Question

(1 point) A is an m×n matrix. Check the true statements below:

Answer 1

Answer:

(B), (C) and (D) are true

Step-by-step explanation:

A) The mapping $x \rightarrow A_x$ is the linear tranformation induced by the matrix A. The image is a vector space generated by the columns of the matrix A, and the range is the dimension of that subspace. So both concepts differ. This statement is False.

B) This statement is True by definition.

C) Col(A) is the set of vectors of R^m that are generated by the columns of A. The columns of A are obtained from multiplying A by the canonical base of $R^n$. if $v \in R^m$ is any vector spanned by the columns of A, lets say $v = c_1 A_1 + c_2 A_2 + ... + c_n A_n$ , where $c_i$ are scalars and $A_i$ is the $i^{th}$ column of A, then Aw = v, with $w = (c_1, c_2, c_3, ... , c_n)$ . Reciprocally, any element of the form Ax is spanned by the columns of A. In fact, $Ax = x_1 A_1 + x_2 A_2 + ... + x_n A_n$ , with  $x = (x_1 , x_2 , ... , x_n)$ . This proves that (c) is True.

D) This is True. Let T be a linear transformation, and let $\lambda \in \mathcal{R}$ , $v, w \in Ker(T)$ . Since T is a linear transformation, it maps 0 to 0, and we have:

1. 0 ∈ ker(T) : T(0) = 0
2. T(v+w) = T(v) + T(w) = 0 + 0 = 0
3. T(λv) = λT(v) = λ*0 = 0

This proves that ker(T) is a vector space, because it is a vector subspace of $R^n$.

E) An mxn matrix can only be multiplied (to its right) by a vector of $R^n$ , thus its null space lives in $R^n$ , not $R^m$. This statement is False.

F) The equation Ax=b being consistent means that that particular element of b is on Col(A) . In order for Col(A) to be $R^m$ , the equation Ax = b needs to be consistent for every element $b \in R^m$ , not just one in particular. A matrix wiht its first column equal to b and the rest of its columns equal to 0, will satisfy that the equation Ax = b is consistent for that b, but Col(A) is <b>, and not $R^m$ . This shows that (F) is False.