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Suppose $CA=I_n$. Show that the equation $Ax = 0$ has only the trivial solution. Explain why $A$ cannot have more columns than rows.

I really don't even know where to begin with this one.

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  • $\begingroup$ In the equation Ax=0, multiply on both sides by C on the left. What do you get? $\endgroup$
    – nk637
    Feb 22 '15 at 5:53
  • $\begingroup$ That would just be $CAx=0$ right? I don't understand how that shows Ax has only the trivial solution. $\endgroup$
    – Sabien
    Feb 22 '15 at 5:57
  • $\begingroup$ But you are given some more information, CA=I. Can you see why x=0 is the only solution now? $\endgroup$
    – nk637
    Feb 22 '15 at 6:00
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    $\begingroup$ Remember that here both x and 0 are vectors i.e. x= (x1, x2, ..., xn)- a column vector and 0= (0,0,...,0)- a row vector. Multiplying the identity matrix I with x just gives you a row vector of the form (x1, x2, ..., xn). So the equation Ix = 0 is really saying (x1,x2...,xn) = (0,0,...,0) as vectors. Thus, x = 0 (as a vector). Is that more clear? $\endgroup$
    – nk637
    Feb 22 '15 at 6:12
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    $\begingroup$ We are given CA=I. So CAx=0 means Ix=0 which gives you that x is the 0 vector. Its important to remember that x and 0 are vectors here and not just numbers. $\endgroup$
    – nk637
    Feb 22 '15 at 6:17
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Hint: Suppose $Ax = 0$. Now multiply both sides on the left by $C$.

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  • $\begingroup$ Yes, you are right, I looked at $A$ as it is a square matrix. $\endgroup$ Feb 22 '15 at 6:02
  • $\begingroup$ You get $C(Ax)=0$ but I'm not seeing how this proves anything. My thought is this but I'm not sure if it's right: That can be rewritten as $CAx = 0$. Since we know $CA$ is equal to the identity, then $x$ has to be $0$. Is that right? $\endgroup$
    – Sabien
    Feb 22 '15 at 6:12
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    $\begingroup$ @Sabien Yes, that's right. $C(Ax) = 0 \implies (CA) x = 0 \implies I_n x = 0 \implies x = 0$. $\endgroup$
    – littleO
    Feb 22 '15 at 6:25
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As roundabout alternatives, you could think of the problem in terms of linear independence (or one-to-oneness) if it is more intuitive for you:

First, note that the columns of $I_n$ are linearly independent ($I_n$ has a pivot position in every column).

Since, $CA = I_n$, then the columns of $A$ must be linearly independent.

To see why this is true, let $B = [\mathbf{b}_1 \dotsb \mathbf{b}_n]$ be a matrix whose columns are linearly dependent.

Since the columns of $B$ are linearly dependent, there exists a vector $\mathbf{b}_j$ in $\{\mathbf{b}_1, \dotsc, \mathbf{b}_n\}$ that can be written as a linear combination of the other vectors in $\{\mathbf{b}_1, \dotsc, \mathbf{b}_n\}$. Because $CB = [C \mathbf{b}_1 \dotsb C\mathbf{b}_j \dotsb C \mathbf{b}_n]$, if $\mathbf{b}_j$ can be written as a linear combination of the other vectors in $\{\mathbf{b}_1, \dotsc , \mathbf{b}_n\}$, then $C\mathbf{b}_j$ can be written as a linear combination of the other vectors in $CB$. So the columns of $CA$ are linearly independent only if the columns of $A$ are linearly independent.

Finally, since the columns of $A$ are linearly independent if and only if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution, $CA = I_n$ only if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.

Equivalently, you could explain it in the context of matrix multiplication as compositions of mappings and one-to-oneness.

Let $T(\mathbf{x})= A\mathbf{x}$, $S(\mathbf{y}) = C\mathbf{y}$, and $R(\mathbf{x}) = S(T(\mathbf{x})) = CA\mathbf{x}$.

If there exists $\mathbf{p} \ne \mathbf{q}$ such that $T(\mathbf{p}) = T(\mathbf{q})$ ($T$ is not one-to-one), then since $R(\mathbf{x}) = S(T(\mathbf{x}))$, $R(\mathbf{p}) = R(\mathbf{q})$ ($R$ is not one-to-one). In other words, $R$ is one-to-one only if $T$ is one-to-one.

In this case, because $R(\mathbf{x}) = CA\mathbf{x} = I_n \mathbf{x} = \mathbf{x}$, $R(\mathbf{u}) = R(\mathbf{v}) \Leftrightarrow \mathbf{u} = \mathbf{v}$, so $R$ is one-to-one and, thus, $T$ is one-to-one.

Then since $T$ is one-to-one if and only if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution, $CA = I_n$ only if $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.

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I think another useful way to look at this problem is to imagine $C $ and $A$ as functions.
You might have seen that if $ g\circ f $ is invertible, then it forces $ f $ to be injective(or one to one).
In our case, $ C $ is $g $ and $f $ is $ A $, and so $ A $ is injective, hence: $ Ker(A)=\{\vec{0}\} $

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