Linear Algebra, solution to homogeneous equation

In summary, the person is trying to solve a problem involving diagonalizing a square matrix but is stuck. They found a solution by using the equation x1+(3/4)r = 0, where r is an arbitrary constant.
  • #1
srfriggen
306
5

Homework Statement



The problem has to do with diagonalizing a square matrix, but the part I'm stuck on is this:

Bx=0, where B is the matrix with rows [000], [0,-4, 0], and [-3, 0, -4].

After performing rref on the augmented matrix Bl0, I get rows [1,0,4/3,0], [0,1,0,0], [0000].

I am trying now to solve for column vector x=[x1,x2,x3]^transposed.



Homework Equations





The Attempt at a Solution





I'm stuck. From every other problem like this I would set x3 = r, where r is an arbitrary constant. x2 has a leading entry and I set that equal to zero, which seems to be the only thing I'm getting correct. I know the answer should be [-4,0,3]^transposed, so x1=-4, x2=0, x3=3 but I can't seem to algebraically get there.

Can someone let me know what I'm missing? Thank you!
 
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  • #2
srfriggen said:

Homework Statement



The problem has to do with diagonalizing a square matrix, but the part I'm stuck on is this:

Bx=0, where B is the matrix with rows [000], [0,-4, 0], and [-3, 0, -4].

After performing rref on the augmented matrix Bl0, I get rows [1,0,4/3,0], [0,1,0,0], [0000].

I am trying now to solve for column vector x=[x1,x2,x3]^transposed.



Homework Equations





The Attempt at a Solution





I'm stuck. From every other problem like this I would set x3 = r, where r is an arbitrary constant. x2 has a leading entry and I set that equal to zero, which seems to be the only thing I'm getting correct. I know the answer should be [-4,0,3]^transposed, so x1=-4, x2=0, x3=3 but I can't seem to algebraically get there.

Can someone let me know what I'm missing? Thank you!





Ok, I now see that what works to get the answer is using the equation, x1+(3/4)r = 0, where r is an arbitrary constant. Setting r = 4, I have x1=-3, x2 = 0, and x3 = 4.

So I've figured out how to get to the solution, but I just don't understand why I should have used the equation x1 + (3/4)r = 0.

I'm going to keep plugging away at the intuition but if someone can help explain it to me I would be grateful. I don't like knowing how to do something without understanding what I'm doing.
 
  • #3
Disregard, I've got it.

I realize that the rref form of B is equal to B*, and B*x yields the correct solutions.
 
  • #5
Mark44 said:
Glad we could help out! :biggrin:



hahahaha! :rofl:
 

1. What is a homogeneous equation in linear algebra?

A homogeneous equation in linear algebra is an equation where the right side is equal to zero. In other words, it is a system of linear equations in which all the constant terms are equal to zero.

2. How do you solve a homogeneous equation in linear algebra?

To solve a homogeneous equation, you must first set up a matrix of coefficients and then use Gaussian elimination or other methods to reduce the matrix to row echelon form. The solutions to the system can then be found by back substitution or other methods.

3. What are the properties of solutions to homogeneous equations?

The solutions to homogeneous equations have a few key properties. First, the zero vector is always a solution. Second, if two vectors are solutions, then any linear combination of those vectors is also a solution. Finally, the number of solutions can be determined by the dimension of the null space of the coefficient matrix.

4. Can a homogeneous equation have infinitely many solutions?

Yes, a homogeneous equation can have infinitely many solutions. This occurs when the dimension of the null space of the coefficient matrix is greater than zero, which means there are infinitely many linearly independent solutions.

5. How is solving a homogeneous equation related to finding the basis of a vector space?

The solutions to a homogeneous equation form a vector space, called the null space. The basis of this vector space is the set of linearly independent solutions, which can be found by solving the homogeneous equation and reducing the coefficient matrix to row echelon form. Therefore, solving a homogeneous equation is closely related to finding the basis of a vector space.

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