From Basis Images to a Transformation
A linear transformation is often introduced by a formula, such as \(T(x,y)=(x+y,2x-y)\). Another useful way to specify one is to say where it sends a small collection of vectors. In \(\mathbb{R}^n\), the standard basis vectors \(\mathbf{e}_1,\dots,\mathbf{e}_n\) provide a particularly direct specification: every input vector is a linear combination of them, with its coordinates as coefficients. Once their images are prescribed, linearity requires the image of every other input to follow from those same coefficients.
We work over \(\mathbb{R}\). The standard basis vector \(\mathbf{e}_j\in\mathbb{R}^n\) has a \(1\) in position \(j\) and \(0\) in every other position. Thus any \(\mathbf{x}=(x_1,\dots,x_n)^\top\in\mathbb{R}^n\) has the coordinate expansion \(\mathbf{x}=x_1\mathbf{e}_1+\cdots+x_n\mathbf{e}_n\). A linear map \(T:\mathbb{R}^n\to\mathbb{R}^m\) preserves linear combinations: for real scalars \(c_1,\dots,c_k\) and vectors \(\mathbf{u}_1,\dots,\mathbf{u}_k\), \(T(\sum_{j=1}^k c_j\mathbf{u}_j)=\sum_{j=1}^k c_jT(\mathbf{u}_j)\).
The prescribed vectors are the columns of the matrix of the construction. If \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\), then multiplying \(\mathbf{A}\) by \(\mathbf{x}\) forms precisely the weighted sum \(x_1\mathbf{v}_1+\cdots+x_n\mathbf{v}_n\). Therefore the map can also be written \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\). This is a map from \(\mathbb{R}^n\) to \(\mathbb{R}^m\), with an \(m\times n\) matrix: the number of columns matches the input coordinates, and the number of entries in each output matches \(m\).
Why the Construction Works
Proof. Let \(\mathbf{x}=(x_1,\dots,x_n)^\top\) and \(\mathbf{y}=(y_1,\dots,y_n)^\top\) be arbitrary inputs, and let \(a,b\in\mathbb{R}\). The \(j\)th coordinate of \(a\mathbf{x}+b\mathbf{y}\) is \(ax_j+by_j\). Applying the proposed rule gives
This holds for all inputs and scalars, so \(T\) is linear. For the standard basis vector \(\mathbf{e}_k\), its \(k\)th coordinate is \(1\), while every other coordinate is \(0\). Hence \(T(\mathbf{e}_k)=\sum_{j=1}^n(\mathbf{e}_k)_j\mathbf{v}_j=\mathbf{v}_k\). This verifies each prescribed image. \(\square\)
Proof. The proposition constructs a linear map with all the prescribed images, so existence holds. To prove uniqueness, suppose \(S:\mathbb{R}^n\to\mathbb{R}^m\) is any linear map satisfying \(S(\mathbf{e}_j)=\mathbf{v}_j\) for every \(j\). For an arbitrary \(\mathbf{x}=(x_1,\dots,x_n)^\top\), its standard-coordinate expansion and linearity give
Thus \(S\) and \(T\) agree on every input, which means they are the same function. There can be no second linear map with the prescribed images. \(\square\)
The hypotheses allow the vectors \(\mathbf{v}_j\) to be zero or linearly dependent. There is no independence requirement: the input basis vectors are independent, but their images under a linear map need not be. The construction still works because it uses the unique standard-coordinate expansion of each input, not an assumption that the output vectors form a basis.
Computing Images by Coordinates
Worked Example: A Map from R² to R³
Construct the linear map \(T:\mathbb{R}^2\to\mathbb{R}^3\) with \(T(\mathbf{e}_1)=(1,2,0)^\top\) and \(T(\mathbf{e}_2)=(-1,1,3)^\top\). Then find \(T((3,-2)^\top)\).
Use the prescribed images as the columns, in the same order as the standard basis vectors:
For \(\mathbf{x}=(x_1,x_2)^\top\), the coordinate formula is
The input \((3,-2)^\top\) has coefficients \(3\) and \(-2\) on \(\mathbf{e}_1\) and \(\mathbf{e}_2\). Thus
Check: Direct matrix multiplication gives \(\begin{bmatrix}1&-1\\2&1\\0&3\end{bmatrix}\begin{bmatrix}3\\-2\end{bmatrix}=\begin{bmatrix}3+2\\6-2\\-6\end{bmatrix}=\begin{bmatrix}5\\4\\-6\end{bmatrix}\), agreeing with the linear-combination calculation. The constructed map also returns the prescribed columns when the inputs are \(\mathbf{e}_1\) and \(\mathbf{e}_2\).
For more than two input coordinates, the same rule applies without any change: each input coordinate weights the corresponding prescribed image. The matrix product is a compact way to calculate the sum, but the linear-combination formula explains why those are the correct weights.
Worked Example: A Map from R³ to R²
Let \(U:\mathbb{R}^3\to\mathbb{R}^2\) be defined by \(U(\mathbf{e}_1)=(2,-1)^\top\), \(U(\mathbf{e}_2)=(0,3)^\top\), and \(U(\mathbf{e}_3)=(-2,4)^\top\). Find a coordinate formula and calculate \(U((1,2,-1)^\top)\).
The three image vectors become the three columns of a \(2\times3\) matrix:
For \(\mathbf{x}=(x_1,x_2,x_3)^\top\), use its three standard coordinates as coefficients:
For \(\mathbf{x}=(1,2,-1)^\top\), the coefficients on the three basis vectors are \(1,2,-1\), so
Check: Multiplying the matrix by the input gives \(\begin{bmatrix}2&0&-2\\-1&3&4\end{bmatrix}\begin{bmatrix}1\\2\\-1\end{bmatrix}=\begin{bmatrix}2+0+2\\-1+6-4\end{bmatrix}=\begin{bmatrix}4\\1\end{bmatrix}\). The output has two coordinates, as required for a map into \(\mathbb{R}^2\).
Dependent Images Are Allowed
It may seem that specifying \(n\) images should require \(n\) independent vectors. That would be necessary if the images themselves were to serve as a basis of \(\mathbb{R}^m\), but that is not the requirement here. They are the columns defining a map. A zero column simply means the corresponding standard basis vector is sent to zero; repeated or dependent columns are also permitted.
Worked Example: Prescribed Images with Dependence
Define \(V:\mathbb{R}^3\to\mathbb{R}^2\) by \(V(\mathbf{e}_1)=(1,1)^\top\), \(V(\mathbf{e}_2)=(2,2)^\top\), and \(V(\mathbf{e}_3)=(0,-1)^\top\). Find \(V((-1,2,4)^\top)\), noting that the first two prescribed images are dependent.
Since \((2,2)^\top=2(1,1)^\top\), those two columns are dependent. This does not prevent the construction. The matrix and coordinate formula are
For the input \((-1,2,4)^\top\), its coordinates give
Check: The matrix product is \(\begin{bmatrix}1&2&0\\1&2&-1\end{bmatrix}\begin{bmatrix}-1\\2\\4\end{bmatrix}=\begin{bmatrix}-1+4\\-1+4-4\end{bmatrix}=\begin{bmatrix}3\\-1\end{bmatrix}\). Also, substituting the three standard basis inputs into the coordinate formula returns \((1,1)^\top\), \((2,2)^\top\), and \((0,-1)^\top\), respectively.
Common Pitfall: Forgetting the Input Coordinates
A frequent error is to add the prescribed images without multiplying each one by the matching input coordinate. For example, in the first worked example the input is \(3\mathbf{e}_1-2\mathbf{e}_2\), not \(\mathbf{e}_1+\mathbf{e}_2\). The incorrect calculation \(T((3,-2)^\top)=T(\mathbf{e}_1)+T(\mathbf{e}_2)=(1,2,0)^\top+(-1,1,3)^\top=(0,3,3)^\top\) ignores both coordinates. The correction is to use \(3T(\mathbf{e}_1)-2T(\mathbf{e}_2)=(5,4,-6)^\top\), as verified by matrix multiplication above.
It is also important to keep the input and output dimensions distinct. For a map \(\mathbb{R}^n\to\mathbb{R}^m\), there are \(n\) prescribed standard-basis images, each with \(m\) entries. Putting those vectors as columns creates an \(m\times n\) matrix. A dimension mismatch in the matrix is a useful warning that either the images were arranged incorrectly or the input and output spaces have been confused.
Check Your Understanding
Use coordinates in the standard basis to set up each map before calculating its action.
- Construct the matrix of the linear map \(\mathbb{R}^2\to\mathbb{R}^2\) with \(T(\mathbf{e}_1)=(2,0)^\top\) and \(T(\mathbf{e}_2)=(-1,3)^\top\). Find \(T((2,1)^\top)\).
- For a map \(S:\mathbb{R}^3\to\mathbb{R}\) with \(S(\mathbf{e}_1)=2\), \(S(\mathbf{e}_2)=0\), and \(S(\mathbf{e}_3)=-1\), find a coordinate formula and calculate \(S((1,-2,3)^\top)\).
- Explain why a prescribed image of \(\mathbf{e}_j\) may be the zero vector without preventing a linear map from being constructed.
- Suppose two linear maps \(\mathbb{R}^n\to\mathbb{R}^m\) agree on every standard basis vector. Explain why they agree on every input.
- A student claims that the images of the standard basis must be linearly independent for the construction to work. Give a counterexample and explain the error.
