Eigenvalues and eigenvectors

Some vectors remain on their span after transformation. Because of linear transformation any vector on that same span remains on that span. These vectors are the eigenvalues with corresponding eigenvalues (scalar).

Since linear transformation of an eigenvector $\overline v$ just scales it, it follows:

$$A \overline v = \lambda \overline v = (\lambda I) \overline v = (A-\lambda I)\overline v = \overline 0$$

This expression is always true for $\overline v = 0$ but this is not allowed. So the only way a multiplication of a matrix with a non-zero vector to be zero is with $det(A-\lambda I)=0$. Transformation associated with that matrix squishes space into a lower dimension? (chapter 5/6)

Exmaple

$$\begin{pmatrix} 3-\lambda & 1 \\ 0 & 2-\lambda \end{pmatrix} = (3-\lambda)(2-\lambda) = 0$$

After finding eigenvalue, we need eigenvectors associated with eigenvalue ($\lambda = 2 \space or \space \lambda = 3$)

• Change matrix A and solve linear system for 0 (homogeneous)

Rotation don’t have Eigenvalues

$$\begin{pmatrix} 0& -1 \\ 1 & 0 \end{pmatrix}$$

Shear

Fixes $î$ and moves $\hat j$ one over

$$\begin{pmatrix} 1& 1 \\ 0 & 1 \end{pmatrix}$$

Only eigenvector are all vectors on $î$ with eigenvalue 1.

Scaled matrices

$$\begin{pmatrix} 2& 0 \\ 0 & 2 \end{pmatrix}$$

All vectors are eigenvectors with eigenvalue 2.

Diagonal matrix

All basis are eigenvectors with the values being their eigenvalues (here: 2 and 2)

Rectabgular matrices

Cant have eigenvalues because not diagonizable

Is vector an eigenvector of matrix B?

check with: $Matrix*vector = scaled \space vector$

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$$\begin{pmatrix} 1& 1 \\ 1 & 1 \end{pmatrix} * \begin{pmatrix} 1 \\ 1 \end{pmatrix}= \begin{pmatrix} 2 \\ 2 \end{pmatrix}=2* \begin{pmatrix} 1 \\ 1 \end{pmatrix} =2v$$

So $v$ is an eigenvector with eigenvalue of 2, since transformation only scales v.

The associated eigenspace is the solution set of the linear system $B-2I_2=0$.

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$$B-2I_2= \begin{pmatrix} 1& 1 \\ 1 & 1 \end{pmatrix} -2* \begin{pmatrix} 1& 0 \\ 0 & 1 \end{pmatrix}= \begin{pmatrix} -1& 1 \\ 1 & -1 \end{pmatrix}$$

We solve the linear system and the resulting solution set is the eigenspace of the eigenvalue 2.

But how to find all eigenvalues?

The following must be true $det(A-\lambda I_n)= 0$

Solve: $det(A-\lambda I_n)= (1-\lambda)^2 = \lambda^2-2\lambda = \lambda(\lambda-2)=0$

The two eigenvalues are 2 and 0, now solve both linear systems with eigenvalues

Eigenvalues Upper triangular matrices?

$$\begin{pmatrix} 2-\lambda& 2&0 \\ 0 & \frac{3}{4}-\lambda & 5 \\ 0&0&-3-\lambda \end{pmatrix} = (2-\lambda)(\frac{3}{4}-\lambda)(-3-\lambda)$$

Eigenvalues: $2,\frac{3}{4},-3$, this works for all upper matrices. Eigenvalues are its diagonal entries.

quadratic formular

$$x_{1,2} = \frac{-b \pm \sqrt(b^2-4ac)}{2a}$$

Diagonalizable

Diagonal matrices have nice properties so it’s convenient to describe a linear mapping by a diagonal matrix.

$$\begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}$$

Desribes the scaling by factor a in $x_1$ direction and by factor d in $x_2$-direction. If a or d is negative then this reflects a reflection.

Linear Mapping that use rotations don’t have eigenvectors

Eigenspace

Eigenspace L is a subspace of $R^n$ associated to the eigenvalue. Since an eigenvalue has infinite eigenvectors associated with the same eigenvalue.

Determination of Eigenvalues and Eigenvectors

$$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$

Eigendecompostion

Sometimes it is very useful to write a matrix 𝐴 as a product of other matrices, say

$A=PDP^{-1}$

such a matrix is called a matrix dececomposition of matrix A. This holds for a linear map described by A and D w.r.t different bases. Goal is to find basis for which matrix D is a diagonal matrix. The matrix $P$ is called transformation matrix. The process of obtaining $P$ and $D$ is called diagonalization.

• $D$ is composed of eigenvalues of $A$
• $P$ is composed of eigenvectors of $A$ and are not unique

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