# [SOLVED]How is M defined?

#### mathmari

##### Well-known member
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Hey!! Let $1\leq n,m\in \mathbb{N}$ and let $\phi, \psi:\mathbb{R}^n\rightarrow \mathbb{R}^m$ be linear maps. Let $\lambda\in \mathbb{R}$.

Show the following:

1. $M(\phi +\psi )=M(\phi )+M(\psi )$
2. $M(\lambda \phi )=\lambda M(\phi )$

What exactly is $M$, it is not defined in this exercise? Is it a matrix? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey mathmari !!

I don't know what M is either. It doesn't look like a matrix. I think it must have been defined somewhere in your text.

#### mathmari

##### Well-known member
MHB Site Helper
I don't know what M is either. It doesn't look like a matrix. I think it must have been defined somewhere in your text.
Can we show the properties if we suppose that $M$ is a map? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Can we show the properties if we suppose that $M$ is a map? $M$ must be a function that maps functions to some vector space so that addition and multiplication with a real scalar are defined.

Suppose we assume $M$ is an arbitrary map.
Can we find a counter example so that $M$ is not linear? Suppose we pick $\phi: x\mapsto 0$ and $\psi: x\mapsto 0$.
And suppose we pick $M: (\mathbb R^n\to\mathbb R^m)\to \mathbb R$ given by $\chi \mapsto 1$.
Are the linearity conditions satisfied? #### mathmari

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MHB Site Helper
$M$ must be a function that maps functions to some vector space so that addition and multiplication with a real scalar are defined.

Suppose we assume $M$ is an arbitrary map.
Can we find a counter example so that $M$ is not linear? Suppose we pick $\phi: x\mapsto 0$ and $\psi: x\mapsto 0$.
And suppose we pick $M: (\mathbb R^n\to\mathbb R^m)\to \mathbb R$ given by $\chi \mapsto 1$.
Are the linearity conditions satisfied? I got stuck right now, how is the map defined? I saw now that the title of this exercise is "Linear Maps and Matrices". So is maybe $M$ related to matrices? #### Klaas van Aarsen

##### MHB Seeker
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I got stuck right now, how is the map defined?
Let's define $L$ as the set of linear functions in $\mathbb R^n\to\mathbb R^m$, so that $\phi, \psi\in L$.
An example of an element $\phi\in L$ is the function given by $\phi(\mathbf x)=\mathbf 0$ that maps any $\mathbf x\in\mathbb R^n$ to the zero element $\mathbf 0\in \mathbb R^m$ yes? Now $M$ must be a function that has an element like $\phi\in L$ as input.
And its output must be something that we can add, and that we can multiply with a real scalar.
Let's define the co-domain of $M$ as $V$ for which addition and scalar multiplication are defined.
Then $M: L\to V$, isn't it? As an example, let's pick again $\phi\in L$ given by $\phi(\mathbf x)=\mathbf 0$.
Then $M(\phi)$ must be an element of $V$ yes? Let $\mathbf v$ be a non-zero element in $V$.
Continuing the example, we can pick $M(\phi)=\mathbf v$, can't we? Moreover, we can choose $M$ such that for any $\chi\in L$ we have $M(\chi)=\mathbf v\ne\mathbf 0$.

If we pick this $M$, and suppose we have some $\phi,\psi\in L$, what will $M(\phi)$, $M(\psi)$, and $M(\phi+\psi)$ be? I saw now that the title of this exercise is "Linear Maps and Matrices". So is maybe $M$ related to matrices?
Since $M$ is applied to a function and not to a vector, it seems to me that it cannot be a matrix.
It is a 'Map' though, and the question asks to prove that it is a Linear Map. #### mathmari

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MHB Site Helper
Since $M$ is applied to a function and not to a vector, it seems to me that it cannot be a matrix.
It is a 'Map' though, and the question asks to prove that it is a Linear Map. I think $M$ is a transformation matrix with respect to the canonical basis. What do we have in this case then? $M$ is then a linear map, or not? Then we get the following:

1. For $x\in \mathbb{R}^n$ we have the following: $$(M(\phi +\psi ))(x)=M(\phi +\psi )(x) \ \overset{\phi, \psi \text{ linear }}{ = } \ M(\phi(x) +\psi(x) ) \ \overset{M \text{ linear }}{ = } \ M(\phi(x)) +M(\psi(x) )=(M(\phi ))(x)+(M(\psi ))(x)$$
2. For $x\in \mathbb{R}^n$ we have the following: $$(M(\lambda \phi ))(x)=M(\lambda \phi )(x) \ \overset{\phi\text{ linear }}{ = } \ M(\lambda \phi(x) ) \ \overset{M \text{ linear }}{ = } \ \lambda M( \phi(x) )=\lambda (M(\phi ))(x)$$

Is everything correct? Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I think $M$ is a transformation matrix with respect to the canonical basis. What do we have in this case then? $M$ is then a linear map, or not? Then we get the following:

1. For $x\in \mathbb{R}^n$ we have the following: $$(M(\phi +\psi ))(x)=M(\phi +\psi )(x) \ \overset{\phi, \psi \text{ linear }}{ = } \ M(\phi(x) +\psi(x) ) \ \overset{M \text{ linear }}{ = } \ M(\phi(x)) +M(\psi(x) )=(M(\phi ))(x)+(M(\psi ))(x)$$
If we treat $M$ as a matrix that represents a linear transformation, and if we take $M(\phi)$ to mean the composition $M\circ\phi$, we get indeed:
$$(M(\phi +\psi ))(x)=(M\circ(\phi +\psi ))(x) = M((\phi +\psi )(x))$$

It does not follow from the linearity condition that:
$$M((\phi +\psi )(x)) \ \overset{\phi, \psi \text{ linear }}{ = } \ M(\phi(x) +\psi(x) )$$
though. That is, functions are added pointwise.
The function $(\phi+\psi)$ is such that for all $x$ we have that $(\phi+\psi)(x)=\phi(x)+\psi(x)$.
So it should be:
$$M((\phi +\psi )(x)) \ \overset{\text{definition of function addition}}{ = } \ M(\phi(x) +\psi(x) )$$

Then we can indeed use that Matrix multiplication is linear to find:
$$M(\phi(x) +\psi(x) ) \ \overset{M \text{ linear }}{ = } \ M(\phi(x)) +M(\psi(x) ) =(M\circ\phi)(x)+(M\circ\psi)(x)=(M(\phi ))(x)+(M(\psi ))(x)$$ 2. For $x\in \mathbb{R}^n$ we have the following: $$(M(\lambda \phi ))(x)=M(\lambda \phi )(x) \ \overset{\phi\text{ linear }}{ = } \ M(\lambda \phi(x) ) \ \overset{M \text{ linear }}{ = } \ \lambda M( \phi(x) )=\lambda (M(\phi ))(x)$$

Is everything correct?
Same here:
$$(\lambda\phi)(x) \ \overset{\text{ definition of scalar multiplication with a function}}{ = } \ \lambda(\phi(x))$$ #### Klaas van Aarsen

##### MHB Seeker
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We have some new information.
Let $\sigma_a: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the reflection on the straight line through the origin, where $a$ describes the angle between the straight line and the positive $x$ axis.

...

Determine the matrix $s_a: = M (\sigma_a)$.
It implies that $M(\phi)$ is the matrix that represents the linear map $\phi: \mathbb R^n\to\mathbb R^m$. #### mathmari

##### Well-known member
MHB Site Helper
We have some new information.

It implies that $M(\phi)$ is the matrix that represents the linear map $\phi: \mathbb R^n\to\mathbb R^m$. Ah ok.. So can we don't that as in post #8 ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Ah ok.. So can we don't that as in post #8 ?
Not quite, because $M(\phi(x))$ is not well defined.
That is, $\phi(x)$ is a vector, and we can't get a matrix from a vector - at least not as intended. I think it must be as follows.

The matrix of a linear map $\chi:\mathbb R^n\to\mathbb R^m$ is:
$$M(\chi) = \begin{pmatrix}\chi(\mathbf e_1)\cdots \chi(\mathbf e_n)\end{pmatrix}$$
So:
\begin{align*}M(\phi+\psi)
&=\Big[(\phi+\psi)(\mathbf e_1)\,\cdots\, (\phi+\psi)(\mathbf e_n)\Big]\\
&=\Big[(\phi(\mathbf e_1)+\psi(\mathbf e_1))\,\cdots\, (\phi(\mathbf e_n)+\psi(\mathbf e_n))\Big]\\
&=\Big[\phi(\mathbf e_1)\,\cdots\, \phi(\mathbf e_n)\Big]
+\Big[\chi(\mathbf e_1)\,\cdots\, \chi(\mathbf e_n)\Big]\\
&=M(\phi)+M(\psi)\\
\end{align*} 