# Difference between revisions of "Eigenvalues and Eigenvectors"

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# Introduction

If you are familiar with matrix multiplication, you know that a matrix times a vector gives out some vector, and most vectors resulting from the product of a vector and a matrix turns out different than the original vector,

i.e. $A\vec{v}=\vec{x}$ where $\vec{v}$ and $\vec{x}$ are different vectors.

What if I told you that there are certain vectors that when multiplied to a matrix, the final vector is the same vector multiplied by a scalar?

i.e. $A\vec{v}=\lambda\vec{v}$, where $\lambda$ is a scalar.

Amazing right? These vectors are called eigenvectors. We can figure out how to compute these vectors from any square (rows and columns are the same) matrix!

Definition: For a square matrix A and a non-zero vector $\vec{v}$, $\vec{v}$is an eigenvector of A if and only if $\lambda\vec{v}=A\vec{v}$, where $\lambda$ is an eigenvalue of A associated with $\vec{v}$.

Before we begin our analysis, here are some concepts you need to be familiar with:

• Determinant
• Is a value attributed to a square matrix.
• Invertibility
• Definition can be found in Matrix
• A square matrix A is not invertible if and only if the determinant of A is zero. Or, $det(A)= 0$ implies that A is not invertible.
• If a square matrix A is invertible, then $A\vec{v} = \vec{k}$ has only one solution $\vec{v}$ for all $\vec{k}$.
• Identity Matrix
• Definition can be found in Matrix

# Eigenvalues

In this section, we will see how certain values will change the entries of an original matrix A so that it can have eigenvectors. Let's try to find the eigenvector given that we know the definition: $A\vec{v}=\lambda\vec{v}$ $A\vec{v}-\lambda\vec{v}=\vec{0}$ $A\vec{v}-\lambda I \vec{v}=\vec{0}$, where I is the identity matrix. $(A-\lambda I)\vec{v}=\vec{0}$

So we have $B\vec{v}=\vec{0}$, where B is the square matrix (A - λI).

If B is invertible, then the only solution $\vec{v}$ to $B\vec{v}=\vec{0}$ is the zero vector. But $\vec{v}$ cannot be zero (non-trivial), so we know that the zero vector cannot be the only solution. Therefore, B is not invertible. Since B is not invertible, we know that its determinant will equal zero!

We have to ask ourselves: what values can make B have a determinant of zero? Well we can take the determinant of $(A-\lambda I)$ (which B represented) and see for what $\lambda$ that determinant will equal zero. Since the only unknown value is λ and the determinant of a n×n matrix is just a nth degree polynomial (we call this the characteristic polynomial), we know this method always works. The values of $\lambda$ are called eigenvalues, and we will work through a simple example.

### Example

Suppose we have $A = \begin{bmatrix} 4 & -5 \\ 2 & -3 \end{bmatrix}$.

To find the eigenvalues of A, we find the determinant of (A - λI): \begin{align} |A-\lambda I| &= \begin{vmatrix} 4-\lambda & -5 \\ 2 & -3-\lambda \end{vmatrix} \\ &= (4-\lambda)(-3-\lambda) + 10 \\ &= \lambda^2 - \lambda - 2 \\ &= (\lambda - 2)(\lambda + 1) = 0 \\ \end{align}

So our eigenvalues are 2 and -1. We write these as λ1 = 2 and λ2 = -1. In general, an n×n matrix will have n eigenvalues because an nth degree polynomial will typically have n solutions (given that there are no repeated solutions).

### The General 2×2 Case

Consider the general 2×2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.

We find the eigenvalues of A by finding the determinant of (A - λI): \begin{align} |A-\lambda I| &= \begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix} \\ &= (a-\lambda)(d-\lambda) - bc \\ &= \lambda^2 - (a+d)\lambda + (ad - bc) = 0 \end{align}

Using the quadratic formula, we see that our eigenvalues are $\lambda_1 = \frac{1}{2} \left( a+d + \sqrt{(a+d)^2 - 4(ad-ac)} \right)$ $\lambda_2 = \frac{1}{2} \left( a+d - \sqrt{(a+d)^2 - 4(ad-ac)} \right)$.

Note that we will typically have two eigenvalues for a 2×2 matrix (except in the case when we have repeating solutions), since we're solving for a second-degree polynomial. Likewise, an n×n matrix will probably have n eigenvalues because an nth degree polynomial will typically have n solutions (once again given that there are no repeated solutions).

### Another way of writing the Characteristic Polynomial

We define the trace of a matrix as the sum of its diagonal entries. So the trace of A (or tr(A)) is a + d. Remember that the determinant of A is ad - bc. We can rewrite the characteristic polynomial of a general 2×2 matrix as \begin{align}0 &= \lambda^2 - (a+d)\lambda + (ad - bc) \\ &= \lambda^2 - \text{tr} (A) \lambda + \text{det} (A) = 0 \end{align}

# Eigenvectors

After finding the eigenvalues of A, we must find the corresponding eigenvectors. Since we already have the equation $(A - \lambda I)\vec{v} = \vec{0}$ available, why don't we use it? Plug one of the value of λ for A into the equation and find vector(s) that satisfy the equation.

We continue our previous example to illustrate this idea.

### Example

Remember that for our matrix $A = \begin{bmatrix} 4 & -5 \\ 2 & -3 \end{bmatrix}$, we have λ1 = 2 and λ2 = -1.

In this case, we will have two eigenvectors, one for each eigenvalue. Note that this is not always true. If for a n×n matrix there are fewer than n eigenvalues (repeated solutions for our polynomial), then at least one eigenvalue will have more than one eigenvector that corresponds to it. However in our case, we have two eigenvalues for a 2×2 matrix, so we know we will definitely have one eigenvector per eigenvalue.

We plug our λ's into the equation $(A - \lambda I)\vec{v} = \vec{0}$ and find our eigenvectors.

λ1 = 2: $\begin{bmatrix} 4-2 & -5 \\ 2 & -3-2 \end{bmatrix} = \begin{bmatrix} 2 & -5 \\ 2 & -5 \end{bmatrix}$ Note that the two rows of the matrix are identical. So what vector, when multiplied by this matrix, gives the zero vector? One that could work is $\vec{v}_1 = \begin{bmatrix} 5 \\ 2 \end{bmatrix}$. This is our first eigenvector.
λ2 = -1: $\begin{bmatrix} 4-(-1) & -5 \\ 2 & -3-(-1) \end{bmatrix} = \begin{bmatrix} 5 & -5 \\ 2 & -2 \end{bmatrix}$ While the two rows of the matrix aren't identical, they are scalar multiples of each other. A vector that works is $\vec{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. This is our second eigenvector.

Thus we have the eigenvalues and the corresponding eigenvectors for A.

### A Property of Eigenvectors

In the previous example, note that there are actually infinitely many eigenvectors. For example, $\vec{v}_2 = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$ works as well. However, all the possible eigenvectors for an eigenvalue are scalar multiples of each other. We defined eigenvalues and eigenvectors by stating $A \vec{v} = \lambda \vec{v}$, where $\vec{v}$ is the eigenvector and λ is the eigenvalue of A.

From here, we see that the vector $c\vec{v}$ for any scalar c is an eigenvector too. \begin{align} A (c\vec{v}) &= c(A \vec{v}) \\ &= c(\lambda \vec{v}) = \lambda (c \vec{v}) \end{align}.

Another way to visualize this is to derive an equation for the eigenvectors in terms of their components. Suppose our eigenvector is $\vec{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$.

Then the equation $(A - \lambda I)\vec{v} = \vec{0}$ can be written as $\begin{bmatrix} a - \lambda & b \\ c & d - \lambda \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

Multiplying this out, we get two equations $(a - \lambda) v_1 + bv_2 = 0$ and $cv_1 + (d - \lambda) v_2 = 0$.

Since they are both equal to 0, we set them equal to each other and solve for v1 and v2. $(a - \lambda) v_1 + bv_2 = cv_1 + (d - \lambda) v_2$ $(a - \lambda) v_1 - cv_1 = (d - \lambda) v_2 - bv_2$ $(a - \lambda - c) v_1 = (d - \lambda - b) v_2$ $\frac{v_1}{v_2} = \frac{d - \lambda - b}{a - \lambda - c}$

Thus the eigenvectors for an eigenvalue are actually a set of vectors. The eigenvector we wrote down is just the basis for this set of vectors.