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Linear Algebra - True or False if $Ay = B$ and $Az=B$ then y = z? Why or why not? you may give a counter example to prove it false.

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Presumably $B=b$. Would you please provide the full context of your question? E.g., could you say what $A$, $b$, $y$, and $z$ are? Hint: Consider the case where $b$ is the zero vector first. –  Jonas Meyer Dec 18 '10 at 2:39
    
It's false. Think of the simplest possible counterexample. –  Qiaochu Yuan Dec 18 '10 at 2:39
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Thank you for giving me permission to do it with a counterexample, but I'll pass. Instead, I'll ask you this: Is there a unique solution to $2x+3y = 0$? –  Arturo Magidin Dec 18 '10 at 2:39
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5 Answers

HINT $\rm\quad A\ y\ =\ A\ z\ \Rightarrow\ y = z\ \ $ is equivalent to $\rm\ A\ x = 0\ \Rightarrow\ x = 0,\ \ for\ \ \ x = y-z\:$

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If we have $Ay = Az$, all we could conclude is $A(y-z) = 0 \Rightarrow y-z \in $Nullspace$(A)$.

To be more explicit in terms of matrices, if $A$ is a square or a skinny matrix, and if $A$ is full rank, then we can conclude that $y=z$. If not, $y$ need not be equal to $z$. All we can say is $y-z \in $Nullspace$(A)$.

A simplest example is $2 + 3 = 1 + 4$ i.e. $[1,1][2,3]^T = [1,1][1,4]^T$

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It is not true if the null space of $A$ is nontrivial: If $v$ is any nonzero vector such that $Av=0$, then $Av=A0$ and, more generally, $Ay$ and $A(y+v)$ are equal, no matter what $y$ is. On the other hand, it should be easy to check now that the question is true if the null space of $A$ is trivial (i.e., if $Av=0$ only for $v=0$).

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If $A=B=0$, then we can find many $y\ne z$ such that the equations are satisfied!

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You'd have y=z only if A is an injective linear map. In fact, that's the definition of being injective: different values in the domain map through A to different values in the codomain or if two values in the domain have identical images through A then they are equal.

In the general case, this does not hold. As an example, consider the operator T on $\mathbb{R}^3$

T(x,y,z) = (0,y,z)

T maps, for example, every triple of the form (a,2,2) to (0,2,2). Therefore, you not only have two, but the whole infinite set {a * (1,2,2) | a $\in$ $\mathbb{R}$} mapping to (0,2,2) through T.

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