Finding Value of Parameter that makes Eigenvalue equals to Zero (Matlab)

Question

Lets say I have a 2x2 matrix

  {b+2 b}
  {-4  5}

How should I approach this problem using Matlab to find the values of b where it gives me the eigenvalue is 0?

Answer 1

As Yvon has eloquently stated, you would simply determine the values of b that would make the determinant of that matrix equal to 0. If you recall from linear algebra theory, the eigenvalues of the matrix can be found by solving this equation:

det(A - lambda*I) = 0

A would be the matrix you are finding the eigenvalues for, lambda would be the eigenvalues of your matrix and I is the identity matrix that has the dimensions of n x n, where n has the same number of rows/columns as A. Note that eigenvalues can only be found with square matrices. In addition, linear algebra theory states that you will have n eigenvalues for the matrix. Because you are expressly stating that one of the eigenvalues is 0 and want to solve for b, this simplifies to:

det(A) = 0

You can actually solve this by hand by computing the determinant of the 2 x 2 matrix by simply doing xz - yw, given that your matrix is of this form:

[x   y]
[w   z]

Therefore, in your case, we have:

(b+2)*5 - (b)*(-4) = 0
5*b + 10 + 4*b = 0
9*b + 10 = 0
b = -10/9

In MATLAB, you can do this symbolically using the symbolic math toolbox:

syms b
A = [b + 2 b; -4 5];
detA = det(A);
x = solve(detA == 0, b);

In MATLAB, x thus gives us:

x =

-10/9

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Sidenote

For a matrix to be invertible, either one of these should be met:

• None of the eigenvalues are equal to 0
• The determinant of the matrix is not equal to 0

Because you are forcing one of the eigenvalues to be 0, what you are essentially doing is determining the value of b that would generate an infinite number of solutions if you were to use this matrix and form a 2 x 2 system of equations. You are also finding the value of b that would not allow this matrix to have an inverse.

To double check, if we substitute b = -10/9 into the matrix, we get:

[ 8/9, -10/9]
[  -4,     5]

Finding the determinant of this matrix is indeed 0. Also, one property of the determinant is that if one of the rows is a multiple of another row, the determinant is automatically 0. We can clearly see this as the first row can be obtained by taking the second row and multiplying it by -2/9.