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I have the following homework question:

True or False: If $B=\{b_1,...,b_n \}$ is a base of $R^n$ and for any $1\le i \le n$ exists $v$ so $Av=b_i$ then $A$ is invertible

I feel this is true but I can't seem to figure out how to prove it. Could someone help me out with this?

Thanks a lot.

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  • $\begingroup$ Does it hold for all $i$? $\endgroup$
    – Ellya
    Commented Apr 8, 2014 at 20:00
  • $\begingroup$ @ellya Yes, it does $\endgroup$
    – Jason
    Commented Apr 8, 2014 at 20:01
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    $\begingroup$ Yes, assuming $A$ is an $n \times n$ matrix. The image contains the basis $B$, so it is surjective. Use the dimension theorem to conclude that $A$ is injective as well. $\endgroup$
    – Ink
    Commented Apr 8, 2014 at 20:06

3 Answers 3

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Let $V$ be a matrix which has each column $i$ as $ v_i$ then $AV=B$, where $B$ has each column $i$ as $ b_i$, since all $b_i$ are linearly independent, the determinant of $B$ is non zero, thus $det (AV)=det (B)\ne0 \Rightarrow det (A)det (V)\ne 0 \Rightarrow det (A)\ne 0$, thus $A$ is invertible.

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  • $\begingroup$ I like this solution a lot. Thanks $\endgroup$
    – Jason
    Commented Apr 8, 2014 at 20:24
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    $\begingroup$ No worries, it's always good to try and do things in a simple way $\endgroup$
    – Ellya
    Commented Apr 8, 2014 at 20:28
  • $\begingroup$ I like your answer for simplicity :) But decided to give the answer to Ant (simply for amount of effort in his answer) Thanks a lot anyway $\endgroup$
    – Jason
    Commented Apr 8, 2014 at 20:50
  • $\begingroup$ yep! seems like mine was a bit cheeky, you'd learn more from his! $\endgroup$
    – Ellya
    Commented Apr 8, 2014 at 20:52
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Call $v_i$ the $i$-th solution, ao $Av_i = b_i$

We want to show that they are linearly independent.

$$a_1v_1 + \dots + a_n v_n = 0$$ Multiplying both sides by $A$, we get

$$a_1 Av_1 + \dots + a_n Av_n = a_1 b_1 + \dots + a_nb_n = 0$$

Since $b_i$ are linearly independent, so are $v_i$

So the rank of $A$ is $n$, and $A$ is invertible.

Let's show this.

If $v_i$ are indipendent, they form a basis. So suppose $$Ax = 0$$ and write $$ x = a_1v_1 + \dots + a_nv_n$$$$ \Rightarrow A(a_1v_1 + \dots + a_nv_n) = 0 \Rightarrow a_1Av_1 + \dots + a_nAv_n = 0 \Rightarrow a_1b_1 + \dots + a_nb_n = 0 \Rightarrow $$$$a_i = 0 \ \forall i$$

So $x = 0$.

This implies that $Ker(A) = \{0\}$ and this implies $rank(A) = n$ (by rouchè-capelli theorem, if you want)

EDIT

Let's do that in another, different way.

Be aware that $A [v_1 ; \dots ; v_n] = Av_1 + \dots + Av_n$, where $v_i$ are column vectors. This will help in a moment.

Now, since $b_i$ is a basis, we can find coefficients $c_{i,k}$ such that $\sum c_{i,k} b_i = e_k$ for every $k$, where $e_k$ is the $k$-th vector of the canonical base.

We then solve the systems $Ax_k = e_k$.

$Ax_k = e_k = c_{1,k}b_1 + \dots + c_{n, k}b_n $. We note that $x_k = c_{1, k}v_1 + \dots + c_{n, k}v_n$ is a solution to the previous system!

If we then put on a matrix $B$ all the vectors $x_k$, thought as column vectors, what will happen?

$$AB = A[x_1; \dots; x_n] = Ax_1 + \dots + Ax_n = e_1 + \dots + e_n = I$$

So $B$ is the inverse of $A$! This not only shows that $A$ has an inverse but also gives a way to calculate it (albeit non efficient) and does not rely on any powerful theorem :)

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  • $\begingroup$ This seems like the solution my teacher had in mind. However I have a question, how does proving that $v_i$ are linearly independent suffice with the rank of $A$ to say that it is invertible? Don't we need to show that the actual vectors of $A$ are linearly independent? $\endgroup$
    – Jason
    Commented Apr 8, 2014 at 20:16
  • $\begingroup$ @Jason I edited and added the explaination :-) $\endgroup$
    – Ant
    Commented Apr 8, 2014 at 20:30
  • $\begingroup$ Very cool, thanks for showing the additional steps :) $\endgroup$
    – Jason
    Commented Apr 8, 2014 at 20:36
  • $\begingroup$ @Jason I also added another way of doing that, which I think it's rather interesting; hope it helps! $\endgroup$
    – Ant
    Commented Apr 8, 2014 at 20:42
  • $\begingroup$ Sweet. That's cool that $B$ is the inverse! Took me a little while to wrap my head around all the linear combinations going on and moving around there ;P I would never have thought of that one on my own. $\endgroup$
    – Jason
    Commented Apr 8, 2014 at 20:49
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Yes. This states that the image of $A$ contains the base $B$, therefore the whole space. So $A$ has rank $n$ and is invertible.

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  • $\begingroup$ Hmm, We have not learned the concept of "Image" yet from a quick look up I did i found it means the vectors I can get by multiplying $A$ by a vector. But I don't see the correlation and how this proves the statement. If you could explain the process I will be grateful. Thanks $\endgroup$
    – Jason
    Commented Apr 8, 2014 at 20:09
  • $\begingroup$ Hard to tell what theory I may use then :) what definition do you have for invertibility? $\endgroup$ Commented Apr 8, 2014 at 20:18
  • $\begingroup$ Well, we defined it as everyone does by $AA^-1 = I$. and then used that as a condition to prove all sorts of things that are enough to determine invertibility. However we did not get to images yet, I assume we will prove what you said when we get there. but up till now we have just learned "det" "inverse" "transpose" and stuff like that on matrices with some relation to the linear equations they can relate to. $\endgroup$
    – Jason
    Commented Apr 8, 2014 at 20:23

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