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Suppose $T:\mathbb R^4\rightarrow\mathbb R^4$ is the transformation induced by the following matrix $A$. Determine whether $T$ is one-to-one and/or onto. If it is not one-to-one, show this by providing two vectors that have the same image under $T$. If $T$ is not onto, show this by providing a vector in $\mathbb R^4$ that is not in the range of $T$.

enter image description here

Please help, I reduced the matrix and I can figure out if it is one to one or onto, but I don't know how to get the image under $T$ or whatever it is asking.

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You reduced the matrix? Show us! –  Brian Fitzpatrick Feb 10 at 1:34
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Note that the second column of $T$ is just $-2$ times the first column. So if you take the vector $T.(-2,0,0,0)^T$ and $T.(0,1,0,0)^T$ then they are the same. Thus your problem is solved.

However, in general, the easiest way to do these problems is to see if you can find two different vectors $x$ and $y$ such that $Tx=Ty=0$.

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so the same image under T is: T.(−2,0,0,0)T and T.(0,1,0,0)T ? –  MindOverMatter Feb 10 at 1:39
    
@MindOverMatter: You should check that :). All you will need is matrix multiplication. The answer is yes. –  voldemort Feb 10 at 1:40
    
How did you figure this out though? I multiplied these vectors through and got T.(−2,0,0,0) = ( -6, 6,4,2)....(0,1,0,0) = (-6, 6,4,2) and that is correct. But I do not under understand how you got this, and I do not understand how you get the images under T if it is NOT one to one. –  MindOverMatter Feb 10 at 1:55
    
@MindOverMatter "I do not understand how you got this" -- do you see that $(-6,6,4,2)$ in your matrix? That's what multiplying by a basis vector does; it isolates a column. Similarly working with $(-2,0,0,0)$ gives the first column multiplied by $-2$. –  tabstop Feb 10 at 2:20
    
Okay, I see it. What about " providing a vector in R4 that is not in the range of T". What does this imply? –  MindOverMatter Feb 10 at 3:28
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