Linear Algebra Questions

The first entry in the product Ax is a sum of products.

If A is an m×n matrix whose columns do not span Rm, then the equation Ax=b is inconsistent for some b in Rm.

The equation Ax=b is consistent if the augmented matrix [A|b] has a pivot position in every row.

If a matrix is in reduced row echelon form, then the first nonzero entry in each row is a 1 and has 0s below it.

If {v1,v2,v3} is an orthogonal basis for W, then multiplying v3 by a scalar c gives a new orthogonal basis {v1,v2,cv3}.

Span{a1,a2} contains only the line through a1 and the origin, and the line through the a2 and the origin.

The general least-squares problem is to find an x that makes Ax as close as possible to b.

If b is in the column space of A, then every solution of Ax=b is a least-squares solution.

If the solution to a system of linear equations is given by (4−2z,−3+z,z), then (4,−3,0) is a solution to the system.

If A=QR, where Q has orthonormal columns, then R=Q^TA.

A least-squares solution of Ax=b is a vector x^ such that ||b−Ax||≤||b−Ax^|| for all x in Rn.

The least-squares solution of Ax=b is the point in the column space of A closest to b.

If a linear system has four equations and seven variables, then it must have infinitely many solutions.

If x is not in a subspace W, then x−projWx is not zero.

If y is in the subspace W and its orthogonal complement W⊥, then y must be the zero vector.

The eigenvalues of a projection matrix are −1 and 1.

If W=Span{x1,x2,x3} with {x1,x2,x3} linearly independent, and if {v1,v2,v3} is an orthogonal set in W, then {v1,v2,v3} is a basis for W.

There exists a real 2×2 matrix with the eigenvalues i and 2i.

The Gram-Schmidt process produces from a linearly independent set {x1,...,xp} an orthogonal set {v1,...,vp} with the property that for each k, the vectors v1,...,vk span the same subspace as that spanned by x1,...,xk.

If A is a projection matrix, then A^2=A.