MHB Solving Linear Transformations with Matrix A and Vectors u1 & u2

[AngrySnorlax]

This is the new topic we are going over in my Linear Algebra class and I am completely lost in how it works. From what I gather from the book, the lecture, and other sources amongst the internet, there are supposed to be a rule set shard with the transformation. The problem being is I cannot seem to find the correct rule set for transformations, unless I am seriously over thinking this whole thing.

Here's the problem:

Let $$T(x) = Ax$$ for the given matrix $A$, and find $T\left(u_1\right)$ and $T\left(u_2\right)$ for the given $u_1$ and $u_2$.

$$A=\left[\begin{array}{c}2 & 1 \\ -3 & 5 \end{array}\right]$$

$$u_1=\left[\begin{array}{c}-4 \\ -2 \end{array}\right]$$

$$u_2=\left[\begin{array}{c}1 \\ -6 \end{array}\right]$$

I'm new to the forum and I'm kind of in a hurry for this particular problem. Anyways, the answer in the back of the back was:

$$T\left(u_1\right)=\left[\begin{array}{c}-10 \\ 2 \end{array}\right]$$

$$T\left(u_2\right)=\left[\begin{array}{c}-4 \\ -33 \end{array}\right]$$

My problem is I have no idea how they got this answer! I've tried everything I could think of, but I simply cannot figure out the process behind this. I just need help figuring out how they came to these answers to hopefully further my understanding on what is going on during these transformation. Any and all help would be greatly appreciated.

 

[Euge]

This is the new topic we are going over in my Linear Algebra class and I am completely lost in how it works. From what I gather from the book, the lecture, and other sources amongst the internet, there are supposed to be a rule set shard with the transformation. The problem being is I cannot seem to find the correct rule set for transformations, unless I am seriously over thinking this whole thing.

Here's the problem:

Let T(x) = Ax for the given matrix A, and find T(u1) and T(u2) for the given u1 and u2.

A=
2 1
-3 5
(yes the above is a 2x2 matrix)

u1=
-4
-2

u2=
1
-6

I am really sorry for the horrible format I typed this in, I'm new to the forum and I'm kind of in a hurry for this particular problem. Anyways, the answer in the back of the back was:

T(u1)=
-10
2

T(u2)=
-4
-33

My problem is I have no idea how they got this answer! I've tried everything I could think of, but I simply cannot figure out the process behind this. I just need help figuring out how they came to these answers to hopefully further my understanding on what is going on during these transformation. Any and all help would be greatly appreciated.

Hi AngrySnorlax,

Have you learned how to multiply matrices yet? Or are you just confused by it?

To get the answers, multiply $A$ with $u_1$ and $A$ with $u_2$:

$$ Au_1 = \begin{bmatrix}2 & 1\\-3 & 5\end{bmatrix}\begin{bmatrix}-4\\-2\end{bmatrix} = \begin{bmatrix}2(-4) + 1(-2)\\(-3)(-4) + 5(-2)\end{bmatrix} = \begin{bmatrix}-10\\2\end{bmatrix}$$

$$ Au_2 = \begin{bmatrix}2 & 1\\-3 & 5\end{bmatrix}\begin{bmatrix}1\\-6\end{bmatrix} = \begin{bmatrix}2(1) + 1(-6)\\(-3)(1) + 5(-6)\end{bmatrix} = \begin{bmatrix}-4\\-33\end{bmatrix}$$

The latter is $T(u_1)$ and the former is $T(u_2)$.

I'm not certain what you mean by the "rule set" of a linear transformation, but let's start with this. Suppose you have a function $T : \Bbb R^n \to \Bbb R^m$. We say that $T$ is a linear transformation if $T(x_1 + x_2) = T(x_1) + T(x_2)$ and $T(rx) = rT(x)$ for all $x_1, x_2, x\in \Bbb R^n$ and $r\in \Bbb R$. It turns out (which I can show you later) that every linear transformation from $\Bbb R^n$ into $\Bbb R^m$ can be expressed in the form $T_A(x) = Ax$, where $A$ is an $m \times n$ matrix and $x$ is an $n \times 1$ matrix (also called a column vector). Conversely, every function from $\Bbb R^n$ to $\Bbb R^m$ of the form $T_A$ is a linear transformation. So we have a one-to-one correspondence between $m \times n$ matrices and linear transformations $\Bbb R^n \to \Bbb R^m$, given by $A \mapsto T_A$.

 

[AngrySnorlax]

Thank you very much for your response! It helped greatly. So I understand it a bit more now, but does this mean you cannot multiply a 3x3 matrix with a 2x2 matrix, or would there be another method behind that? This isn't for the homework, its more out of curiosity.

 

[Euge]

Thank you very much for your response! It helped greatly. So I understand it a bit more now, but does this mean you cannot multiply a 3x3 matrix with a 2x2 matrix, or would there be another method behind that? This isn't for the homework, its more out of curiosity.

That's right -- at least, not in the usual way. Standard multiplication of matrices is defined as follows. Given an $m \times n$ matrix $A = [a_{ij}]$ and an $n \times k$ matrix $B = [b_{ij}]$, then product $AB = [c_{ij}]$ is defined as the $m \times k$ matrix such that $c_{ij}$ is the dot product of the $i^{th}$ row of $A$ and the $j^{th}$ column of $B$, i.e.,

$$c_{ij} = \begin{bmatrix}a_{i1} & a_{i2} & \cdots & a_{in}\end{bmatrix}\begin{bmatrix}b_{1j}\\b_{2j}\\ \vdots \\ b_{nj}\end{bmatrix} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}.$$

Using the sigma notation the relationship can be written

$$c_{ij} = \sum_{k = 1}^n a_{ik}b_{kj}.$$

Notice that there are as many columns of $A$ as there are rows of $B$. This is necessary to define $AB$.

There is however a notion of a "product" between matrices with no restriction on size. This is known as the tensor product of matrices. Given an $m \times n$ matrix $A = [a_{ij}]$ and a $k \times p$ matrix $B = [b_{ij}]$, then tensor product of $A$ with $B$, written $A \otimes B$, is the matrix

$$\begin{bmatrix}a_{11}B & a_{12}B & \cdots & a_{1n}B\\
a_{21}B & a_{22}B & \cdots & a_{2n}B\\ \vdots & \vdots & {} & \vdots\\ a_{m1}B & a_{m2}B & \cdots & a_{mn}B\end{bmatrix}$$

The above form is called a block matrix. Each block $a_{ij}B$ has the same size as $B$, i.e., $k \times p$; in particular, they all have $kp$ entries. Since there are $m$ rows and $n$ columns of these blocks, it follows that the above matrix has $mk$ rows and $np$ columns. Thus $A \otimes B$ has size $mk \times np$.

 

[Deveno]

Thank you very much for your response! It helped greatly. So I understand it a bit more now, but does this mean you cannot multiply a 3x3 matrix with a 2x2 matrix, or would there be another method behind that? This isn't for the homework, its more out of curiosity.

Yes (usually). The rule is this:

An $m \times n$ matrix multiplied by an $n \times p$ matrix will give an $m \times p$ matrix. The rows of the first matrix, and the columns of the second matrix "zip up", and so these numbers have to MATCH (just like two halves of a zipper need the same number of teeth to "mesh").

Explicitly, to get the $i,j$-th entry of the product matrix, we take the $i,k$-th entry of the first matrix, and multiply this by the $k,j$-the entry of the second matrix. We do this for every $k$, as $k$ runs from 1 to $n$, and then add them all.

So, for example:$\begin{bmatrix}\ast&\ast&\ast\\ 1&-2&4\\ \ast&\ast&\ast\\ \ast&\ast&\ast \end{bmatrix}\begin{bmatrix}3&\ast\\ 4&\ast\\ 2&\ast \end{bmatrix} = \begin{bmatrix}\ast&\ast\\3&\ast\\ \ast&\ast\\ \ast&\ast\end{bmatrix}$

Since $(1)(3) + (-2)(4) + 4(2) = 3 - 8 + 8 = 3$.