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Let A be a 4×3 matrix and let c=2a1+a2+a3 (a) If N(A) = {0}, what can you conclude about the solutions to the linear system Ax = c?
The "column space" of a matrix is the same thing for the columns and the simplest way to do that take the transpose, so that columns become rows, and reduce that. The null space of linear transformation, A, is the space of all vectors, v, such that Av= 0. Since row reduction does not change that, it is sufficient to look at the row reduced matrix:
Otherwise, there must be some element x_0\in X which is not in the null space of f, and by taking a scalar multiple of it, we may assume that f(x_0)=1. If x is any element of X then x = f(x)x_0 + (x-f(x)x_0), and x-f(x)x_0\in N(f). So every element of X is a sum of something in the 1-dimensional space spanned by x_0 and something in N(f). You ...
Dear Colleagues, Could you please help me in solving the problem: If Z is an (n-1)-dimensional subspace of an n-dimensional vector space X, show that Z is the null space of a suitable linear functional f on X, which is uniquely determined to within a scalar multiple. Regards, Raed.
You have to start by producing a set of vectors that span the subspace. The nullspace of your matrix is the solution space of "Ax=0", so the problem boils down to solving this linear system. A= \begin{bmatrix}1&-1&1 \\ 2&0&1+i \\ 0&1+i&-1\end{bmatrix} \begin{bmatrix}x \\ y \\ z\end{bmatrix}
if b is a non-zero vector in the null space of A T, then A T b=0, that's what it means to be in the null space. So now suppose for contradiction that there does exist a solution to Ax=b then A T Ax = A T b = 0 So either x is 0 or A T A is 0. A T A is |A| 2 and is only 0 if A=0. So there are no non-trivial solutions to Ax=b.
That is, the null space of A is always a subspace of the null space of BA. Now, the other way. Suppose BAu= 0. The B(Au)= 0 and since B is invertible, Au= 0 so we have that the null space of BA is a subspace of the null space of A. That proves that A and BA have exactly the same null spaces so of course they are of the same dimension.
Let (e_n) be a total orthonormal sequence in a seperable Hilbert Space H and define the right shift operator to be the linear operator T:H \\longrightarrow H such that Te_n=e_{n+1} for n=1,2,\\cdots. Explain the name. Find the range, null space, norm and Hilbert Adjoint operator of T.
Nov 26, 2010. #2. First you have to reduce your matrix to row echelon form. The nonzero rows of this reduced matrix form a basis for r o w (A) . What are they? Now the columns of this reduced matrix wih leading 1's identify the pivot columns of your original matrix. And these form a basis for c o l (A) .
A non-square matrix does NOT have an inverse, of course. Your matrix a, above, maps R 3 into R 2 so obviously, its null space is non-trivial. In this case, its null space is the one dimensional subspace spanned by <1, -2, 1>. That also means that, given a two by two matrix c, there exist an infinite number of two by three matrices, b, such that ...