Q.If $P = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ and $Q = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$, prove that $PQ = \begin{bmatrix} xa & 0 & 0 \\ 0 & yb & 0 \\ 0 & 0 & zc \end{bmatrix} = QP$.

Concept: Diagonal Matrix Multiplication — the product of two diagonal matrices is commutative and is obtained by multiplying corresponding diagonal entries.

Step 1: Write $P$ and $Q$ as diagonal matrices. Their product $PQ$ is defined by row‑of‑$P$ times column‑of‑$Q$.

Step 2: Compute $PQ$ entry‑wise. For the $(1,1)$ entry: first row of $P$ is $[x,0,0]$, first column of $Q$ is $[a,0,0]^T$, so $x \cdot a + 0 + 0 = xa$. Similarly, $(2,2)$ entry: $y \cdot b = yb$, and $(3,3)$ entry: $z \cdot c = zc$. All off‑diagonal entries are zero because each row of $P$ and each column of $Q$ have only one non‑zero term, and they ne …

The key insight is that diagonal matrices are the simplest matrices to multiply. Because all off-diagonal entries are zero, each row of $P$ has only one non-zero entry, and each column of $Q$ has only one non-zero entry. This makes the dot products in matrix multiplication trivial — they collapse to a single multiplication of the corresponding diagonal elements.

Let’s walk through it step by step.

1. Recall the definition of matrix multiplication. If $A$ is $3 \times 3$ and $B$ is $3 \times 3$, then the entry $(i,j)$ of $AB$ is the dot product of row $i$ of $A$ with column $j$ of $B$:

2. Apply this to $P$ and $Q$.

   $P$ has entries $P_{11}=x$, $P_{22}=y$, $P_{33}=z$, and all other $P_{ij}=0$.

   $Q$ has entries $Q_{11}=a$, $Q_{22}=b$, $Q_{33}=c$, and all other $Q_{ij}=0$.

   For the product $PQ$, look at entry $(1,1)$:

For entry $(1,2)$:

Every off-diagonal entry works the same way — the only non-zero term in the sum would require both $P_{ik}$ and $Q_{kj}$ to be non-zero, but that never happens when $i \neq j$ because the non-zero entries of $P$ and $Q$ are only on the diagonal.

3. Generalise the pattern.

   For any $i$ and $j$:

   • If $i = j$, then $(PQ)_{ii} = P_{ii} Q_{ii}$ (since $k=i$ is the only term that survives).
   • If $i \neq j$, then every term $P_{ik} Q_{kj}$ is zero — either $P_{ik}=0$ (when $k \neq i$) or $Q_{kj}=0$ (when $k \neq j$), and $i \neq j$ means no single $k$ can equal both $i$ and $j$.

   So the product is exactly:

4. Now check $QP$. …

Method: Multiplying diagonal matrices (and why they commute)

Use this when both matrices are diagonal.

Steps

Step 1: Apply the row-times-column rule.

For diagonal $P$ and $Q$, the $(i,j)$ entry of $PQ$ is a sum in which only the aligned diagonal terms survive; every off-diagonal entry is $0$.

Step 2: Multiply corresponding diagonal entries.

The $(i,i)$ entry of $PQ$ is the product of the $i$-th diagonal entries: …

Common Mistakes

Mistake 1: Adding the diagonal entries instead of multiplying them.

Why it's wrong: the $(i,i)$ entry of $PQ$ is the product of the two $i$-th diagonal entries ($xa$, $yb$, $zc$), not their sum. Correct approach: multiply corresponding diagonal entries.

Mistake 2: Assuming this proves all matrices commute.

Why it's wrong: $PQ=QP$ holds here because the matrices are diagonal; general matrices do not commute. Correct approach: state that commutativity follows from the diagonal structure, not as a universal rule. …