1.2. Multiplying matrices on the left

Multiplying on the left by an elementary matrix is part of the following general principle:

Proposition. If you start with a matrix $A$ and multiply on the left by $M$, then in $MA$ every row is a linear combination of rows of $A$. Moreover, just which linear combinations depends on $M$ and not on $A$.

To invent an $M$ so that $MA$ produces particular linear combinations of the rows of $A$, look at the case $A = I$, which gives $MI = M$.

For example, if you want a matrix $M$ whose effect when multiplied on the left is to subtract 2 times row 1 from row 2, then make $M$ by starting with the identity matrix of the right size and subtracting 2 times row 1 from row 2. In the $3 \times 3$ case this would be $M = \left[\begin{array}{rrr}1&0&0\ -2&1&0\ 0&0&1\end{array}\right]$.

In other words, to get $M$, do to the identity matrix what you want $M$ to do to other matrices, when $M$ is used as a multiplier on the left.

As you know, many religions and cultures have a version of the Golden Rule, which in older English is ``Do unto others what you would have others do unto you.'' So it is said that the Golden Rule for matrix multiplication (on the left) is

``Do unto others what you do unto $I$.''

Problem V-4. (a) Suppose you want to scale the rows of a $3 \times 3$ matrix $A$ by 10, 5, and 2 respectively. By what matrix should you multiply $A$ on the left to accomplish this?

(b) What should you do if instead you want to scale the columns of $A$ by the same three factors?

Problem V-5. A permutation matrix is a matrix obtained by permuting the rows of an identity matrix, or equivalently, a matrix that has zero entries except for a single 1 in each row and column. An example is $P = \small\left[\begin{array}{cccc}0&0&1&0\ 0&0&0&1\ 1&0&0&0\ 0&1&0&0\end{array}\right]$. What does $\tau _ P$ do to $\left[\begin{array}{c}x _ 1\ x _ 2\ x _ 3\ x _ 4\end{array}\right]$?

(Method: Use the Golden Rule. Of course, this can be computed directly too; the point is that you can tell what happens more intuitively, without doing a computation.)

Problem V-6. For $N = \left[\begin{array}{cc}0&0\ 1&0\end{array}\right]$,

(a) Calculate $NH$ where $H = \left[\begin{array}{ccccccc}H&E&L&P&I&A& M\ S&I&N&K&I&N&G\end{array}\right]$.

(b) If you wanted to produce that effect on $H$ by multiplying on the left, how would you know to use $N$?

(c) Find $N^2$ in your head, without writing anything down except the answer.

(d) Find $N _ 4^4$, where $N _ 4 = \left[\begin{array}{cccc}0&0&0&0\ 1&0&0&0\\ 0&1&0&0\ 0&0&1&0\end{array}\right]$.

Note. If $M$ is a square matrix such that $M^k = 0$ for some $k$, then $M$ is said to be nilpotent (meaning ``nothing-power''). Nilpotent matrices turn out to be important building blocks in the theory of ``similar'' matrices.

$N = \left[\begin{array}{cc}0&0\ 1&0\end{array}\right]$ and its transpose are the simplest nilpotent matrices other than the zero matrix. They often provide counterexamples to statements you might think should be true.

Example: From experience with numbers, you might think $A^2 = 0 \Rightarrow A = 0$, but that's not true for matrices and $N$ is a counterexample.

Problem V-7. Show that if $v$ and $w$ are orthogonal column vectors in $\mathbb{R}^n$ (see Problem V-), then the $n \times n$ matrix $w v ^ t$ is nilpotent.

Proposition. Any invertible matrix $M$ is a product of elementary matrices.

Proof. Starting with an $n \times n$ invertible matrix $M$ and row-reducing to row-reduced echelon form (RREF), since $M$ is invertible and therefore has rank $n$, the RREF must also have rank $n$ and so is $I$. Therefore the net effect is $E _ k \dots E _ 1 M = I$. Then $M = E^{-1} _ 1 \dots E^{-1} _ k$, and inverses of elementary matrices are elementary.

Corollary. Any product $MA$ with $M$ invertible can be achieved by starting with $A$ and performing elementary row operations.

Corollary. To invert $A$ (assuming it is invertible), start with $[A\vert I]$ and row-reduce until you reach the form $[I\vert M]$. Then $M = A^{-1}$.

Proof. Row-reducing by a product of elementary matrices equalling some $M$ results in $[MA\vert MI]$, so if $MA = I$ then the second half of the matrix is $M = A^{-1}$.

Problem V-8. Find the inverse of $\left[\begin{array}{ccc}1&2&3\ 0&1&4\ 0&0&1\end{array}\right]$.

Here is a handy shortcut for finding $A^{-1} B$, something that often occurs.

Proposition. If $A$ is an invertible $n \times n$ matrix and $B$ is an $m \times n$ matrix, the row-reduced echelon form of the augmented matrix $[A\vert B]$ is $[I\vert A^{-1} B]$.

Proof. Going from $A$ to $I$ means the effect of the row-reduction is to multiply by $A^{-1}$, and the Golden Rule of matrix multiplication says $B$ is affected the same way.