Having to do copious calculations by hand when preparing for an exam, I came to realize that there was an alternative way of interpreting a matrix multiplication. This new insight would allow me to instantly guess the following product *without ever doing any numerical multiplication*:

\[\begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & 6 \\

7 & -8 & 0

\end{bmatrix}

\begin{bmatrix}

0 & 0 & 0 \\

0& 1 & 0 \\

1& 0 & 0

\end{bmatrix}

=

\begin{bmatrix}

3 & 2 & 0 \\

6 & 5 & 0 \\

0 & -8 & 0

\end{bmatrix}\]

Was there a way to have known that the first column of the product would be the third column of the first matrix?

**MATRIX TIMES VECTOR**

**Observation:** If has $n$ columns and $v$ is an -vector, then

\[Av=\begin{bmatrix}

c & c & c & \cdots \\

o & o & o & \cdots \\

l & l & l & \cdots \\

u & u & u & \cdots \\

m & m & m & \cdots \\

n & n & n & \cdots \\

\end{bmatrix}

\begin{bmatrix}

a \\

b \\

c \\

\vdots

\end{bmatrix}

=

a* \begin{bmatrix}

1^{st} \\

c \\

o \\

l \\

u \\

m \\

n

\end{bmatrix}

+b* \begin{bmatrix}

2^{nd} \\

c \\

o \\

l \\

u \\

m \\

n

\end{bmatrix}

+c* \begin{bmatrix}

3^{rd} \\

c \\

o \\

l \\

u \\

m \\

n

\end{bmatrix}+ \cdots \]

That is, the product is a linear combination of the columns with coefficients coming from the vector.

Examples:

1.

\[\begin{bmatrix}

1 & -5 & 3 \\

0 & 0.5 & 2\\

1 & -8 & 0

\end{bmatrix}

\begin{bmatrix}

2\\

-1 \\

0

\end{bmatrix}

= 2* \begin{bmatrix} 1st\\ col \end{bmatrix} + (-1)*\begin{bmatrix} 2nd\\ col \end{bmatrix} + 0 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =

2 \begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix} -\begin{bmatrix}

-5 \\

0.5 \\

-8

\end{bmatrix}+0 = \begin{bmatrix}

7 \\

-0.5 \\

10

\end{bmatrix}\]

2. What vector to multiply a matrix into, so that we get the second column of as the product?

Answer: \[ \begin{bmatrix}

0 \\

1 \\

0 \\

\vdots

\end{bmatrix}. \]

It is not only the practical calculations that benefit from this new method. Some lemmas and theorems of linear algebra become more obvious and accessible. For example:

Ax=0 has a nontrivial solution iff the columns of A are linearly dependent (as vectors).

dimension of {Ax: x ranging over all column vectors} = number of independent columns of A

Given and , has a solution iff can be written as a linear combination of columns of .

**MATRIX TIMES MATRIX**

In computing

\[AB=\begin{bmatrix}

c & c & c & \cdots \\

o & o & o & \cdots \\

l & l & l & \cdots \\

u & u & u & \cdots \\

m & m & m & \cdots \\

n & n & n & \cdots \\

\end{bmatrix}

\begin{bmatrix}

c & c & c & \cdots \\

o & o & o & \cdots \\

l & l & l & \cdots \\

u & u & u & \cdots \\

m & m & m & \cdots \\

n & n & n & \cdots \\

\end{bmatrix}

, \ \]

To get the j’th column of the product, ignore all other columns of B except the j’th, and then do A*[column j of B], using the above method. Do this for all columns and you will have the product .

Examples:

1.

\[\begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

1 & 0 & 1 \\

0& 1 & 1 \\

2& 3 & 0

\end{bmatrix} \]

The **first **column will be

\[ \begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

1 \\

0 \\

2

\end{bmatrix} =

1*\begin{bmatrix}

1 \\

4 \\

7

\end{bmatrix} + 2* \begin{bmatrix}

3 \\

-2 \\

0

\end{bmatrix} = \begin{bmatrix}

7\\

0 \\

7

\end{bmatrix} \]

The **second** column will be

\[ \begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

0 \\

1 \\

3

\end{bmatrix} =

1*\begin{bmatrix}

2 \\

5 \\

-4

\end{bmatrix} + 3* \begin{bmatrix}

3 \\

-2 \\

0

\end{bmatrix} = \begin{bmatrix}

11 \\

-1 \\

-4

\end{bmatrix} \]

Column 3 is computed similarly, and the full result is:

\[\begin{bmatrix}

1 & 2 & 3 \\

4 & 5 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

1 & 0 & 1 \\

0& 1 & 1 \\

2& 3 & 0

\end{bmatrix} = \begin{bmatrix}

7 & 11 & 3 \\

0 & -1 & 9 \\

7 & -4 & 3

\end{bmatrix} \]

Note: Most of the calculations above are doable in mind, and I have written them out to explain the procedure only.

2. Multiply \[A= \begin{bmatrix}

2 & 33 & 0 \\

0 & -1 & 9 \\

4 & -14 & 13

\end{bmatrix} \] into a matrix so that the columns 2 and 3 of A are interchanged.

Answer: \[ B=\begin{bmatrix}

1 & 0 & 0 \\

0 & 0 & 1 \\

0 & 1 & 0

\end{bmatrix} \]

3. Find the 3rd column of the product

\[\begin{bmatrix}

10 & 2 & 3 \\

4 & 0 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix}

1 & 0 & 1 \\

0& 1 & 1 \\

-2& 0 & -1

\end{bmatrix} \]

Answer:

thrid column of AB = A * third column of B. So

\[ \begin{bmatrix}

10 & 2 & 3 \\

4 & 0 & -2 \\

7 & -4 & 0

\end{bmatrix}

\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}= \begin{bmatrix} 9 \\ 6 \\ 3 \end{bmatrix} \]

(Please verify!)

4. Is the following true or false? *“If we change arrays in the second column of B, then in the product AB only the arrays in the second column may change — all other arrays on other columns will remain the same.”*

Answer: True.

Whether you are convinced that computing AB one column at a time is more practical than the conventional array-by-array multiplication or not, one cannot ignore its usefulness in helping one to “see” the linear algebra behind matrices and their multiplication. So, next time you encounter a sparse matrix save yourself tens of multiplications by zero!

**HOMEWORK**

Can you guess and prove a row-wise version of the above?