Norm of a Linear Transformation .... Another question ....

[Math Amateur]

I am reading Hugo D. Junghenn's book: "A Course in Real Analysis" ...

I am currently focused on Chapter 9: "Differentiation on $\mathbb{R}^n$"

I need some help with the proof of Proposition 9.2.3 ...

Proposition 9.2.3 and the preceding relevant Definition 9.2.2 read as follows:

In the above proof Junghenn let's $\mathbf{a}_i = ( a_{i1}, a_{i2}, \ ... \ ... \ , a_{in} )$

and then states that$T \mathbf{x} = ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )$ where $\mathbf{x} = ( x_1, x_2, \ ... \ ... \ x_n )$(Note: Junghenn defines vectors in $\mathbb{R}^n$ as row vectors ... ... )Now I believe I can show $T \mathbf{x}^t = [a_{ij} ]_{ m \times n } \mathbf{x}^t = ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )^t$ ...

... ... as follows:$T \mathbf{x}^t = [a_{ij} ]_{ m \times n } \mathbf{x}^t = \begin{pmatrix} a_{11} & a_{12} & ... & ... & a_{1n} \\ a_{21} & a_{22} & ... & ... & a_{2n} \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ a_{m1} & a_{m2} & ... & ... & a_{mn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ . \\ . \\ x_n \end{pmatrix}$$= \begin{pmatrix} a_{11} x_1 + a_{12} x_2 + \ ... \ ... \ + a_{1n} x_n \\ a_{21} x_1 + a_{22} x_2 + \ ... \ ... \ + a_{2n} x_n \\ ... \\ ... \\ a_{m1} x_1 + a_{m2} x_2 + \ ... \ ... \ + a_{mn} x_n \end{pmatrix}$$= \begin{pmatrix} \mathbf{a}_1 \cdot \mathbf{x} \\ \mathbf{a}_2 \cdot \mathbf{x} \\ . \\ . \\ \mathbf{a}_n \cdot \mathbf{x} \end{pmatrix}$

$= ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )^t$

So ... I have shown$T \mathbf{x}^t = [a_{ij} ]_{ m \times n } \mathbf{x}^t = ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )^t$...How do I reconcile or 'square' that with Junghenn's statement that$T \mathbf{x} = ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )$ where $\mathbf{x} = ( x_1, x_2, \ ... \ ... \ x_n )$(Note: I don't think that taking the transpose of both sides works ... ?)
Hope someone can help ...

Peter

 

[Math Amateur]

I am reading Hugo D. Junghenn's book: "A Course in Real Analysis" ...

I am currently focused on Chapter 9: "Differentiation on $\mathbb{R}^n$"

I need some help with the proof of Proposition 9.2.3 ...

Proposition 9.2.3 and the preceding relevant Definition 9.2.2 read as follows:
View attachment 221403
View attachment 221404

In the above proof Junghenn let's $\mathbf{a}_i = ( a_{i1}, a_{i2}, \ ... \ ... \ , a_{in} )$

and then states that$T \mathbf{x} = ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )$ where $\mathbf{x} = ( x_1, x_2, \ ... \ ... \ x_n )$(Note: Junghenn defines vectors in $\mathbb{R}^n$ as row vectors ... ... )Now I believe I can show $T \mathbf{x}^t = [a_{ij} ]_{ m \times n } \mathbf{x}^t = ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )^t$ ...

... ... as follows:$T \mathbf{x}^t = [a_{ij} ]_{ m \times n } \mathbf{x}^t = \begin{pmatrix} a_{11} & a_{12} & ... & ... & a_{1n} \\ a_{21} & a_{22} & ... & ... & a_{2n} \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... \\ a_{m1} & a_{m2} & ... & ... & a_{mn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ . \\ . \\ x_n \end{pmatrix}$$= \begin{pmatrix} a_{11} x_1 + a_{12} x_2 + \ ... \ ... \ + a_{1n} x_n \\ a_{21} x_1 + a_{22} x_2 + \ ... \ ... \ + a_{2n} x_n \\ ... \\ ... \\ a_{m1} x_1 + a_{m2} x_2 + \ ... \ ... \ + a_{mn} x_n \end{pmatrix}$$= \begin{pmatrix} \mathbf{a}_1 \cdot \mathbf{x} \\ \mathbf{a}_2 \cdot \mathbf{x} \\ . \\ . \\ \mathbf{a}_n \cdot \mathbf{x} \end{pmatrix}$

$= ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )^t$

So ... I have shown$T \mathbf{x}^t = [a_{ij} ]_{ m \times n } \mathbf{x}^t = ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )^t$...How do I reconcile or 'square' that with Junghenn's statement that$T \mathbf{x} = ( \mathbf{a}_1 \cdot \mathbf{x}, \mathbf{a}_2 \cdot \mathbf{x}, \ ... \ ... \ , \mathbf{a}_n \cdot \mathbf{x} )$ where $\mathbf{x} = ( x_1, x_2, \ ... \ ... \ x_n )$(Note: I don't think that taking the transpose of both sides works ... ?)
Hope someone can help ...

Peter

Proposition 9.2.3 does not seem to be displayed ... so I am reloading the image... as follows

Peter

 

[fresh_42]

What you have written is correct. The mistake is not your calculation, but the interpretation of the text.
$T \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ which says nothing about basis, coordinates or matrices. If we want to write it as such, with matrices and vectors, we have to make a choice:

Case 1: $$ x \longmapsto x \cdot T = (x_1,\ldots ,x_n) \cdot \begin{bmatrix} a_{11} &\ldots &a_{1m} \\ \vdots && \vdots \\a_{n1} & \ldots & a_{nm} \end{bmatrix} = [x \cdot \begin{bmatrix}a_{1i} \\ \vdots \\ a_{ni}\end{bmatrix}]_i= \text{ row times column }$$

Case 2: $$ x \longmapsto T \cdot x = \begin{bmatrix} a_{11} &\ldots &a_{1n} \\ \vdots && \vdots \\a_{m1} & \ldots & a_{mn} \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix}= [(a_{i1}, \ldots , a_{in}) \cdot x ]_i= \text{ row times column }$$

In both cases we have "row times column", because this is how matrices are multiplied. It doesn't change.

In the first case, we have $T \in \mathbb{M}(n,m,\mathbb{R})$, i.e.$T$ is represented by an $(n \times m)-$ matrix, if we apply row vectors from the left on the matrix of $T$. Some authors prefer this convention as $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ has the same order as $\mathbb{M}(n,m,\mathbb{R})\,.$

In the second case, we have $T \in \mathbb{M}(m,n,\mathbb{R})$, i.e.$T$ is represented by an $(m \times n)-$ matrix, if we apply column vectors from the right on the matrix of $T$. This is the more common notations, although $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ and $\mathbb{M}(m,n,\mathbb{R})$ reversed the order of $n$ and $m$.

The difference is not $T$, because we haven't changed the linear transformation at all. The difference is how we write it in matrix form, how $T$ is represented by a matrix.

Now Junghenn writes it the same way as you did: vectors on the right, but as column. He did not choose row vectors here, in fact he didn't even care about it. On the right $x$ is automatically a column, and $x^\tau$ a row. Even the notation $\mathbf{a_i} \cdot \mathbf{x}$ is row times column, so it should have been $\mathbf{a_i}^\tau \cdot \mathbf{x}$, resp. vice versa if really written as rows with probably hundreds of ${}^\tau$ throughout the book, but as it is clear, I'd say it's a forgivable sloppiness here. As long as you write $T\cdot x$ then it better should be meant as a column, for otherwise it would have to be $T\cdot x^\tau$. However, here's the crux: As long as I write $Tx$ I have no matrices nor vector components, only a linear transformation on an element of a vector space, so I don't have to bother, how you like to represent them in coordinates: the transposition mark isn't needed! But the proof is about coordinates, so the point at which it switches from a general concept to a coordinate based concept isn't quite clear and some transposition marks may have been forgotten.

Long story short:

• your calculation is right
• it's always row times column
• we have a choice whether $x \mapsto T(x)$ is represented by "coordinates of $T$" times "coordinates of $x$" or by "coordinates of $x$" times "coordinates of $T$"
• $x \mapsto T \cdot x$ in coordinates means $x$ is a column vector
• strictly speaking we have a little sloppy notation as the vector product is $a_i \cdot x := \langle a_i ,x \rangle = a_i^\tau \cdot x$ with columns
• $(T\cdot x)^\tau = x^\tau T^\tau \neq x^\tau T$

And now I hope I haven't confused $n,m$, left, right, row, column, ${}^\tau,{}^{no\,\, \tau}$ or anything else which can be confused. Btw. this is one reason why I don't like coordinates. Much too much business about nothingness!

 

[Math Amateur]

Thanks fresh_42 ...

Most helpful ... would never have understood that from the text ...

THanks again ...

Peter

 

[fresh_42]

Thanks fresh_42 ...

Most helpful ... would never have understood that from the text ...

THanks again ...

Peter

I just saw, that I did confuse the notation a bit the way I wrote the vectors in the cases - of course - and I hope I corrected it now without making even more mistakes. Now the rule for notation is: Case 1 means the vectors which are not explicitly written, i.e. only by $"x"$ are rows, and in case 2 they are columns.

 

[WWGD]

Thanks fresh_42 ...

Most helpful ... would never have understood that from the text ...

THanks again ...

Peter

Yes, you want the product $VW$ (with $V,W$ as $n \times 1$, resp. $1 \times k$ matrices) to be a scalar. Then you want $VW$ to be a $1 \times 1$ matrix. As Fresh pointed out, matrix dimension is given by $(m \times n )(n \times p)$= $m \times p$.

 

[fresh_42]

A good trick is to use " Dimensionality": you want your product of matrices ( incl. vectors as $n \times 1$ matrices) to be a scalar. Then you want $VW$ to be a $1 \times 1$ matrix.

... and $(n \times m) \cdot (m \times p) = (n \times p)$.

 

[WWGD]

... and $(n \times m) \cdot (m \times p) = (n \times p)$.

Yes, sorry, I was about to edit that in.