How can I prove that a set is a basis for $\Bbb R^n$?

Given $V :\Bbb R^n \to \Bbb R^n$ is a linear transformations with a standard matrix $A = [V]$.

Let $B = \left[\vec{b_1}, ..., \vec{b_n}\right] \in M_{n\times n}(\Bbb R)$ with a $\operatorname{rank}(B) = n$

So, how would I prove that the set of vectors $D = \{\vec{b_1}, ... ,\vec{b_n}\}$ is a basis for $\Bbb R^n$

and that $B\left( \left[A\vec{b_1}\right]_D,\cdots, \left[A\vec{b_n}\right]_D\right) = AB$

I have tried to go over this question countless times and am completely lost.

• That second statement is not true. – Fryie Mar 8 '16 at 2:16
• Why is the second statement not true? Is there a counter-example that would disprove it? – AlgebraQuestions Mar 8 '16 at 2:59
• I think I just misinterpreted your notation. It's probably true after all, if by the brackets and the $D$ you mean change of basis. – Fryie Mar 8 '16 at 3:45
• @Fryie ; I understood as far as the std basis changing, but didn't get how the inverse made its way into the original equation – AlgebraQuestions Mar 8 '16 at 15:51

Since $B$ has full rank, all its column vectors are linearly independent. But in a space of dimension $n$ any set of $n$ linearly independent vectors is a basis.

For your second statement, it is equivalent to $B(B^{-1}AB)$, since $B^{-1}$ is the change of basis matrix from the standard basis to $D$. Then, just use associativity.

• So the part $[A\vec{b_1}]$ is the standard basis that is changed to D, and then we get the equation : $B(B^{-1})$? – AlgebraQuestions Mar 8 '16 at 15:35
• The part in the parentheses in your statements is equal to $AB$, but expressed in terms of the basis $D$. Since the basis $D$ consists of the column vectors of $B$, converting $AB$ to the basis $D$ is equivalent to multiplying it by $B^{-1}$ on the left. This is a general result in linear algebra about changing bases that you can prove to yourself by just setting up systems of linear equations. – Fryie Mar 8 '16 at 17:13
• Oh, okay. Thank you so much for the help :) – AlgebraQuestions Mar 8 '16 at 17:54

$\operatorname{rank} B=n$ is equivalent to saying that the unique solution to the linear system $$B\left(\begin{matrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{matrix}\right)=0 \Leftrightarrow x_1\vec{b_1}+x_2\vec{b_2}+...+x_n\vec{b_n}=0$$ is the zero vector: $$(x_1,x_2,...x_n)=(0,0,...,0)$$ and this condition is equivalent to the linear independence of the columns of $B$.

• You mean $rank(B) = n$ – Fryie Mar 8 '16 at 1:50
• of course. thank you for mentioning. – KonKan Mar 8 '16 at 1:51
• Unrelatedly, is there a way in MathJax to properly typeset "rank"? – Fryie Mar 8 '16 at 1:53
• not one that i know of. – KonKan Mar 8 '16 at 1:54
• Apparently, the person who edited the OP's post found out. :) it shows up in the edit history – Fryie Mar 8 '16 at 1:59