Visual walkthrough — Matrices — review, operations, types

We assume only one thing: a matrix is a machine that moves arrows. We build everything else.

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Step 1 — An arrow is just two numbers

WHAT. Draw graph paper. A vector $\mathbf{x}$ is an arrow from the origin (the centre cross) to a point. It is fully described by two numbers: how far right ($x_1$) and how far up ($x_2$).

WHY. Before a machine can move an arrow, we need a way to name an arrow with numbers. That naming is the whole reason coordinates exist. See Dot Product and Vectors for the arrow-as-numbers idea.

PICTURE. The red arrow below reaches the point $(3,1)$: three steps right, one step up.

$\mathbf{x}=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\underset{\text{steps right}}{\underbrace{x_1}}\underset{\text{steps up}}{\underbrace{x_2}}$

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Step 2 — A matrix moves an arrow: what "apply $B$" means

WHAT. A $2\times 2$ matrix $B$ takes the arrow $\mathbf{x}$ and produces a new arrow $B\mathbf{x}$. The rule for the new arrow's first number is: walk along row 1 of $B$, multiply each entry by the matching entry of $\mathbf{x}$, add them up. Same for row 2.

$B\mathbf{x}=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}\stackrel{\text{new right-amount}}{\overbrace{b_{11}{x}_1+b_{12}{x}_2}}\\ \underset{\text{new up-amount}}{\underbrace{b_{21}{x}_1+b_{22}{x}_2}}\end{array}\right]$

Term by term in the top slot: $b_{11}$ says "how much of the old right becomes new right", $b_{12}$ says "how much of the old up leaks into new right". Every entry is a mixing amount.

WHY row-dot-vector? Because a matrix is meant to be linear: doubling the input doubles the output, and inputs add. The only formula built from the input that obeys that is a weighted sum of $x_1,x_2$ — exactly a dot product of a row with the arrow. Any curvier rule would break linearity. This is the machine in Linear Transformations.

PICTURE. Below, the pale-yellow arrow is $\mathbf{x}=(1,1)$; $B$ shears it (leans it rightward) into the chalk-blue arrow $B\mathbf{x}$.

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Step 3 — Two machines in a row: the question that forces multiplication

WHAT. Suppose we first apply $B$, then apply $A$ to the result. So the arrow's journey is $\mathbf{x}\stackrel{B}{\to }B\mathbf{x}\stackrel{A}{\to }A(B\mathbf{x}).$

The question: is there a single matrix — call it $AB$ — that does both jumps at once? If yes, its entries are not free to choose; they are whatever makes $(AB)\mathbf{x}=A(B\mathbf{x})$ true for every arrow $\mathbf{x}$.

WHY this and not entrywise? We could invent a product by multiplying matching entries (that is the "Hadamard" idea in the parent's mistake box). But that product answers no useful question — it does not describe "do $B$ then $A$". We want the product that chains machines, because chaining is what transformations, and the systems in Systems of Linear Equations, actually need.

PICTURE. The two-hop journey: pink arrow $\to$ (after $B$) blue arrow $\to$ (after $A$) yellow arrow. We are hunting for one machine that jumps pink straight to yellow.

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Step 4 — Grind it out: the row·column formula appears on its own

WHAT. Let us compute $A(B\mathbf{x})$ honestly and watch the definition of $AB$ fall out. Write $\mathbf{y}=B\mathbf{x}$, so $y_k={\sum }_j{b}_{kj}{x}_j$ (Step 2's rule, with $k$ the row index, $j$ summed over the input slots).

Now apply $A$. The $i$-th number of $A\mathbf{y}$ is row $i$ of $A$ dotted with $\mathbf{y}$:

$[A\mathbf{y}{]}_i={\sum }_k{a}_{ik}{y}_k={\sum }_k{a}_{ik}(\underset{y_k}{\underbrace{\sum _j{b}_{kj}{x}_j}}).$

Swap the order of the two sums (allowed — it is just adding the same numbers in a different order):

$[A(B\mathbf{x}){]}_i={\sum }_j(\underset{\text{call this }(AB{)}_{ij}}{\underbrace{\sum _k{a}_{ik}{b}_{kj}}})x_j.$

Term by term: $a_{ik}$ is "how much slot $k$ feeds output $i$"; $b_{kj}$ is "how much input $j$ feeds slot $k$". Their product summed over the middle index $k$ counts every path from input $j$ to output $i$. That total is $(AB{)}_{ij}$.

WHY the inner index must match. The letter $k$ runs over the columns of $A$ and the rows of $B$ simultaneously. If $A$ has $p$ columns but $B$ has a different number of rows, the sum ${\sum }_k$ has mismatched terms — it is meaningless. Hence the inner dimensions must agree.

PICTURE. Entry $(AB{)}_{ij}$ = the highlighted row of $A$ sliding along the highlighted column of $B$; each aligned pair is multiplied, the pile is summed.

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Step 5 — Do a real one, entry by entry

WHAT. Take $A=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$, $B=\left[\begin{array}{cc}4& 0\\ 1& 5\end{array}\right]$. Build $AB$ one slot at a time.

$(AB{)}_{11}=\underset{\text{row1⋅col1}}{\underbrace{1\cdot 4}}+2\cdot 1=6,(AB{)}_{12}=1\cdot 0+2\cdot 5=10,$ $(AB{)}_{21}=0\cdot 4+3\cdot 1=3,(AB{)}_{22}=0\cdot 0+3\cdot 5=15.$

$AB=\left[\begin{array}{cc}6& 10\\ 3& 15\end{array}\right].$

WHY line them up like this. Slot $(2,1)$ uses row 2 of $A$ and column 1 of $B$ — the row index copies from $A$, the column index copies from $B$. That bookkeeping is the whole skill.

PICTURE. The $(2,1)$ slot lit up: row 2 of $A$ against column 1 of $B$, giving $0\cdot 4+3\cdot 1=3$.

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Step 6 — Order matters: compute $BA$ and watch it disagree

WHAT. Same $A,B$, but now $B$ acts second:

$BA=\left[\begin{array}{cc}4& 0\\ 1& 5\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}4\cdot 1+0\cdot 0& 4\cdot 2+0\cdot 3\\ 1\cdot 1+5\cdot 0& 1\cdot 2+5\cdot 3\end{array}\right]=\left[\begin{array}{cc}4& 8\\ 1& 17\end{array}\right].$

Compare: $AB=\left[\begin{array}{cc}6& 10\\ 3& 15\end{array}\right]\ne \left[\begin{array}{cc}4& 8\\ 1& 17\end{array}\right]=BA$.

WHY they differ. $AB$ means "do $B$, then $A$"; $BA$ means "do $A$, then $B$". These are different journeys — socks then shoes vs shoes then socks. The algebra simply records that the geometry is different.

PICTURE. Two journeys of the same starting arrow: one path (shear-then-scale) lands somewhere different from the other (scale-then-shear).

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Step 7 — The degenerate cases: zero, identity, and vanishing products

WHAT. Three limiting machines you must never be surprised by.

• Identity $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ leaves every arrow untouched: $I\mathbf{x}=\mathbf{x}$. Row 1 picks out $x_1$, row 2 picks out $x_2$ — pure "do nothing".
• Zero matrix $0$ crushes every arrow to the origin: $0\mathbf{x}=0$.
• Zero divisors. Two nonzero machines can compose to the zero machine: $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$ The right machine keeps only the "up" part of an arrow; the left machine keeps only the "right" part. Do them in sequence and nothing survives.

WHY this can't happen with ordinary numbers. For plain numbers $ab=0$ forces $a=0$ or $b=0$. Matrices break this because a machine can throw away a whole direction (it is singular, $\det =0$; see Determinant and Inverse of a Matrix). Two machines that discard complementary directions leave nothing.

PICTURE. Left machine flattens all arrows onto the horizontal line; right machine flattens onto the vertical line. An arrow squashed vertical-first, then horizontal, ends at the origin.

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The one-picture summary

Everything above collapses to a single claim: the row·column rule is not a convention, it is the unique fingerprint of "chain the machines."

Recall Feynman retelling

Picture arrows on graph paper. A matrix is a little machine that grabs each arrow and moves its tip — but only in a "fair" way: double the arrow, double the move; add arrows, add the moves. That fairness forces each output number to be a weighted mix of the input numbers — a dot of a row with the arrow.

Now put two machines in a line: first $B$, then $A$. Is one super-machine the same as both? Yes — and when you patiently follow one arrow through, the numbers of that super-machine turn out to be row of $A$ times column of $B$, summed down the middle. Nobody chose that formula; it fell out of "do them in sequence."

Because "do $B$ then $A$" is a different trip from "do $A$ then $B$" — socks-then-shoes versus shoes-then-socks — $AB$ and $BA$ usually disagree. And because a machine can flatten a whole direction to nothing, two nonzero machines can multiply to the do-nothing-but-crush zero machine. The identity is the lazy machine that touches nothing at all.

Connections

• Linear Transformations — the machines these matrices are.
• Dot Product and Vectors — every product entry is one of these.
• Systems of Linear Equations — chaining and inverting these machines solves them.
• Determinant and Inverse of a Matrix — which machines flatten a direction (singular).
• Eigenvalues and Eigenvectors — arrows a machine only stretches.