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Let A and B be 4 × 4 matrices such that $$AB=I_4$$ let $x_1, x_2, x_3, x_4, x_5$ be vectors in $\mathbb R^4$ that span $\mathbb R^4$, and let $y_1, y_2, y_3$ be vectors in $\mathbb R^4$ that are linearly independent. Prove (accurately) that

  • the vectors $Ax_1, Ax_2, Ax_3, Ax_4, Ax_5$ also span $\mathbb R^4$,
  • the vectors $By_1, By_2, By_3$ are also linearly independent
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  • $\begingroup$ $A$ and $B$ are clearly invertible, and that's all we are told about them. So what do you know about invertible matrices? $\endgroup$ – Arthur Oct 15 '19 at 9:18
  • $\begingroup$ would it be that Ax = 0 has only the solution x = 0 $\endgroup$ – baked goods Oct 15 '19 at 9:28
  • $\begingroup$ @EmanWong That is indeed one property that invertible matrices have. Have you tried anything so far? Do you understand the definitions involved, i.e. what it means for vectors to "span $\Bbb R^4$" and what it means for vectors to be linearly independent? Have you tried using these definitions to start a proof? If so, then where are you getting stuck? $\endgroup$ – Ben Grossmann Oct 15 '19 at 9:41
  • $\begingroup$ @Omnomnomnom I'm just not sure how to even go about proving both of these statements. I'm not sure what condition am i trying to show in order to show that the vectors span $ R^4$ or are linearly independent. $\endgroup$ – baked goods Oct 15 '19 at 22:37
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First, it's important to note that if $AB = I_4$, then (since $A,B$ are square matrices) we have $B = A^{-1}$ and $A = B^{-1}$, which is to say that $$ AB = BA = I_4. $$ For more on that, see this post.

Let's go through the first problem. We are trying to show that $Ax_1,\dots,Ax_5$ span $\Bbb R^4$. By definition, that means the following:

For any vector $b \in \Bbb R^4$, there exist coefficients $c_i$ such that $c_1 Ax_1 + \cdots + c_5 Ax_5 = b$.

With that in mind, we can construct the proof as follows. Begin with a vector $b \in \Bbb R^4$; our goal will be to show that satisfactory coefficients $c_i$ exist. Before writing a formal proof, it is useful to note that $$ c_1 Ax_1 + \cdots + c_5 Ax_5 = b \iff\\ A[c_1 x_1 + \cdots + c_5 x_5] = b \iff\\ c_1 x_1 + \cdots + c_5 x_5 = A^{-1}b $$

That is, we have come to the following insight: if there exist coefficients such that $c_1 x_1 + \cdots + c_5 x_5 = A^{-1}b$, then these same coefficients will satisfy $c_1 Ax_1 + \cdots + c_5 Ax_5 = b$, which is ultimately what we want.

Thus, one proof would be as follows: argue that there exist coefficients $c_i$ for which $c_1 x_1 + \cdots + c_5 x_5 = A^{-1}b$ (how do we know that this is true?). Then, present the series of equations above to conclude that these coefficients also satisfy $c_1 Ax_1 + \cdots + c_5 Ax_5 = b$. Because $b$ was an arbitrary vector in $\Bbb R^4$, this is exactly the conclusion we want: the vectors $Ax_1,\dots,Ax_5$ span $\Bbb R^4$.


If we want to avoid the result that I refer to in the beginning (desirable in the infinite-dimensional case for instance), then the problem is a bit trickier but the idea is essentially the same.

First, select $c_1,\dots,c_5$ such that $$ c_1x_1 + \cdots + c_5 x_5 = Bb. $$ Now, multiply both sides of the equation (from the left) by $A$ to find that $$ A(c_1x_1 + \cdots + c_5 x_5) = A(Bb) \implies\\ c_1(Ax_1) + \cdots + c_5(Ax_5) = (AB)b \implies\\ c_1(Ax_1) + \cdots + c_5(Ax_5) = b $$ so that once again, we can conclude that $Ax_1,\dots,Ax_5$ spans $\Bbb R^4$.

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