Composition of two change of basis formulas

Question

Suppose we have basis $B_1,B_2,B_3$ and let $P$ be the change of basis formula from $B_1 \to B_2$ and $Q$ change of basis from $B_2 \to B_3$. Show $PQ$ is change of basis formula from $B_1$ to $B_3$.

Attempt

Write $B_1 = \{ v_1,...,v_n \}, B_2 = \{ u_1,...,u_n \}, B_3 = \{ w_1,...,w_n \}$. We know that if for $i=1,...,n$, we have

$$ u_i = a_{i1} v_1 + a_{i2} v_2 + ... + a_{in} v_n $$

then $P$ is the matrix $[ a_{ji} ]$. Next, write

$$ w_i = b_{i1} u_1 + b_{i2} u_2 + ... + b_{in} u_n , \; \; \; i=1,...,n $$

So, $Q = [b_{ji}]$. Notice we can write

$$ w_i = b_{i1}( a_{11} v_1 + a_{12} v_2 + ... + a_{1n} v_n ) + b_{i2}( a_{21} v_1 + ... + a_{2n}v_n) + ... + b_{in} (a_{n1} v_1 + a_{n2} v_2 + ... + a_{nn} v_n ) $$

Which we rearrange and write

$$ w_i = v_1 ( \sum_{j=1}^n b_{ij} a_{1j}) + v_2 ( \sum_{j=1}^n b_{ij} a_{2j} ) + ... + v_n ( \sum_{j=1}^n b_{ij} a_{nj}) $$

i=1,...n.

this means that PQ is change of basis from $B_1 \to B_3$. IS this a correct argument?

Answer 1

The last big formula in your proof should be

$$ w_i = v_1 ( \sum_{j=1}^n b_{ij} a_{j1}) + v_2 ( \sum_{j=1}^n b_{ij} a_{j2} ) + ... + v_n ( \sum_{j=1}^n b_{ij} a_{jn}) $$

instead, the rest of it is correct. (I have used the definition of the change-of-basis matrix from Schaum's Outline: Linear Algebra 4th Ed.)

PS. A detailed canonical answer is required to address all the concerns.

Your proof is already sufficiently detailed, it uses only rearrangements of equalities of sums of coefficients.