Solution Space - Linear Algebra

Question

For a matrix:\begin{bmatrix}-1&2&3&-3&6&7\\ 1&-1&-2&2&-5&-6\\ -1&1&2&-1&2&4\\ -2&2&4&-2&4&8\\\end{bmatrix}

To solve for whether a specific vector is inside the solution space, you take the reduced-row echelon form and times it by the vector. If it equals 0, it is in the solutionspace.

Can you please explain why this is the case?

Answer 1

I'm not 100% certain I fully understand the question, but I believe I have the general idea. (Correct me if I'm wrong).

I take it by solution space you're referring to the set of vectors that map to the kernel of this matrix, or - in other words - are solutions to the homogeneous case of this matrix:

\begin{bmatrix}-1&2&3&-3&6&7 & | \space0\\ 1&-1&-2&2&-5&-6 & | \space0\\ -1&1&2&-1&2&4 & | \space0\\ -2&2&4&-2&4&8 & | \space0\\\end{bmatrix}

There are a few ways to go about this.

(1) Simply multiply by your vector [4 -3 2 -1 -1 1]^T. If the 0 vector is the product, then it's in the solution space.

(2) Reduce the matrix to row echelon form, and multiply it again. Again, if the 0 vector is the product, then it's in the solution space.

Note that (1) & (2) are equivalent. The only difference is that you reduced it.

However, if you want to formally prove that the vector is in the solution space, which is what I think you're asking, then you need to use the row echelon form of the matrix. (Note that 1 and 2 are sufficient ways to show that the vector is in the solution space).

The row echelon form of this matrix is

\begin{bmatrix}1&0&-1&0&-1&-3\\ 0&1&1&0&-2&-1\\ 0&0&0&1&-3&-2\\ 0&0&0&0&0&0\\\end{bmatrix}

As you can see, the matrix has rank 3, and therefore has 3 free variables. From here you can write a basis for the null space and construct a matrix composed of the basis vectors. Finally, just show that the vector you mentioned is in the range of this matrix.

Answer 2

First of all, if you multiply the matrix by a vector and you get zero, the vector is in the solution space by definition.

Now, if you multiply the vector by a reduced form of the matrix, the vector is in the solution space.

Go back to the motivation of a reducing a matrix. Whenever you reduce a matrix, you are solving a system like here or here.