MIT 18.06 Lecture 21: Eigenvalues and Eigenvectors

Linear Algebra

MIT 18.06

Eigenvalues

Eigenvectors

Understanding eigenvalues and eigenvectors: the key to understanding how matrices transform space

Author

Chao Ma

Published

November 8, 2025

An eigenvector of A is a vector such that $Ax = \lambda x$, meaning A only scales (not rotates) x.

\[ Ax \parallel x\\ Ax=\lambda x \]

For a singular matrix, $\lambda=0$ and x is in the null space.

Special Matrices

Projection Matrix

If P is the projection matrix onto plane A, then:

Any vector on the plane is an eigenvector of P with eigenvalue 1
Any vector perpendicular to plane A is also an eigenvector of P with eigenvalue 0

Permutation Matrix

For a permutation matrix, it permutes the items in the vector into another order. Vectors that have the same items remain unchanged after multiplication.

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

$x_1=\begin{bmatrix}1\\1\end{bmatrix}$, $\lambda=1$

Since A reverses the direction of the vector, if we choose x with the same magnitudes but opposite signs, we obtain another eigenvector corresponding to eigenvalue $-1$:

$x_2=\begin{bmatrix}-1\\1\end{bmatrix}$, $\lambda=-1$

Key Facts

Any $n \times n$ matrix has n eigenvalues (counting algebraic multiplicities, possibly complex)
For a real symmetric (or Hermitian) matrix, all eigenvalues are real, and the eigenvectors corresponding to distinct eigenvalues are orthogonal
The sum of the eigenvalues equals the trace of the matrix:

\[ \operatorname{tr}(A) = \sum_{i=1}^n \lambda_i \]

The product of the eigenvalues equals the determinant of the matrix:

\[ \det(A) = \prod_{i=1}^n \lambda_i \]

Solving $Ax=\lambda x$

Method

Rewrite the eigenvalue equation:

\[ (A-\lambda I)x=\mathbf{0} \]

For non-trivial solutions to exist, $A-\lambda I$ must be singular, so x is in the null space. This requires:

\[ \det (A-\lambda I)=0 \]

This is called the characteristic equation.

Example

Find the eigenvalues and eigenvectors of:

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

Finding $\lambda$:

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

Solving: $(3-\lambda)^2=1$, so $3-\lambda=\pm 1$

\[ \lambda_1=4\\ \lambda_2=2 \]

Finding eigenvectors:

For $\lambda_1=4$:

\[ A-4I=\begin{bmatrix}-1&1\\1&-1\end{bmatrix}\\ x_1=\begin{bmatrix}1\\1\end{bmatrix} \]

For $\lambda_2=2$:

\[ A-2I=\begin{bmatrix}1&1\\1&1\end{bmatrix}\\ x_2=\begin{bmatrix}-1\\1\end{bmatrix} \]

Effect of Shifting the Matrix

If we add $nI$ to matrix A, the eigenvectors remain unchanged while the eigenvalues increase by n:

\[ (A+3I)x=Ax+3x=\lambda x+3x=(\lambda+3)x \]

So the new eigenvalues are $\lambda+3$.

Complex Eigenvalues

Rotation Matrix

For a $90°$ rotation matrix:

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

From the key facts, we know:

\[ \lambda_1+\lambda_2=0\\ \lambda_1\lambda_2=1 \]

Computing the characteristic equation:

\[ \det(A- \lambda I)=\begin{vmatrix}-\lambda &-1\\1& - \lambda \end{vmatrix}=\lambda^2+1=0 \]

\[ \lambda_1=i\\ \lambda_2=-i \]

Both eigenvalues are complex numbers, reflecting the fact that rotation cannot be represented by simple scaling along real directions.

Triangular Matrix

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

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

\[ \lambda_1=\lambda_2=3 \]

For a triangular matrix, the determinant of $A-\lambda I$ is determined by the diagonal only, so the eigenvalues are simply the diagonal entries.

Finding the eigenvector:

\[ (A-3I)x=\begin{bmatrix}0&1\\0&0\end{bmatrix}x=\mathbf{0} \]

Since $0x_1+1x_2=x_2=0$, we must have $x_2=0$ and $x_1$ is free.

\[ x=c\begin{bmatrix}1\\0\end{bmatrix} \]

In this case, we have a repeated eigenvalue $\lambda=3$, but only one independent eigenvector. This indicates that the matrix is not diagonalizable.

--- title: "MIT 18.06 Lecture 21: Eigenvalues and Eigenvectors" author: "Chao Ma" date: "2025-11-08" categories: [Linear Algebra, MIT 18.06, Eigenvalues, Eigenvectors] description: "Understanding eigenvalues and eigenvectors: the key to understanding how matrices transform space" --- {{< video https://www.youtube.com/watch?v=cdZnhQjJu4I >}} **An eigenvector of A is a vector such that $Ax = \lambda x$, meaning A only scales (not rotates) x.** $$ Ax \parallel x\\ Ax=\lambda x $$ For a singular matrix, $\lambda=0$ and x is in the null space. ## Special Matrices ### Projection Matrix If P is the projection matrix onto plane A, then: - Any vector on the plane is an eigenvector of P with eigenvalue 1 - Any vector perpendicular to plane A is also an eigenvector of P with eigenvalue 0 ### Permutation Matrix For a permutation matrix, it permutes the items in the vector into another order. Vectors that have the same items remain unchanged after multiplication. $$ A=\begin{bmatrix}0&1\\1&0\end{bmatrix} $$ - $x_1=\begin{bmatrix}1\\1\end{bmatrix}$, $\lambda=1$ Since A reverses the direction of the vector, if we choose x with the same magnitudes but opposite signs, we obtain another eigenvector corresponding to eigenvalue $-1$: - $x_2=\begin{bmatrix}-1\\1\end{bmatrix}$, $\lambda=-1$ ## Key Facts - Any $n \times n$ **matrix** has n eigenvalues (counting algebraic multiplicities, possibly complex) - For a **real symmetric** (or Hermitian) matrix, all eigenvalues are **real**, and the eigenvectors corresponding to distinct eigenvalues are **orthogonal** - The **sum** of the eigenvalues equals the **trace** of the matrix: $$ \operatorname{tr}(A) = \sum_{i=1}^n \lambda_i $$ - The **product** of the eigenvalues equals the **determinant** of the matrix: $$ \det(A) = \prod_{i=1}^n \lambda_i $$ ## Solving $Ax=\lambda x$ ### Method Rewrite the eigenvalue equation: $$ (A-\lambda I)x=\mathbf{0} $$ For non-trivial solutions to exist, $A-\lambda I$ must be singular, so x is in the null space. This requires: $$ \det (A-\lambda I)=0 $$ This is called the **characteristic equation**. ### Example Find the eigenvalues and eigenvectors of: $$ A=\begin{bmatrix}3&1\\1&3\end{bmatrix} $$ **Finding $\lambda$:** $$ \det(A-\lambda I)=\begin{vmatrix}3-\lambda & 1\\ 1&3-\lambda\end{vmatrix}=(3-\lambda)^2-1=0 $$ Solving: $(3-\lambda)^2=1$, so $3-\lambda=\pm 1$ $$ \lambda_1=4\\ \lambda_2=2 $$ **Finding eigenvectors:** For $\lambda_1=4$: $$ A-4I=\begin{bmatrix}-1&1\\1&-1\end{bmatrix}\\ x_1=\begin{bmatrix}1\\1\end{bmatrix} $$ For $\lambda_2=2$: $$ A-2I=\begin{bmatrix}1&1\\1&1\end{bmatrix}\\ x_2=\begin{bmatrix}-1\\1\end{bmatrix} $$ ## Effect of Shifting the Matrix If we add $nI$ to matrix A, the eigenvectors remain unchanged while the eigenvalues increase by n: $$ (A+3I)x=Ax+3x=\lambda x+3x=(\lambda+3)x $$ So the new eigenvalues are $\lambda+3$. ## Complex Eigenvalues ### Rotation Matrix For a $90°$ rotation matrix: $$ A=\begin{bmatrix}0&-1\\1&0\end{bmatrix} $$ From the key facts, we know: $$ \lambda_1+\lambda_2=0\\ \lambda_1\lambda_2=1 $$ Computing the characteristic equation: $$ \det(A- \lambda I)=\begin{vmatrix}-\lambda &-1\\1& - \lambda \end{vmatrix}=\lambda^2+1=0 $$ $$ \lambda_1=i\\ \lambda_2=-i $$ Both eigenvalues are complex numbers, reflecting the fact that rotation cannot be represented by simple scaling along real directions. ## Triangular Matrix $$ A=\begin{bmatrix}3&1\\0&3\end{bmatrix} $$ $$ \det(A-\lambda I)=\begin{vmatrix}3-\lambda&1\\0&3-\lambda\end{vmatrix}=(3-\lambda)^2=0 $$ $$ \lambda_1=\lambda_2=3 $$ For a triangular matrix, the determinant of $A-\lambda I$ is determined by the diagonal only, so the eigenvalues are simply the diagonal entries. **Finding the eigenvector:** $$ (A-3I)x=\begin{bmatrix}0&1\\0&0\end{bmatrix}x=\mathbf{0} $$ Since $0x_1+1x_2=x_2=0$, we must have $x_2=0$ and $x_1$ is free. $$ x=c\begin{bmatrix}1\\0\end{bmatrix} $$ In this case, we have a **repeated eigenvalue** $\lambda=3$, but only one independent eigenvector. This indicates that the matrix is not diagonalizable.