# Matrices of Linear Maps Review

We will now review some of the recent content regarding matrices of linear maps.

- Recall from the Solution Spaces to Systems of Linear Equations page that a system of $m$ linear equations in $n$ unknowns could be represented succinctly as a linear transformation$T \in \mathcal L (\mathbb{F}^n, \mathbb{F}^m)$ as:

(1)

\begin{align} \quad T((x_1, x_2, …, x_n)) = \left ( \sum_{k=1}^{n} a_{1,k}x_k, \sum_{k=1}^{n} a_{2,k}x_k, …, \sum_{k=1}^{n} a_{m,k}x_k \right ) \end{align}

- A
**Homogenous Linear System**has the form $T(x_1, x_2, …, x_n) = 0$ which is equivalent to:

(2)

\begin{align} \sum_{k=1}^{n} a_{1,k}x_k = a_{1,1}x_1 + a_{1,2}x_2 + … + a_{1,n}x_n = 0 \\ \sum_{k=1}^{n} a_{2,k}x_k = a_{2,1}x_1 + a_{2,2}x_2 + … + a_{2,n}x_n = 0\\ \vdots \quad \quad \quad \quad\quad \quad\quad \quad \\ \sum_{k=1}^{n} a_{m,k}x_k = a_{m,1}x_1 + a_{m,2}x_2 + … + a_{m,n}x_n = 0\\ \end{align}

- We already know that homogenous linear systems always have a solution – namely the trivial solution $(x_1, x_2, …, x_n) = (0, 0, …, 0)$. However, we proved that if the number of unknowns $n$ is greater than the number of equations $m$ then a nontrivial solution is guaranteed to exist.

- We then looked at
**Nonhomogenous Linear Systems**which have the form $T(x_1, x_2, …, x_n) = (b_1, b_2, …, b_m)$ where not all of the $b$‘s are equal to zero. This is equivalent to:

(3)

\begin{align} \sum_{k=1}^{n} a_{1,k}x_k = a_{1,1}x_1 + a_{1,2}x_2 + … + a_{1,n}x_n = b_1 \\ \sum_{k=1}^{n} a_{2,k}x_k = a_{2,1}x_1 + a_{2,2}x_2 + … + a_{2,n}x_n = b_2\\ \vdots \quad \quad \quad \quad\quad \quad\quad \quad \\ \sum_{k=1}^{n} a_{m,k}x_k = a_{m,1}x_1 + a_{m,2}x_2 + … + a_{m,n}x_n = b_m\\ \end{align}

- We then proved that if the number of equations $m$ is greater than the number of unknowns $n$ then there exists a vector $b = (b_1, b_2, …, b_m)$ for which there exists no solution to the nonhomogenous linear system from above.

- On the The Matrix of a Linear Map we said that if $V$ and $W$ are both finite-dimensional vector spaces with bases $B_V = \{ v_1, v_2, …, v_n \}$ and $B_W = \{ w_1, w_2, …, w_m \}$ respectively and if $T \in \mathcal L (V)$ then the
**Matrix of $T$**with respect to the bases $B_V$ and $B_W$ is denoted $\mathcal M (T, B_V, B_W)$ where $T(v_k) = a_{1,k}w_1 + a_{2,k}w_2 + … + a_{m, k} w_m$ for each $k = 1, 2, …, n$ is given by:

(4)

\begin{align} \quad \mathcal M (T, B_V, B_W) = \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix} \end{align}

- In other words, the matrix of $T$ with respect to $B_V$ and $B_W$ can be obtained by applying the linear map $T$ to each basis vector. The columns of $\mathcal M (T, B_V, B_W)$ will be the values of the coefficients of the linear combination of the basis vectors from $W$ which equal to $T$ when applied to basis vectors in $V$.