# Linear Algebra - Suppose $CA=I_n$. Show that the equation $Ax = 0$ has only the trivial solution.

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.

• In the equation Ax=0, multiply on both sides by C on the left. What do you get? Feb 22 '15 at 5:53
• That would just be $CAx=0$ right? I don't understand how that shows Ax has only the trivial solution. Feb 22 '15 at 5:57
• But you are given some more information, CA=I. Can you see why x=0 is the only solution now? Feb 22 '15 at 6:00
• 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? Feb 22 '15 at 6:12
• 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. Feb 22 '15 at 6:17

Hint: Suppose $Ax = 0$. Now multiply both sides on the left by $C$.

• Yes, you are right, I looked at $A$ as it is a square matrix. Feb 22 '15 at 6:02
• 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? Feb 22 '15 at 6:12
• @Sabien Yes, that's right. $C(Ax) = 0 \implies (CA) x = 0 \implies I_n x = 0 \implies x = 0$. Feb 22 '15 at 6:25

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.

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: $$Ker(A)=\{\vec{0}\}$$