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# [section1] Matrices, dot products and matrix multiplication - tanzer.trail1.post2

edited February 2020

Prev: Numbers and vectors

Here is a 2-by-3 matrix:

$\begin{bmatrix} 1 & 2 & 3 \\ -10 & 0 & 7 \end{bmatrix}$

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1.
edited February 2020

Here is the transpose of that matrix:

$\begin{bmatrix} 1 & -10 \\ 2 & 0 \\ 3 & 7 \end{bmatrix}$

Comment Source:Here is the _transpose_ of that matrix: \$\begin{bmatrix} 1 & -10 \\\\ 2 & 0 \\\\ 3 & 7 \end{bmatrix} \$
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2.
edited February 2020

The transpose turned a 2-by-3 matrix into a 3-by-2 matrix.

The transpose turns rows into columns and columns into rows.

You can picture it as a reflection of the matrix across a 45-degree axis going from top-left to bottom-right.

Comment Source:The transpose turned a 2-by-3 matrix into a 3-by-2 matrix. The transpose turns rows into columns and columns into rows. You can picture it as a reflection of the matrix across a 45-degree axis going from top-left to bottom-right.
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3.
edited February 2020

If $$M$$ is a matrix, then its transpose is written $$M^T$$.

For example:

$\begin{bmatrix} 1 & 2 & 3 \\ -10 & 0 & 7 \end{bmatrix}^T = \begin{bmatrix} 1 & -10 \\ 2 & 0 \\ 3 & 7 \end{bmatrix}$

Comment Source:If \$$M\$$ is a matrix, then its transpose is written \$$M^T\$$. For example: \$\begin{bmatrix} 1 & 2 & 3 \\\\ -10 & 0 & 7 \end{bmatrix}^T = \begin{bmatrix} 1 & -10 \\\\ 2 & 0 \\\\ 3 & 7 \end{bmatrix} \$
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4.
edited February 2020

Taking the transpose of the transpose gives you back the original matrix:

${(M^T)}^T = M$

A row-vector is a matrix with just one row.

A column-vector is a matrix with just one column.

Comment Source:Taking the transpose of the transpose gives you back the original matrix: \${(M^T)}^T = M\$ A _row-vector_ is a matrix with just one row. A _column-vector_ is a matrix with just one column.
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5.
edited February 2020

The transpose of a row vector is a column vector:

$\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}^T = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$

Comment Source:The transpose of a row vector is a column vector: \$\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}^T = \begin{bmatrix} 1 \\\\ 2 \\\\ 3 \end{bmatrix} \$
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6.

And the transpose of a column vector is a row vector:

$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}^T = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$

Comment Source:And the transpose of a column vector is a row vector: \$\begin{bmatrix} 1 \\\\ 2 \\\\ 3 \end{bmatrix}^T = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \$
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7.
edited February 2020

The dot product is an operation that takes two vectors in and outputs a single number (a scalar).

Here is how it works:

$\begin{bmatrix} 1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 10 & 20 \end{bmatrix} = 1 \cdot 10 + 2 \cdot 20 = 50$

Comment Source:The _dot product_ is an operation that takes two vectors in and outputs a single number (a scalar). Here is how it works: \$\begin{bmatrix} 1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 10 & 20 \end{bmatrix} = 1 \cdot 10 + 2 \cdot 20 = 50 \$
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edited February 2020

It will prove convenient to express the dot product in a form where its left input is a row vector, and its right input is a column vector.

So we would write:

$\begin{bmatrix} 1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 10 \\ 20 \end{bmatrix} = 1 \cdot 10 + 2 \cdot 20 =50$

Comment Source:It will prove convenient to express the dot product in a form where its left input is a row vector, and its right input is a column vector. So we would write: \$\begin{bmatrix} 1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 10 \\\\ 20 \end{bmatrix} = 1 \cdot 10 + 2 \cdot 20 =50 \$
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9.
edited February 2020

Matrix multiplication is an operation that takes two matrices for inputs, and outputs a product matrix.

We may write $$A \times B = C$$, where the variables now stand for matrices.

If $$A$$ is a row vector of length n, and $$B$$ is column vector of length n, then the matrix multiplication $$A \cdot B$$ is precisely defined to the dot product $$A \cdot B$$ of $$A$$ and $$B$$.

The notation coincides; matrix multiplication is a strict generalization of the dot product, to the case of general matrices.

Comment Source:_Matrix multiplication_ is an operation that takes two matrices for inputs, and outputs a product matrix. We may write \$$A \times B = C\$$, where the variables now stand for matrices. If \$$A\$$ is a row vector of length n, and \$$B\$$ is column vector of length n, then the matrix multiplication \$$A \cdot B\$$ is precisely defined to the dot product \$$A \cdot B\$$ of \$$A\$$ and \$$B\$$. The notation coincides; matrix multiplication is a strict generalization of the dot product, to the case of general matrices.
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Now there is a compatibility condition on the matrix product $$A \cdot B$$, which causes this product to be defined only of the shapes of $$A$$ and $$B$$ are compatible.

Rather than just telling you the condition, here I'll explain how derive it from your understanding of the dot product.

Comment Source:Now there is a compatibility condition on the matrix product \$$A \cdot B\$$, which causes this product to be defined only of the shapes of \$$A\$$ and \$$B\$$ are compatible. Rather than just telling you the condition, here I'll explain how derive it from your understanding of the dot product.
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11.

Suppose $$A$$ is row vector, and $$B$$ is a column vector, and we wish to form their dot product $$A \cdot B$$.

Now clearly this will only work if the number of columns in $$A$$ equals the number of rows in $$B$$.

And that is the general compatibility requirement for $$A$$ and $$B$$, in order for the product of $$A$$ and $$B$$ to be defined:

$ncols(A) = nrows(B)$

Comment Source:Suppose \$$A\$$ is row vector, and \$$B\$$ is a column vector, and we wish to form their dot product \$$A \cdot B\$$. Now clearly this will only work if the number of columns in \$$A\$$ equals the number of rows in \$$B\$$. And _that_ is the general compatibility requirement for \$$A\$$ and \$$B\$$, in order for the product of \$$A\$$ and \$$B\$$ to be defined: \$ncols(A) = nrows(B)\$
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12.
edited February 2020

Now we'll define matrix multiplication, in general, by bootstrapping from the concept of dot products.

Comment Source:Now we'll define matrix multiplication, in general, by bootstrapping from the concept of dot products.
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edited February 2020

Say we want to multiply $$A$$ and $$B$$, and that $$ncols(A) = nrows(B) = k$$.

Suppose that $$A$$ has m rows, and $$B) has n columns. So \(A$$ is m-by-k, and $$B$$ is k-by-n.

The key will be to picture $$A$$ as a matrix of row-vectors, and $$B$$ as a matrix of column vectors.

Comment Source:Say we want to multiply \$$A\$$ and \$$B\$$, and that \$$ncols(A) = nrows(B) = k\$$. Suppose that \$$A\$$ has m rows, and \$$B) has n columns. So \\(A\$$ is m-by-k, and \$$B\$$ is k-by-n. The key will be to picture \$$A\$$ as a matrix of row-vectors, and \$$B\$$ as a matrix of column vectors.
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edited February 2020

The key is to picture $$A$$ as a matrix of row-vectors and $$B$$ as a matrix of column vectors.

Here are the rows of A:

$A = \begin{bmatrix} \text{---} & a_1 & \text{---} \\ \text{---} & a_2 & \text{---} \\ & \vdots & \\ \text{---} & a_m & \text{---} \end{bmatrix}$

Here are the columns of B:

$B = \begin{bmatrix} \vert & \vert & & \vert \\ b_1 & b_2 & ... & b_n \\ \vert & \vert & & \vert \end{bmatrix}$

Comment Source:The key is to picture \$$A\$$ as a matrix of row-vectors and \$$B\$$ as a matrix of column vectors. Here are the rows of A: \$A = \begin{bmatrix} \text{---} & a_1 & \text{---} \\\\ \text{---} & a_2 & \text{---} \\\\ & \vdots & \\\\ \text{---} & a_m & \text{---} \end{bmatrix} \$ Here are the columns of B: \$B = \begin{bmatrix} \vert & \vert & & \vert \\\\ b_1 & b_2 & ... & b_n \\\\ \vert & \vert & & \vert \end{bmatrix} \$
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15.
edited February 2020

Then the product matrix $$A \cdot B$$ consists of all possible dot products of a row from $$A$$ with a column from $$B$$:

$A \cdot B = \begin{bmatrix} \text{---} & a_1 & \text{---} \\ \text{---} & a_2 & \text{---} \\ & \vdots & \\ \text{---} & a_m & \text{---} \end{bmatrix} \cdot \begin{bmatrix} \vert & \vert & & \vert \\ b_1 & b_2 & ... & b_n \\ \vert & \vert & & \vert \end{bmatrix} = \begin{bmatrix} a_1 \cdot b_1 & a_1 \cdot b_2 & \text{---} & a_1 \cdot b_n \\ a_2 \cdot b_1 & a_2 \cdot b_2 & \text{---} & a_2 \cdot b_n \\ a_m \cdot b_1 & a_m \cdot b_2 & \text{---} & a_m \cdot b_n \end{bmatrix}$

Comment Source:Then the product matrix \$$A \cdot B\$$ consists of all possible dot products of a row from \$$A\$$ with a column from \$$B\$$: \$A \cdot B = \begin{bmatrix} \text{---} & a_1 & \text{---} \\\\ \text{---} & a_2 & \text{---} \\\\ & \vdots & \\\\ \text{---} & a_m & \text{---} \end{bmatrix} \cdot \begin{bmatrix} \vert & \vert & & \vert \\\\ b_1 & b_2 & ... & b_n \\\\ \vert & \vert & & \vert \end{bmatrix} = \begin{bmatrix} a_1 \cdot b_1 & a_1 \cdot b_2 & \text{---} & a_1 \cdot b_n \\\\ a_2 \cdot b_1 & a_2 \cdot b_2 & \text{---} & a_2 \cdot b_n \\\\ a_m \cdot b_1 & a_m \cdot b_2 & \text{---} & a_m \cdot b_n \end{bmatrix} \$
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16.

Let's work through an example.

Comment Source:Let's work through an example.
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17.
edited February 2020

Suppose $$A$$ is 3-by-2:

$\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}$

And $$B$$ is 2-by-4:

$\begin{bmatrix} 10 & 20 & 30 & 40 \\ 50 & 60 & 70 & 80 \end{bmatrix}$

Comment Source:Suppose \$$A\$$ is 3-by-2: \$\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\\\ 5 & 6 \end{bmatrix} \$ And \$$B\$$ is 2-by-4: \$\begin{bmatrix} 10 & 20 & 30 & 40 \\\\ 50 & 60 & 70 & 80 \end{bmatrix} \$
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18.
edited February 2020

We can form the product $$A \cdot B$$, because:

$ncols(A) = nrows(B) = 2$

Comment Source:We can form the product \$$A \cdot B\$$, because: \$ncols(A) = nrows(B) = 2\$
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19.
edited February 2020

Here is $$A$$ in row format:

$A = \begin{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \\ \begin{bmatrix} 3 & 4 \end{bmatrix} \\ \begin{bmatrix} 5 & 6 \end{bmatrix} \end{bmatrix}$

And $$B$$ in column format:

$B = \begin{bmatrix} \begin{bmatrix} 10 \\ 50 \end{bmatrix} && \begin{bmatrix} 20 \\ 60 \end{bmatrix} && \begin{bmatrix} 30 \\ 70 \end{bmatrix} && \begin{bmatrix} 40 \\ 80 \end{bmatrix} \end{bmatrix}$

Comment Source:Here is \$$A\$$ in row format: \$A = \begin{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \\\\ \begin{bmatrix} 3 & 4 \end{bmatrix} \\\\ \begin{bmatrix} 5 & 6 \end{bmatrix} \end{bmatrix} \$ And \$$B\$$ in column format: \$B = \begin{bmatrix} \begin{bmatrix} 10 \\\\ 50 \end{bmatrix} && \begin{bmatrix} 20 \\\\ 60 \end{bmatrix} && \begin{bmatrix} 30 \\\\ 70 \end{bmatrix} && \begin{bmatrix} 40 \\\\ 80 \end{bmatrix} \end{bmatrix} \$
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20.
edited February 2020

So we have row vectors $$a_1 = \begin{bmatrix} 1 & 2 \end{bmatrix}$$, $$a_2 = \begin{bmatrix} 3 & 4 \end{bmatrix}$$, $$a_3 = \begin{bmatrix} 5 & 6 \end{bmatrix}$$,

and column vectors $$b_1 = \begin{bmatrix} 10 \\ 50 \end{bmatrix}$$, $$b_2 = \begin{bmatrix} 20 \\ 60 \end{bmatrix}$$, $$b_3 = \begin{bmatrix} 30 \\ 70 \end{bmatrix}$$, $$b_4 = \begin{bmatrix} 30 \\ 80 \end{bmatrix}$$.

So the product matrix $$A \cdot B$$ is the 3-by-4 matrix of dot products $$a_i \cdot b_j$$.

Comment Source:So we have row vectors \$$a_1 = \begin{bmatrix} 1 & 2 \end{bmatrix}\$$, \$$a_2 = \begin{bmatrix} 3 & 4 \end{bmatrix}\$$, \$$a_3 = \begin{bmatrix} 5 & 6 \end{bmatrix}\$$, and column vectors \$$b_1 = \begin{bmatrix} 10 \\\\ 50 \end{bmatrix} \$$, \$$b_2 = \begin{bmatrix} 20 \\\\ 60 \end{bmatrix} \$$, \$$b_3 = \begin{bmatrix} 30 \\\\ 70 \end{bmatrix} \$$, \$$b_4 = \begin{bmatrix} 30 \\\\ 80 \end{bmatrix} \$$. So the product matrix \$$A \cdot B\$$ is the 3-by-4 matrix of dot products \$$a_i \cdot b_j\$$.
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edited February 2020

$A \cdot B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \cdot \begin{bmatrix} 10 & 20 & 30 & 40 \\ 50 & 60 & 70 & 80 \end{bmatrix} = \begin{bmatrix} 110 & 140 & 170 & 200 \\ 230 & 300 & 370 & 440 \\ 350 & 460 & 570 & 680 \end{bmatrix} = \begin{bmatrix} a_1 \cdot b_1 & a_1 \cdot b_2 & a_1 \cdot b_3 & a_1 \cdot b_4 \\ a_2 \cdot b_1 & a_2 \cdot b_2 & a_2 \cdot b_3 & a_2 \cdot b_4 \\ a_3 \cdot b_1 & a_3 \cdot b_2 & a_3 \cdot b_3 & a_3 \cdot b_4 \end{bmatrix}$

Comment Source:\$A \cdot B = \begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\\\ 5 & 6 \end{bmatrix} \cdot \begin{bmatrix} 10 & 20 & 30 & 40 \\\\ 50 & 60 & 70 & 80 \end{bmatrix} = \begin{bmatrix} 110 & 140 & 170 & 200 \\\\ 230 & 300 & 370 & 440 \\\\ 350 & 460 & 570 & 680 \end{bmatrix} = \begin{bmatrix} a_1 \cdot b_1 & a_1 \cdot b_2 & a_1 \cdot b_3 & a_1 \cdot b_4 \\\\ a_2 \cdot b_1 & a_2 \cdot b_2 & a_2 \cdot b_3 & a_2 \cdot b_4 \\\\ a_3 \cdot b_1 & a_3 \cdot b_2 & a_3 \cdot b_3 & a_3 \cdot b_4 \end{bmatrix} \$
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22.
edited February 2020
Comment Source:. * * * [Prev: Numbers and vectors](https://forum.azimuthproject.org/discussion/2475/section1-linear-algebra-numbers-and-vectors-tanzer-trail1-post1)