I'm trying to know what is the matrix representation of a composition of linear transformations: $V\to E \to W \qquad T_1: TV=E \quad T_2:TE=W $

Also $\dim V=n \qquad \dim E=m \qquad \dim W=k$

Where I define

$\beta=\{ a_1,a_2,\dots,a_n\}$ Basis of V.

$\gamma=\{ b_1,b_2,\dots,b_m\}$ Basis of E.

$\delta=\{ d_1,d_2,\dots,d_k\}$ Basis of W.


$$T_2 \circ T_1(a_1)=T_2\left(\sum\limits_{m}\alpha_{m1}b_m\right)=\sum\limits_{k}\beta_{k1}d_k$$

$$T_2 \circ T_1(a_2)=T_2\left(\sum\limits_{m}\alpha_{m2}b_m\right)=\sum\limits_{k}\beta_{k2}d_k$$ $$\vdots$$

$$T_2 \circ T_1(a_n)=T_2\left(\sum\limits_{m}\alpha_{mn}b_m\right)=\sum\limits_{k}\beta_{kn}d_k$$

So the matrix representation of this transformations is $\left[T_2\circ T_1\right]_\gamma^\delta = \left[T_2\right]_\gamma^\delta\left[T_1\right]_\beta^\gamma$ $$\left[T_1\right]_\beta^\gamma= \begin{bmatrix} \alpha_{11}&\alpha_{12}& \ldots & \alpha_{1n} \\ \alpha_{21}&\alpha_{22}& \ldots & \alpha_{2n} \\ \vdots&\vdots& & \vdots\\ \alpha_{m1}&\alpha_{m2}& \ldots & \alpha_{mn} \\ \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} $$

$$\left[T_2\right]_\gamma^\delta= \begin{bmatrix} \beta_{11}&\beta_{12}& \ldots & \beta_{1m} \\ \beta_{21}&\beta_{22}& \ldots & \beta_{2m} \\ \vdots&\vdots& & \vdots\\ \beta_{k1}&\beta_{k2}& \ldots & \beta_{km} \\ \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \\ \vdots \\ d_k \end{bmatrix} $$

So I know the matrix should be of $k$ rows and $n$ columns. But I got stuck on what should be the matrix representation of $T_2 \circ T_1$ using the definitions of the matrix representation $T_1$ and $T_2$.

I was reading this document http://aleph0.clarku.edu/~djoyce/ma130/composition.pdf, but still don't grasp the idea completely.


3 Answers 3


There is a trick which I find helps in calculating matrix representations, similarity transformations, diagonalization, etc...: Write vectors to the left, coordinates to the right (to be consistent with standard notations). More precisely:

We consider as above basis vectors $(\vec{a}_i)$ in $V$ (normally one doesn't put arrows but it may help reading). Any vector $\vec{x}\in V$ may be written: $$ \vec{x}= \sum_i \vec{a}_i x_i,$$ where $(x_i)$ are the coordinates in this basis. Note that I write the basis vectors to the left, coordinates to the right. Acting with $T_1$ on $\vec{a}_i$ we get (again basis to the left): $$ T_1 \vec{a}_i = \sum_j \vec{b}_j \alpha_{ji}$$ Then by linearity (and exhanging order of summation): $$ T_2 T_1 a_i = \sum_j \left( T_2 \vec{b}_j \right) \alpha_{ji} = \sum_j \left( \sum_k \vec{d}_k \beta_{kj} \right) \alpha_{ji}= \sum_k \vec{d}_k \left( \sum_j \beta_{kj} \alpha_{ji} \right) $$ from which you may read off the matrix representation (the term in parentheses to the right, thus answering the question). Returning to the distinction between basis and coordinates, let's act with $T_1$ upon the (general) vector $\vec{x}$ given above, using linearity:

$$ T_1 \vec{x} = \sum_i (T \vec{a}_i) x_i = \sum_i \left( \sum_j \vec{b}_j \alpha_{ji} \right) x_i = \sum_j \vec{b}_j \left( \sum_i\alpha_{ji} x_i \right) .$$ The matrix is multiplied to the right on basis vectors, but to the left on coordinates. The point is that using this order the indices that are summed over are always adjacent. This reduces errors quite drastically.


Picture is best way to understand this. See Kuldeep Singh's Linear Algebra: Step by Step (2013) p 410.

enter image description here


Quite simple - you multiply the matrices: $T_2 T_1$

It works because matrix multiplication is associative, so:

$$(T_2 T_1) x = T_2 (T_1 x)$$

Now, $T_1x$ is a vector in $E$, and $T_2$ transforms it into $W$.

Or, abusing notation a bit:

$$(T_2T_1)V = T_2(T_1V) = T_2E = W$$


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