5.1 Eigenvectors and Eigenvalues

Eigenvector

Nonzero vector $x$ such that

where $\lambda$ is the eigenvalue corresponding to the eigenvector $x$.

Given an eigenvalue $\lambda$, we can rearrange the above equation and solve for its corresponding eigenvector as a solution to the homogeneous system:

Eigenspace

The set of all solutions $x$ to $(A-\lambda_i I)x = 0$ is called the eigenspace of $A$ corresponding to an eigenvalue $\lambda_i$, which includes $\vec 0$ and all the eigenvectors $x$ corresponding to $\lambda_i$.

Theorem 1

The eigenvalues of a triangular matrix are the entries on its main diagonal.

Theorem 2

If $v_1,\dots,v_r$ are eigenvectors that correspond to distinct eigenvalues $\lambda_1,\dots,\lambda_r$ of an $n\times n$ matrix $A$, then the set $\{v_1,\dots,v_r\}$ is linearly independent.

Intuitively, this makes sense because we can’t have two different stretching factors $\lambda_1$ and $\lambda_2$ along the same eigenvector $x$.

Questions

Is $\lambda=3$ and eigenvalue of $A = \begin{bmatrix} 1 & 2 & 2\\ 3 & -2 & 1\\ 0 & 1 & 1 \end{bmatrix}$? If so, find one corresponding eigenvector.

Starting with the definition of an eigenvector:

We rearrange the equation

Now, $\lambda=3$ is an eigenvalue only if the following system has a non-zero solution $x$:

We find a solution by row-reducing the matrix $A-3I$

This system has only the trivial solution $x = \vec 0$, which is not a valid eigenvector by definition. Thus, $3$ is not an eigenvalue of $A$.

Find a basis for the eigenspace corresponding to $\lambda = -2,5$ of the matrix $A = \begin{bmatrix} 1 & 4\\ 3 & 2 \end{bmatrix}$

The eigenspace is the set of all solutions $x$ to the homogeneous equation $(A-\lambda_i I)x=\vec 0$. Using the given eigenvalues:

So the eigenspace corresponding to $\lambda = -2$ is all vectors along $\begin{bmatrix} -1\\1 \end{bmatrix}$

Similarly, for $\lambda = 5$

True or False: To find the eigenvalues of $A$, reduce $A$ to echelon form.

False. Row operations change the characteristic polynomial, thus changing the eigenvalues. We find eigenvalues by solving for the roots of the characteristic polynomial $\det(A-\lambda I)=0$

5.2 The Characteristic Equation

The eigenvalues of matrix $A$ are all scalars $\lambda_i$ such that the homogeneous system

has a nontrivial solution $x \neq \vec 0$. This problem is equivalent to finding all $\lambda_i$ such that the matrix $A-\lambda I$ is not invertible (by Invertible Matrix Theorem the homogeneous system only has a nontrivial solution when the matrix is not invertible). We know that a matrix is not invertible iff:

The above equation is called the characteristic equation. The algebraic multiplicity of an eigenvalue $\lambda_i$ is its multiplicity as a root of the characteristic equation.

Theorem 4

If $n \times n$ matrices $A$ and $B$ are similar, then they have the same characteristic polynomial and hence the same eigenvalues with the same multiplicities.

Note: Similarity is not the same as row equivalence.

Questions

Find the characteristic polynomial and eigenvalues of $A = \begin{bmatrix}7 & -2 \\ 2 & 3\end{bmatrix}$

Given the above characteristic polynomial, we solve for eigenvalues by finding the roots.

We find that $A$ has the eigenvalue $\lambda=5$ with algebraic multiplicity (exponent) $2$.

List the eigenvalues of $A$ with their multiplicities.

By Theorem 1, we know that the eigenvalues of a triangular matrix are the entries on its diagonal. Thus, $\lambda = 1,1,-4,5$.

Find $h$ in the matrix $A$ such that the eigenspace of $\lambda=5$ is $2$-dimensional.

The eigenspace of an eigenvalue is the set of all solutions $x$ to $(A-\lambda_i I)x = 0$.

If $h=6$, then this homogeneous system has $2$ free variables, thus making it $2$-dimensional.

Use a property of determinants to show that $A$ and $A^\top$ have the same characteristic polynomial.

Let $B = A-\lambda I$:

5.3 Diagonalization

Theorem 5: The Diagonalization Theorem

An $n \times n$ matrix $A$ is diagonalizable iff $A$ has $n$ linearly independent eigenvectors.

where the $n$ eigenvectors form the $P$ matrix, and $D$ is a diagonal matrix with the eigenvalues of $A$.

Example: Diagonalize $A = \begin{bmatrix}1 & 3 & 3\\ -3 & -5 & -3\\ 3 & 3 &1\end{bmatrix}$

First, we find the eigenvalues of $A$ as the roots of the characteristic polynomial:

Then we find the eigenvalues corresponding to the two eigenvalues by plugging them into the homogeneous equation $(A-\lambda I)x = 0$ where the nonzero solutions $x$ to this homogeneous system are eigenvectors corresponding to the eigenvalue. By the Diagonalization Theorem, there must be exactly $n$ linearly independent eigenvectors for a matrix to be diagonalizable.

Now we construct $P$ and $D$:

Theorem 6

An $n \times n$ matrix with $n$ distinct eigenvalues is diagonalizable because it guarantees $n$ linearly independent eigenvectors. This is a sufficient condition for a matrix to be diagonalizable. However, it doesn’t mean that a diagonalizable matrix must have $n$ distinct eigenvalues.

Questions

Diagonalize $A=\begin{bmatrix}2 & 3 \\4 & 1\end{bmatrix}$

We find diagonal matrix $D$ such that its entries are the two eigenvalues, and a matrix $P$ such that its $2$ columns are the $2$ linearly independent eigenvectors corresponding to the eigenvalues.

We find the eigenvalues as roots to the characteristic polynomial:

The matrix $A$ has $2$ distinct eigenvalues, which is sufficient for diagonalization by Theorem 6 because it guarantees $2$ linearly independent eigenvectors which guarantees that the matrix is diagonalizable by the Diagonalization Theorem.

We find the two eigenvectors as solutions to the homogeneous equations of their corresponding eigenvalues

Diagonalize $A=\begin{bmatrix} 0 & -4 & -6\\ -1 & 0 & -3\\ 1 & 2 & 5 \end{bmatrix}$, given $\lambda=2,1$

We’re given the eigenvalues but not their multiplicities which is a bit annoying but it’s ok. Let’s just find the eigenvectors as the solution to the homogeneous system

We have $3$ linearly independent eigenvectors, thus the matrix is diagonalizable by the Diagonalization Theorem:

True or False: If $A$ is diagonalizable, then $A$ is invertible.

False. A diagonalizable matrix may have $\lambda =0$ as an eigenvalue, which would make it non-invertible.

5.4 Eigenvectors and Linear Transformations

Theorem 8: Diagonal Matrix Represenation

Suppose $A$ is diagonalizble as $A=PDP^{-1}$. The columns of $P$ form a basis for $\R^n$, thus $D$ is the $P$-coordinate matrix $[T]_P$ for the transformation $x \mapsto Ax$.

Informally, this theorem says that a matrix diagonalization is expressing the matrix as a change of coordinates into the eigenspace, scaling along the axes (eigenvectors), and changing back into the original coordinate space. Where $D$ expresses the transformation $T$ in the eigenbasis $P$.

Thus, we can generalize $T(x) = Ax$ to non-standard basis $\mathcal{B}$:

Proof

We have the eigenbasis $P=\begin{bmatrix}b_1 & \dots & b_n\end{bmatrix}$. We can show that the matrix $D$ is the transformation $T$ in the eigenbasis $P$:

First, we remember that, in the standard basis, the columns $a_i$ of a linear transformation matrix $A$ are defined as $T(e_i)$:

We can then extend this definition beyond the standard basis, to the eigenbasis $P$:

Then, again using the definition $T(x) = Ax$:

The change-of-coordinates matrix $P_{\mathcal{E} \leftarrow P}$ is just $P$, so $P_{P\leftarrow \mathcal{E} } = P^{-1}$

Factoring out both matrices:

The Matrix of a Linear Transformation

Given the coefficients $[x]_\mathcal{B}$ of the basis we can find the coefficients of the transformed $T(x)$ as :

$[T(x)]_\mathcal{B}$:

where $A$ contains the transformed basis vectors

This is just a generalization of $A = [T(e_1)…T(e_n)]$ in the standard basis. Notice that $A$ is not a change-of-coordinates matrix because both sides are coefficients of $\mathcal{B}$-basis vectors.

Note: The set of all matrices similar to matrix $A$ is equivalent to the set of all matrix representations of the transformation $x \mapsto Ax$.

Questions

Let $T: P_2 \to P_2$ by $T(p) = p(0) - p(1)t+p(2)t^2$.

a. Show that $T$ is a linear transformation

First, let’s understand the problem. This linear transformation takes in a polynomial definition with a parameter $t$, for example $p(t) = t$. Then, we can plug this polynomial into the transformation as

To show that $T$ is a linear transformation, we show that it is covered under vector addition and scalar multiplication.

To show vector addition, we show that $T(p+q) = T(p)+T(q)$:

To show scalar multiplication, we show that $T(cp) =cT(p)$:

b. Find $T(p)$ when $p(t) = -2+t$. Is $p$ an eigenvector of $T$?

First, we know that $T(p)$ has the terms $p(0),p(1),p(2)$, so we compute them individually:

Then, we just plug these terms in:

Thus, we find that $p(t) = -2+t$ is an eigenvector of this linear transformation with an eigenvalue of $1$.

c. Find the matrix for $T$ relative to the basis $\{1,t,t^2\}$ for $P_2$

We know that the transformation matrix of a linear transformation is given by the transformed standard basis vectors:

Thus, using the standard basis vectors in $P_2$:

Let $\mathcal{B} = \{b_1,b_2,b_3\}$ be a basis for a vector space $V$. Find $T(2b_1-b_2+4b_3)$ when $T$ is a linear transformation from $V$ to $V$ whose matrix relative to $\mathcal{B}$ is

By Theorem 1, $[T]_\mathcal{B}$ is just the generalization of the transformation matrix $A$ to non-standard basis $\mathcal{B}$ such that

Thus:

Let $T: \R^2\to\R^2$ by $T(x)=Ax$. Find a basis $\mathcal{B}$ for $\R^2$ with the property that $[T]_\mathcal{B}$ is diagonal.

By The Diagonal Matrix Representation Theorem, if a linear transformation $T: x \mapsto Ax$ is defined by a diagonalizable matrix $A=PDP^{-1}$:

Where $P^{-1}$ is the change-of-coordinates matrix from the standard basis into the eigenbasis.

Thus, this problem is equivalent to diagonalizing the matrix $A$ where $P$ is the eigenbasis referred to as $\mathcal{B}$.

To diagonalize $A$, we first find its eigenvalues as the roots to the characteristic polynomial

We then find the eigenvectors as solutions to the homogeneous system

Thus, the diagonalization:

Let $T: P_3 \to P_3$ by $T(p)=p(0)-p(1)t-p(1)t^2+p(0)t^3$

a. Find $T(p)$ when $p(t) = 1+t+t^2+t^3$. Is $p$ an eigenvector of $T$? If $p$ is an eigenvector, what is its eigenvalue?

$T(p)$ has the terms $p(0)$ and $p(1)$, so let’s compute them individually:

Now, plug them into $T(p)$:

Obviously, this polynomial is not an eigenvector of $T$.

b. Find $T(p)$ when $p(t)=t+t^2$. Is $p$ an eigenvector of $T$? If $p$ is an eigenvector, what is its eigenvalue?

Similarly

Thus, (t+t^2) is an eigenvector of $T$ with an eigenvalue of $\lambda = -2$

True or False: If $A$ is similar to $B$, then $A^2$ is similar to $B^2$.

True.

5.5 Complex Eigenvalues

A complex scalar eigenvalue $\lambda\in C^n$ satisfies $\det{(A - \lambda I)} = 0$ iff there is a nonzero vector $x$ in $C^n$ such that $Ax=\lambda x$.

Example: The matrix $A=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$rotates the plane counterclockwise and has no eigenvectors in $\R^2$.

Theorem 9

Let $A$ be a real $2\times 2$ matrix with a complex eigenvalue $\lambda = a -bi$ and an associated eigenvector $v\in C^2$, then

where

Scaling Factor

We compute the scaling factor $r$ of an imaginary eigenvalue $\lambda = a+bi$:

Example: List the eigenvalues $A = \begin{bmatrix}3&3\\-3&3\end{bmatrix}$. Give the scaling factor $r$ and angle of rotation $\phi$

We find the eigenvalues as the roots of the characteristic polynomial

We then find the angle of rotation $\phi$ using the formula:

Questions

4, 10, 16