Find a Basis for the Range of a Linear Transformation of Vector Spaces of Matrices

Vector Space Problems and Solutions

Problem 682

Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.\]

Find a basis for the range of $T$.

 
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Solution.

For any matrix $M \in V$ we can write $T(M)$ as a sum
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = a \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & -3 \end{bmatrix} + b \begin{bmatrix} 1 & 0 \\ 2 & 0 \\ 2 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 0 & -3 \\ -1 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & 2 \\ -1 & 0 \\ 0 & 0 \end{bmatrix}.\]

From this, we see that any element in the range of $T$ can be written as a linear sum of four elements
\[\mathbf{v}_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & -3 \end{bmatrix} , \quad \mathbf{v}_2 = \begin{bmatrix} 1 & 0 \\ 2 & 0 \\ 2 & 0 \end{bmatrix},\] \[\mathbf{v}_3 = \begin{bmatrix} 0 & 0 \\ 0 & -3 \\ -1 & 0 \end{bmatrix} , \quad \mathbf{v}_4 = \begin{bmatrix} 0 & 2 \\ -1 & 0 \\ 0 & 0 \end{bmatrix}.\]

This means that the set $\{ \mathbf{v}_1 , \mathbf{v}_2 , \mathbf{v}_3 , \mathbf{v}_4 \}$ is a basis of $W$ as long as the four vectors are linearly independent. To check this, we will show that the coordinate vectors for the $\mathbf{v}_i$, relative to the standard basis, are linearly independent. The standard basis is composed of the matrices
\[\mathbf{e}_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \end{bmatrix},\] \[\mathbf{e}_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_5 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{bmatrix} , \, \mathbf{e}_6 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{bmatrix}.\]


Relative to the standard basis $B = \{ \mathbf{e}_1 , \mathbf{e}_2 , \mathbf{e}_3 , \mathbf{e}_4 , \mathbf{e}_5 , \mathbf{e}_6 \}$, the coordinate vectors for the $\mathbf{v}_i$ are
$$ [ \mathbf{v}_1 ]_{B} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ -3 \end{bmatrix} , \, [ \mathbf{v}_2 ]_{B} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ 0 \\ 2 \\ 0 \end{bmatrix} , \, [ \mathbf{v}_3 ]_{B} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ -3 \\ -1 \\ 0 \end{bmatrix} , \, [ \mathbf{v}_4 ]_{B} = \begin{bmatrix} 0 \\ 2 \\ -1 \\ 0 \\ 0 \\ 0 \end{bmatrix} . $$

To show that these are linearly independent, we put these vectors in order and obtain the matrix
\[ \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & -3 & 0 \\ 0 & 2 & -1 & 0 \\ -3 & 0 & 0 & 0 \end{bmatrix} . \]

To show linear independence of the columns, it suffices to show that this matrix has rank $4$. To do this, we will row-reduce it.

\begin{align*}
\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & -3 & 0 \\ 0 & 2 & -1 & 0 \\ -3 & 0 & 0 & 0 \end{bmatrix} \xrightarrow[\frac{-1}{3} R_6]{ \substack{ \frac{1}{2} R_2 \\[4pt] \frac{-1}{3} R_4 } } \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & -1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{ R_1 \leftrightarrow R_6 } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & -1 & 0 \\ 1 & 1 & 0 & 0 \end{bmatrix} \\[6pt] \xrightarrow{ R_6 – R_1 } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix} \xrightarrow[R_5 – 2 R_6 + R_4]{ R_3 – 2 R_6 + R_2} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix} .
\end{align*}


The four columns of this matrix are clearly linearly independent, and so it has rank $4$. Thus the original matrix has rank $4$ as well, and so the set $\{ \mathbf{v}_1 , \mathbf{v}_2 , \mathbf{v}_3 , \mathbf{v}_4 \}$ is linearly independent and is a basis of $W$.


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