrix has \(n\) eigenvalues but some of them repeat?

\[ A = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \]

\[ A = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \]

\[ |A - \lambda I| = (\lambda - 2)^2 \]

\(\lambda = 2\) is the only eigenvalue

\[ A = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \] \(\lambda = 2\) is the only eigenvalue of \(A\)

\[ \begin{aligned} (A - 2I)\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= 0 \quad \quad\\\\ \end{aligned} \]

\[ A = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \] \(\lambda = 2\) is the only eigenvalue of \(A\)

\[ \begin{aligned} (A - 2I)\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= 0\\\\ \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1\\v_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix} \end{aligned} \]

\[ A = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \] \(\lambda = 2\) is the only eigenvalue of \(A\)

\[ \begin{aligned} (A - 2I)\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= 0\\\\ \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1\\v_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix}\\\\ v_2 &= 0 \end{aligned} \]

\[ A = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \] \(\lambda = 2\) is the only eigenvalue of \(A\)

\[ \begin{aligned} (A - 2I)\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= 0\\\\ \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1\\v_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix}\\\\ v_2 &= 0 \end{aligned} \] \(\text{span} \left( \begin{bmatrix}1\\0\end{bmatrix} \right)\) is the eigenspace corresponding to \(\lambda = 2\)

\[ A = \begin{bmatrix} 2 & 1\\ 0 & 2 \end{bmatrix} \] \(\lambda = 2\) is the only eigenvalue of \(A\)

\[ \begin{aligned} (A - 2I)\begin{bmatrix} v_1\\ v_2 \end{bmatrix} &= 0\\\\ \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1\\v_2 \end{bmatrix} &= \begin{bmatrix} 0\\ 0 \end{bmatrix}\\\\ v_2 &= 0 \end{aligned} \] \(\text{span} \left( \begin{bmatrix}1\\0\end{bmatrix} \right)\) is the eigenspace corresponding to \(\lambda = 2\)

Since we don’t have two linearly independent eigenvectors, \(A\) is not diagonalizable