Diagonalizing Symmetric Matrices: Definition & Examples

Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we define symmetric and diagonal matrices. We then use eigenvalues and eigenvectors to form a very special matrix which is then used to diagonalize a symmetric matrix.

Getting to There from Here

Symmetric matrices appear often in math, science and engineering. In this lesson, we start with a symmetric matrix and show how to get a diagonal matrix. But first, some definitions.

Some Definitions

A diagonal matrix, D, has numbers along the main diagonal and zeros everywhere else. A symmetric matrix, A, has equal numbers in the off-diagonal locations.

The task is to find a matrix P which will let us convert A into D.

Once we get the matrix P, then D = Pt AP. The transpose of P is written as Pt.

This is a lot of terminology to absorb all at once. Let's work through the process step-by-step with actual examples of finding P and Pt.

The Steps for Diagonalizing a Symmetric Matrix

Step 1: Find the eigenvalues of A.

Here's a typical symmetric matrix:


A=[6.8 2.4;2.4 8.2]


See the same number, 2.4, in the off-diagonal locations?

We are going to play with the equation A - λI.

For now, think of λ (lambda) as being a variable like x. And the ''I'' matrix is the identity matrix which is a special diagonal matrix having 1's along the main diagonal.

Substituting for A and I in A - λI:


A-lambda_I_substitution


Multiply through the I matrix by λ


A-lambda_I_multiply_through_the_I_by_lambda


Subtract the two matrices


A-lambda_I_subtract_the_two_matrices


Take the determinant of the resulting matrix


take_the_determinant_of_the_resulting_matrix


Expand the two factors enclosed in parentheses on the right-hand side.


the_determinant_expanding_the_two_factors


Continue to simplify


the_determinant_continuing_to_simplify


The right-hand-side is almost ready to be factored. Just reorganize the terms.


ready_to_be_factored


This is factored as


(lambda-10)(lambda-5)


Now, we set det(A - λI) to 0 and solve for λ. We get


(lambda-10)(lambda-5)=0


Either of the factors (λ - 10) or (λ - 5) could be zero. If (λ - 10) = 0, then λ = 10. We call this λ1. The other possibility is (λ - 5) = 0 which means λ2 = 5. The λ1 and λ2 are the eigenvalues of A.

Step 2: Find the eigenvectors.

A matrix has dimensions. This is the number of rows and number of columns. For example, a 3x2 matrix has 3 rows and 2 columns. The matrix, A, is a 2x2 matrix. If either the number of rows or the number of columns of a matrix is one, we call this matrix a vector. The vectors we will use have 2 rows and 1 column.

What if multiplying a matrix by a certain vector gives the same result as multiplying this vector by an eigenvalue? This special vector is called an eigenvector. We are looking for the eigenvector, v1, which goes with the eigenvector, λ1. The words ''which goes with'' are commonly replaced with ''associated with''. In other words, we are looking for the eigenvector, v1, associated with the eigenvalue, λ1, satisfying


Av_1=lambda_1 v_1


For now, we don't know the numbers in v1. No problem. We will use the letters a and b.

Substituting


[8 2.4;2.4 8.2][a; b]=10[a; b]


Multiplying the matrix times the vector produces two equations. The first equation is


6.8a+2.4b=10a


Bringing all terms to the left:


6.8a+2.4b-10a=0


Simplifying


-32.a+2.4b=0


The second of the two equations is


2.4a+8.2b=10b


Bringing all the terms to the left-hand-side


2.4a+8.2b-10b=0


Simplifying


2.4a-1.8b=0


Multiplying the matrix times the vector gave us two equations:

  • -3.2a + 2.4b = 0
  • 2.4a - 1.8b = 0

Plotting b vs a gives a straight line for each equation. And, the straight lines are the same straight line! Unlike two lines crossing at one point giving a unique solution for a and b, these lines have an infinite number of points in common. The best we can do is to select one of the points and use it to relate a and b.


a=3 and b=4 solves both equations
a=3_and_b=4_solves_both_equations


Try substituting 3 for a and 4 for b in each equation to verify these numbers work. Thus, the eigenvector is


the_vector_v1=[3; 4]


Note, as a practical matter, we could have chosen any point on the line other than the point at the origin. The numbers 3 and 4 are nice because they are whole numbers. But we could have let a = 1 which would give b = 4/3. Both equations are satisfied with this choice as well. Later we will normalize the eigenvector. The normalized eigenvector is unique regardless of which point we choose on the line. The point at the origin provides no information because it says zero times any number is a solution.

To find the other eigenvector, use the second eigenvalue.


Av_2=lambda_2 v_2


As before, we substitute for A and λ with the idea of finding the numbers for the eigenvector, v2.


[6.8 2.4;2.4 8.2][a;b]^t=5[a;b]^t


As before, we get two equations and simplify. The first result is


1.8a+2.4b=0


and the second is


2.4a+3.2b=0


Once again, we have two equations with no unique answer. Two values that work are a = -4 and b = 3.

Thus, the eigenvector, associated with λ = 5 is


v2=[-4 3]^t


Step 3: Normalize the eigenvectors.

Next, we make the length of each eigenvector equal to 1. This is called normalizing.

We find the length of the vector, v1, by taking the square root of the sum of 3 squared and 4 squared.


length_of_v1_=5


v1 surrounded by a pair of vertical lines means ''the length of v1''.

The length of v1 is 5.

To normalize v1, we divide v1 by its length.


u1=v1/length(v1)


Just to be clear, the normalized version of v1 is written as u1.

Substituting we get


u1=[3;4]^t/5


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