[Math] How to solve a matrix multiplication that results in zero? F*x = 0

linear algebramatricesvectors

I would like to solve a system of linear equations, all of which equal to zero. Is there a way to do this using linear algebra without decomposing into separate equations.

The equation is simple, F is a known matrix, lets say a 3×3

x is a vector of unknown variables, for example, x,y,z

0 is a zero vector, in this example 0,0,0

The equation is F*x = 0

What is the best approach to solving this?

If I did inv(F)*F*x = inv(F)*0 the answer I get is x = 0, y = 0, z = 0

Whats the best approach to solving this problem using matrices? Like adding an identity matrix to both sides?

Best Answer

Consider the equation $A\textbf{x} = \textbf{0}$ with $A$ an invertible matrix. Then clearly you can multiply the inverse matrix to both sides and get

$$ A^{-1}A\textbf{x} = A^{-1}\textbf{0} \implies \textbf{x} = \textbf{0} $$

Thus, the zero vector is the unique solution to the equation $A\textbf{x} = \textbf{0}$. So, if your matrix is invertible, then you have already solved the problem.

Assuming that the matrix $A$ is not invertible, then the above manipulation does not work, because $A^{-1}$ doesn't exist. To solve the equation in this case you can write down the augmented matrix $\left [ A \ | \ \textbf{0} \right ]$ and perform Gaussian elimination to find all solutions.

Related Solutions

[Math] Can the product of two non-zero symmetric matrices be anti-symmetric

Of course, it's not possible in dimension $1$. If the dimension $d$ is greater than $2$, then let $A=(a_{l, r})_{1\leq l,r\leq d}$ and $B=(b_{l, r})_{1\leq l,r\leq d}$, with $a_{1,1}=1$, $b_{2,2}= 1$ and all the other entries are $0$. Then $A$ and $B$ are non-zero, symmetric and $AB=0$, which is skew-symmetric.

[Math] How to combine two matrix equations into one

I'll show you the code for doing this with square matrices, and leave the generalization to rectangular matrices to you. First we'll generate a data matrix S:

>> n = 4;
>> S=reshape(1:n*n, [n n]);

Now define the shift-by-one matrix on vectors:

>> E = sparse(2:n, 1:n-1, 1, n, n);

and the corresponding central difference matrix:

>> D = (E' - E) / 2;

Now you can define X and Y gradient matrices, which act on matrices (suitably reshaped as vectors) by:

>> I = speye(n);                      % identity matrix
>> Dx = kron(I,D);
>> Dy = kron(D,I);

and use them in the following way:

>> Sx = reshape( Dx * S(:), [n n]);
>> Sy = reshape( Dy * S(:), [n n]);

i.e. whenever you want to calculate a gradient, you first reshape the matrix S into a vector, apply the appropriate derivative operator, and then reshape the result back into a matrix.

Note that I've defined E, D etc as sparse matrices, for computational efficiency. If you want to see what any of these matrices look like, you can convert them to dense matrices using the function full, for example:

>> full(D)
ans =
         0    0.5000         0         0
   -0.5000         0    0.5000         0
         0   -0.5000         0    0.5000
         0         0   -0.5000         0

Best Answer

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[Math] Can the product of two non-zero symmetric matrices be anti-symmetric

[Math] How to combine two matrix equations into one

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