MLF | Lecture | Week-4

Lecture Outline

Linear Transformations
Eigenvectors and Eigenvalues
Characteristic Polynomial
Computing Eigenvectors
Diagonalization
Tests for Diagonalizability
Fibonacci
Orthogonally Diagonalizable Matrices

Characteristic Polynomial

How do we find the eigenvalues of a matrix \(A\)?

Characteristic Polynomial

\[ \begin{aligned} Ax &= \lambda x \end{aligned} \]

Characteristic Polynomial

\[ \begin{aligned} Ax &= \lambda x\\\\ Ax &= \lambda Ix\\\\ \end{aligned} \]

Characteristic Polynomial

\[ \begin{aligned} Ax &= \lambda x\\\\ Ax &= \lambda Ix\\\\ Ax - \lambda I x &= 0 \end{aligned} \]

Characteristic Polynomial

\[ \begin{aligned} Ax &= \lambda x\\\\ Ax &= \lambda Ix\\\\ Ax - \lambda I x &= 0\\\\ (A - \lambda I)x &= 0 \end{aligned} \]

Characteristic Polynomial

Let \(x\) be an eigenvector with eigenvalue \(\lambda\)
Then \((A - \lambda I)x = 0\)
\(x\) is in the nullspace of \(A - \lambda I\)
The linear transformation corresponding to \(A - \lambda I\) is not one-one
The linear transformation is not invertible
\(|A - \lambda I| = 0\)

Example

\[ A = \begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix} \]

Example

\[ A = \begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= 0\\\\ \end{aligned} \]

Example

\[ A = \begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= 0\\\\ \begin{vmatrix} 3 - \lambda & 1\\ 0 & 2 - \lambda \end{vmatrix} &= 0 \end{aligned} \]

Example

\[ A = \begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= 0\\\\ \begin{vmatrix} 3 - \lambda & 1\\ 0 & 2 - \lambda \end{vmatrix} &= 0\\\\ (3 - \lambda)(2 - \lambda) &= 0 \end{aligned} \]

Example

\[ A = \begin{bmatrix} 3 & 1\\ 0 & 2 \end{bmatrix} \]

\[ \begin{aligned} |A - \lambda I| &= 0\\\\ \begin{vmatrix} 3 - \lambda & 1\\ 0 & 2 - \lambda \end{vmatrix} &= 0\\\\ (3 - \lambda)(2 - \lambda) &= 0\\\\ \lambda &= 2, 3 \end{aligned} \]

Characteristic Polynomial

\[ |A - \lambda I| = (\lambda_1 - \lambda) \cdots (\lambda_n - \lambda) \]

Properties

The sum of the eigenvalues is equal to the trace of the matrix.
The product of the eigenvalues is equal to the determinant of the matrix.

Properties

The sum of the eigenvalues is equal to the trace of the matrix.
The product of the eigenvalues is equal to the determinant of the matrix.

\(\lambda_1 + \cdots + \lambda_n = \text{tr}(A)\)
\(\lambda_1 \cdots \lambda_n = |A|\)

Properties

\[ |A - \lambda I| = (\lambda_1 - \lambda) \cdots (\lambda_n - \lambda) \]

If we set \(\lambda = 0\), we have:

Properties

\[ |A - \lambda I| = (\lambda_1 - \lambda) \cdots (\lambda_n - \lambda) \]

If we set \(\lambda = 0\), we have:

\[ |A| = \lambda_1 \cdots \lambda_n \]

Properties

\[ |A - \lambda I| = (\lambda_1 - \lambda) \cdots (\lambda_n - \lambda) \]

If we set \(\lambda = 0\), we have:

\[ |A| = \lambda_1 \cdots \lambda_n \]

The determinant of a matrix is equal to the product of its eigenvalues.