ASK linear algebra

Question

Concept of linear independent

A. The linear system Ax=b has unique solution for b, then the columns of A are linearly independent

B. The column of the change-of-coordinate matrix P are linearly independent

C. The subset of a linearly-dependent vector set is linearly-dependent

D. If A is diagonalization, then A has linearly-independent columns

E. If AT*A(AT means transpose of A) is invertible, then A has linearly-independent columns

My answer is (A)(B)(E)

(A) A is invertible, so it is true

(B) true

(C) S={(1,0),(2,0)} and subset of S={(1,0)},so it is wrong

(D) A is diagonalization, does not mean A is invertible, so it is wrong

(E) determine of AT*A is not zero, so determine of A is not zero too

Is my answer correct?

Answer 1

For (A), it depends on what the author means by "$Ax=b$ has unique solution (for $b$?)", but the point is that when $A$ is $m\times n$ with $m>n$, we may select $b\notin \operatorname{col}A$, in which case $[\forall x,\forall y,((Ax=b\land Ay=b)\rightarrow x=y)]$ is true and $[\exists z, Az=b]$ is false.

The rest seems fine.