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Let's take a rectangular matrix A with rank=r=n

$$A=\begin{bmatrix}4&0\\0&5\\0&0\end{bmatrix}$$

Infinitely many left inverses B exist of the form

$$B=\begin{bmatrix}1/4&0&b13\\0&1/5&b23\end{bmatrix}$$

Coming to the system of linear equations

$$Ax=\begin{bmatrix}b1\\b2\\b3\end{bmatrix}$$

This can have solutions only if b3=0.

But if I take $$b=\begin{bmatrix}4\\5\\5\end{bmatrix}$$

And try to solve Ax=b by multiplying by B (letting b12=b23=0) on both sides I get

$$BAx=Bb$$ $$x=Bb$$ $$x=\begin{bmatrix}1\\1\end{bmatrix}$$

Subsitituting this x back in Ax=b gives $$Ax=\begin{bmatrix}4\\5\\0\end{bmatrix}=\begin{bmatrix}4\\5\\5\end{bmatrix}$$

This implies $$0=5$$

I am just trying to understand why I am getting such a contradiction. I know that for $$b=\begin{bmatrix}4\\5\\5\end{bmatrix}$$ Ax=b is unsolvable, but what went wrong in the algebraic manipulations I did to show 0=5. Was it like a divide by zero thingy?

hgz
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    Let $B$ ($2\times 3$ matrix) be a left inverse of $A$. If $Ax=b,$ then $x=Bb$ (True). However $BAx=Bb\Leftrightarrow B(Ax-b)=0$ does not imply $Ax-b=0$, since the linear transformation $$B:{\mathbb R}^3\rightarrow {\mathbb R}^2$$ has nontrivial kernel. – Pythagoras Jul 24 '21 at 05:46

1 Answers1

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You're getting a contradiction (absurdity) because you start with an absurdity.

$$A\mathbf x=\begin{bmatrix}4&0\\0&5\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4x\\5y\\0\end{bmatrix}=\mathbf b$$

ie, $b_3$ must be $0$ for a solution to exist. Since you take a $\mathbf b$ not of that form, you get an absurdity implying that no solution $\mathbf x$ exists for that choice of $\mathbf b$

Consider the following analogy: let us consider the map $f\colon x\mapsto x^2$ on $\Bbb R$ and find a real solution $x$ to the equation $f(x)=-1$, ie, $x^2+1=0$ which is an absurdity for real $x$, ergo no real solution $x$ exists. Indeed, solutions exists if the map was on $\Bbb C$, namely $x=\pm i$

One more example: Let us try to find a solution to the equation $x+1=x$ where $x\in\Bbb C$. But this gives us $x+1-x=0$, ie, $1=0$, an absurdity, ie, no solution exists on $\Bbb C$. However, if $x\in R$ where $R=(\{0\},+,\cdot)$ is the trivial (zero) ring where $1$ and $0$ mean the same element, it has a solution, namely the only element of $R$

Prasun Biswas
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  • Hmm.. okay thanks a lot Prasun Biswas, that made a lot of sense. I think I made a logical fallacy. I am self-teaching myself mathematics, could you point me out to a catalogue of the most common logical fallacies in mathematics – hgz Jul 24 '21 at 05:28
  • @Steveaustin: Unfortunately, I'm not aware of such a catalogue online, but I did find this with a quick search: https://math.stackexchange.com/questions/1611814/examples-of-fallacies-in-arithmetic-and-or-algebra – Prasun Biswas Jul 24 '21 at 05:33