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I have a question regarding linear transformation. There is a question that asks, Let $A$ be a 6×5 matrix. What must $a$ and $b$ be in order to define $T: \Bbb{R}^a\to \Bbb{R}^b$ by $T(x) = Ax$?

I know that the answer is $a=5$ and $b=6$. I am thinking that since there are 6 vectors in $A$ with five variables each, then the transformation will be a plane in $\Bbb{R}^6$ because there are 6 vectors in $\Bbb{R}^5$. Is that right? Another question is, what if one of the rows in A is all zeros? Will it still produce a plane in $\Bbb{R}^6$?

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    You need to ask, if $A$ is a $6\times 5$ matrix, what sort of an object must $x$ be for the product $Ax$ to be defined (make sense)? Also ask, what sort of an object will $Ax$ be? – 2017-02-10
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    by defined do you mean if it is consistent? for A to be consistent, than the last row should have all zeros, and its solution most be 0. Ax=b, and b will have 6 entries, the last one being 0. is that right? – 2017-02-10
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    Matrix multiplication is defined only when the matrices being multiplied have certain sizes. Specifically the product of $A$ and $B$ is defined if $A$ is $m\times n$ and $B$ is $n\times p$. The issue of multiplication being defined does not have to do with whether the matrices contain zero entries, just with their sizes. – 2017-02-10
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    i dont think we have covered vector multiplication yet. in my book it says that "a transformation T from Rn to Rm is a rule that assigns to each vector x in Rn a vector T(x) in Rm". so each row is a vector, and a 6×5 matrix has 6 rows a 5 columns, the columns being the variables. So when i multiply the variables by the rows, how does that result in a R6? – 2017-02-10
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    The result of applying an $b\times a$ matrix $A$ to a vector $x$ of length $a$ is a vector of length $b$. Each row of matrix $A$ gets multiplied by the vector $x$ of length $a$, resulting in a scalar for that row. There are $b$ rows, so the result of the multiplication is a vector of length $b$, in $\Bbb{R}^b$. Now, one thing that may be bugging you is that the image of $A$, i.e., the set of $Ax$'s for all possible $x$'s, is not going to be *all* of $\Bbb{R}^6$, indeed for some $A$ could be rather small. But don't worry: we write $f:X\to Y$ even when the image of $f$ is smaller than $Y$. – 2017-02-10
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    aaa things are clearer now!! thank you!!!!!!! – 2017-02-10

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