# Prove matrices multiplied by vectors span a space, and matrices multiplied by vectors are linearly independent

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
• $A$ and $B$ are clearly invertible, and that's all we are told about them. So what do you know about invertible matrices? – Arthur Oct 15 '19 at 9:18
• would it be that Ax = 0 has only the solution x = 0 – baked goods Oct 15 '19 at 9:28
• @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? – Ben Grossmann Oct 15 '19 at 9:41
• @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. – baked goods Oct 15 '19 at 22:37

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 $$\Bbb R^4$$.