MHB Linear transformation and its matrix

[Guest2]

1. Show that the map $\mathcal{A}$ from $\mathbb{R}^3$ to $\mathbb{R}^3$ defined by $\mathcal{A}(x,y,z) = (x+y, x-y, z)$ is a linear transformation. Find its matrix in standard basis.

2. Find the dimensions of $\text{Im}(\mathcal{A})$ and $\text{Ker}(\mathcal{A})$, and find their basis for the linear transformation $\mathcal{A}$ on $\mathbb{R}^3$ defined by $\mathcal{A} (x,y,z) = (x-2z, y+z, 0)$.

1. If let $u = (x,y,z)$, and $v = (x', y', z')$. Then

$\mathcal{A}(u+v) = (x+y+x'+y', x-y+x'-y', z+z')= (x+y, x-y, z)+(x'+y', x'-y', z') = \mathcal{A}(u)+\mathcal{A}(v).$

Moreover, $ \mathcal{A}(\lambda u) = (\lambda x +\lambda y, \lambda x- \lambda y, \lambda z) = \lambda (x+y, x-y, z) = \lambda \mathcal{A}(u)$

Hence $\mathcal{A}$ is a linear transformation. However, I can't find its matrix.

 

[Deveno]

A partial answer:

$M_{\mathcal{A}} = \begin{bmatrix}1&1&0\\1&-1& \ast\\ \ast&\ast&\ast\end{bmatrix}$.

Perhaps you can finish.

 

[Guest2]

A partial answer:

$M_{\mathcal{A}} = \begin{bmatrix}1&1&0\\1&-1& \ast\\ \ast&\ast&\ast\end{bmatrix}$.

Perhaps you can finish.

$M_{\mathcal{A}} = \begin{bmatrix}1&1&0\\1&-1& 0\\ 0& 0& 1\end{bmatrix}$ I think.

So for question $2$ the concerned matrix is

$M_{\mathcal{A}} = \begin{bmatrix}1&0&-2\\0&1& 1\\ 0& 0& 0\end{bmatrix} $

Solving we get $y = -z, ~ x = 2z,$ and $z$ is free. So $\text{Ker}\mathcal{A} = \left\{(2z,-z,z): z \in \mathbb{R}\right\}$

How do I find $\text{Im}\mathcal{A}$? Is it true that we just take the columns of $M_{\mathcal{A}}$?

 

[Deveno]

$M_{\mathcal{A}} = \begin{bmatrix}1&1&0\\1&-1& 0\\ 0& 0& 1\end{bmatrix}$ I think.

So for question $2$ the concerned matrix is

$M_{\mathcal{A}} = \begin{bmatrix}1&1&-2\\0&1& 1\\ 0& 0& 0\end{bmatrix} $

Solving we get $y = -z, ~ x = 3z,$ and $z$ is free. So $\text{Ker}\mathcal{A} = \left\{(3z,-z,z): z \in \mathbb{R}\right\}$

How do I find $\text{Im}\mathcal{A}$? Is it true that we just take the columns of $M_{\mathcal{A}}$?

If the matrix for #2 was as you say, we would would have:

$\begin{bmatrix}1&1&-2\\0&1&1\\0&0&0\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}x+y-2z\\y+z\\0\end{bmatrix}$

suggesting $\mathcal{A}(x,y,z) = (x+y-2z,y+z,0)$, which is not the case.

 

[Guest2]

If the matrix for #2 was as you say, we would would have:

$\begin{bmatrix}1&1&-2\\0&1&1\\0&0&0\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}x+y-2z\\y+z\\0\end{bmatrix}$

suggesting $\mathcal{A}(x,y,z) = (x+y-2z,y+z,0)$, which is not the case.

Yeah, I made a mistake. It should have said:

$M_{\mathcal{A}} = \begin{bmatrix}1&0&-2\\0&1& 1\\ 0& 0& 0\end{bmatrix} $

Solving we get $y = -z, ~ x = 2z,$ and $z$ is free. So $\text{Ker}\mathcal{A} = \left\{(2z,-z,z): z \in \mathbb{R}\right\}$, which has dimension $1$.

Apparently, the dimension of $\text{Im}(\mathcal{A}) = \text{rank}(\mathcal{A})$, so in this case dimension of $\text{Im}(\mathcal{A}) = 2$.

 

[Deveno]

Yeah, I made a mistake. It have said:

$M_{\mathcal{A}} = \begin{bmatrix}1&0&-2\\0&1& 1\\ 0& 0& 0\end{bmatrix} $

Solving we get $y = -z, ~ x = 2z,$ and $z$ is free. So $\text{Ker}\mathcal{A} = \left\{(2z,-z,z): z \in \mathbb{R}\right\}$, which has dimension $1$.

Apparently, the dimension of $\text{Im}(\mathcal{A}) = \text{rank}(\mathcal{A})$, so in this case dimension of $\text{Im}(\mathcal{A}) = 2$.

It is also the case for this example that $\dim(\text{Im}(\mathcal{A})) = 3 - \dim(\text{ker}(\mathcal{A}))$.

Do you understand why $\dim(\text{Im}(\mathcal{A})) = \text{rank}(\mathcal{A})$? Does this suggest a way to define the rank of a linear transformation, without using a matrix?

Your amended answer looks much better. ^^

 

[Guest2]

It is also the case for this example that $\dim(\text{Im}(\mathcal{A})) = 3 - \dim(\text{ker}(\mathcal{A}))$.

Do you understand why $\dim(\text{Im}(\mathcal{A})) = \text{rank}(\mathcal{A})$? Does this suggest a way to define the rank of a linear transformation, without using a matrix?

Your amended answer looks much better. ^^

My understanding is that the image is the space spanned by the columns. So the rank then would be the dimension of the space spanned by the columns? I would say my understanding of why $\dim(\text{Im}(\mathcal{A})) = \text{rank}(\mathcal{A})$ is actually shaky at best.

Also, I wonder whether one can determine whether a transformation is linear by consider the corresponding matrix? If so, what property are we looking for in the matrix?

 

[Deveno]

Any mapping from $\Bbb R^n \to \Bbb R^m$ that can be represented as (left) multiplication by a matrix IS a linear transformation.

This is because, for two $n$-vectors $x,y$ expressed as $n \times 1$ matrices, with $A$ an $m \times n$ matrix:

$A(x + y) = Ax + Ay$ (matrix multiplication DISTRIBUTES over matrix addition).

$A(rx) = (rA)(x) = r(Ax)$ (scalar multiplication can be regarded as multiplication by the $n \times n$ matrix $rI_n$ on the right of $A$, and the left of $x$, and as the $m \times m$ matrix $rI_m$ on the left of $A$).

You are correct about the image of $A$ being spanned by the columns of $A$. This is immediate from the fact that $A$ is linear, and that if we write the $j$-th standard basis vector as a column vector ($n \times 1$ matrix), then $Ae_j$ gives us the $j$-th column of $A$.

(Recall that the image of $A$ is determined by the images of the basis vectors, since:

$A(v) = A(c_1e_1 +\cdots + c_ne_n) = c_1A(e_1)+\cdots +c_nA(e_n)$).

Thus the columns of $A$ SPAN the image of $A$. They may not be linearly independent, but if we find a maximally linearly independent subset of the columns, they form a BASIS.

The size of such a basis IS the rank of $A$. This follows from the fact that performing row-reduction amounts to multiplying on the left by an INVERTIBLE matrix, so it doesn't change the dimension of the image of $A$ (invertible matrices preserve dimensions), and performing column-reduction amounts to multiplying on the right of $A$ by an invertible matrix (it is easily checked that row-reduction elementary operations and the equivalent for columns are reversible).

Since the goal of both these types of operations is to maximize the number of zero rows (or columns), after we have done so, the number of non-zero rows (or columns) is the dimension of the image we originally sought.

In a nutshell: column rank = row rank, which we simply call "rank". Row-reduction, and column-reduction are two different paths to the same place: the dimension of the "range".

But it gets even better: if we start with a space with $n$-dimensions, we have the RANK-NULLITY theorem:

$\dim(\text{ker}(A)) + \text{rank}(A) = n$.

So we only ever need to find one of:

a) The dimension of the kernel (= null space)
b) The rank

This is a real time-saver.

 

[Guest2]

That was most helpful, thank you very much!

The size of such a basis IS the rank of $A$. This follows from the fact that performing row-reduction amounts to multiplying on the left by an INVERTIBLE matrix, so it doesn't change the dimension of the image of $A$ (invertible matrices preserve dimensions), and performing column-reduction amounts to multiplying on the right of $A$ by an invertible matrix (it is easily checked that row-reduction elementary operations and the equivalent for columns are reversible).

This demystifies the whole idea of row reduction for me. I've often wondered in passing why it works.