I have some problems understanding the answer in my book for this question:
Let $A$ be any matrix and $k$ be a scalar. Prove that $(kA)^T=kA^T$
Books answer:
"Since k is a scalar (a $1\times 1$ matrix) we have $k^T = k$
therefore $(kA)^T=A^Tk^T=A^Tk=kA^T$"
Now, I understand the proof but I don't think it is valid, here are my reservations:
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How can you view $k$ as a scalar and a matrix at the same time? The operation $kA$ shouldn't even be valid if $k$ is a matrix if $A$ has more than 1 column right?
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And if you can't view $k$ as a matrix then $k^T$ is not defined since the author only has defined the transpose for matrices.
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The author has only defined scalar multiplication for a scalar $k$ and a matrix $A$ as $kA$ and not $Ak$. Doesn't he have to make this definition before using $A^Tk$?
Well, the main thing I am confused about is (1.) and (2.),
(3.) is maybe just a formality I guess.
Edit: Here comes the definitions of matrix transpose and scalar multiplication provided by the book.
Transpose: "We define the transpose of a matrix as:
$A=(a_ij)$ implies $A^T = (a_ji)$ [The subscript ij changes to ji]"
Scalar multiplication he define by showing matrices which I don't know how to do on a computer, but basically he states that multiplying a matrix $A$ with a scalar $k$ is written as $kA$ and corresponds to multiplying each entry of $A$ by $k$.
Best Answer
I agree, this is somewhat an abuse of notation; if we define $k$ as a $1\times 1$ matrix then we do indeed run into problems if we try to use the usual definition of matrix multiplication. And again, unless we define scalar multiplication with a matrix separately to matrix multiplication we do not have that the operation commutes in general (i.e. we would have $[k,\mathbf{A}]\neq 0$ in general).
I would argue that a better way to prove the result would be something like the following:
$$(k\mathbf{A})^{T}_{ij}=(k\mathbf{A})_{ji}=k\mathbf{A}_{ji}$$
Thus we have that:
$$(k\mathbf{A})^{T}=k\mathbf{A}^{T}$$
As required.