Forming Standard Matrices with Standard Basis Vectors

We will now look at one method for forming the standard matrix of a linear transformation with standard basis vectors.

# Forming Standard Matrices with Standard Basis Vectors

 Definition: The standard vectors in $\mathbb{R}^n$ denoted $\vec{e_1}, \vec{e_2}, ..., \vec{e_n}$ are defined to be the $n$-component vectors such that for any $\vec{e_i}$, the $i^{th}$ component is 1 and every other component in $\vec{e_i}$ is 0.

For example, in $\mathbb{R}^4$, the standard basis vectors are $e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, $e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, $e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, and $e_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$.

Now recall the following two properties that defined a transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ to be linear for any vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ and some scalar $k$:

• 1. $T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$.
• 2. $T(k\vec{u}) = kT(\vec{u})$.

We can extend property 1 to apply to multiple vectors, that is $T(v_1 + v_2 + ... + v_n) = T(v_1) + T(v_2) + ... + T(v_n)$. Now if take all of the standard basis vectors and apply our transformation $T$ on them such that we get $T(e_1)$, $T(e_2)$, …, $T(e_n)$, we cant thus form $A$ by making it that matrix whose successive columns are these transformations, that is:

(1)
\begin{align} A = [T(e_1) \mid T(e_2) \mid \: ... \: \mid T(e_n)] \end{align}

Now suppose that $x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$. We know that $Ax$ is a linear combination, and thus:

(2)
\begin{align} Ax = x_1T(e_1) + x_2T(e_2) + ... + x_nT(e_n) \\ Ax = T(x_1e_1) + T(x_2e_2) + ... + T(x_ne_n) \\ Ax = T(x_1e_1 + x_2e_2 + ... + x_ne_n) \\ Ax = T(x) \end{align}