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 $T$.

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 $W$.

The post Find a Basis for the Range of a Linear Transformation of Vector Spaces of Matrices first appeared on Problems in Mathematics.

Let $V$ be the vector space of all $2\times 2$ real matrices.
Let $S=\{A_1, A_2, A_3, A_4\}$, where
\[A_1=\begin{bmatrix}
1 & 2\\
-1& 3
\end{bmatrix}, A_2=\begin{bmatrix}
0 & -1\\
1& 4
\end{bmatrix}, A_3=\begin{bmatrix}
-1 & 0\\
1& -10
\end{bmatrix}, A_4=\begin{bmatrix}
3 & 7\\
-2& 6
\end{bmatrix}.\] Then find a basis for the span $\Span(S)$.

Solution.

Let $B=\{E_{11}, E_{12}, E_{21}, E_{22}\}$ be the standard basis of $V$, where
\[E_{11}=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}, E_{12}=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}, E_{21}=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}, E_{22}=\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.\] We find the coordinate vectors of $A_1, A_2, A_3, A_4$ with respect to the basis $B$.

For the matrix $A_1$, we can write it as a linear combination of basis vectors of $B$ as follows.
We have
\[A_1=1\cdot E_{11}+2E_{12}+(-1)E_{21}+3E_{22}.\] Hence the coordinate vector of $A_1$ with respect to the basis $B$ is
\[[A_1]_B=\begin{bmatrix}
1 \\
2 \\
-1 \\
3
\end{bmatrix}.\] Similarly, we obtain
\[[A_2]_B=\begin{bmatrix}
0 \\
-1 \\
1 \\
4
\end{bmatrix}, [A_3]_B=\begin{bmatrix}
-1 \\
0 \\
1 \\
-10
\end{bmatrix}, [A_4]_B=\begin{bmatrix}
3 \\
7 \\
-2 \\
6
\end{bmatrix}.\]

Let
\begin{align*}
T&=\{[A_1]_B, [A_2]_B, [A_3]_B, [A_4]_B\}\\[6pt] &= \left \{ \,\begin{bmatrix}
1 \\
2 \\
-1 \\
3
\end{bmatrix}, \begin{bmatrix}
0 \\
-1 \\
1 \\
4
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1 \\
-10
\end{bmatrix}, \begin{bmatrix}
3 \\
7 \\
-2 \\
6
\end{bmatrix} \, \right\}.
\end{align*}
Then $T$ is a subset in $\R^4$.

We find a basis of $\Span(T)$ by the leading $1$-method.
We form a matrix whose columns are vectors in $T$ and reduce it by elementary row operations as follows.

\begin{align*}
\begin{bmatrix}
1 & 0 & -1 & 3 \\
2 &-1 & 0 & 7 \\
-1 & 1 & 1 & -2 \\
3 & 4 & -10 & 6
\end{bmatrix}
\xrightarrow{\substack{R_2-2R_1 \\ R_3+R_1 \\ R_4-3R_1}}
\begin{bmatrix}
1 & 0 & -1 & 3 \\
0 &-1 & 2 & 1 \\
0 & 1 & 0 & 1 \\
0 & 4 & -7 & -3
\end{bmatrix}\\[6pt] \xrightarrow{R_2 \leftrightarrow R_3}
\begin{bmatrix}
1 & 0 & -1 & 3 \\
0 & 1 & 0 & 1 \\
0 &-1 & 2 & 1 \\
0 & 4 & -7 & -3
\end{bmatrix}
\xrightarrow{\substack{R_3+R_2 \\R_4-4R_2}}
\begin{bmatrix}
1 & 0 & -1 & 3 \\
0 &1 & 0 & 1 \\
0 & 0 & 2 & 2 \\
0 & 0 & -7 & -7
\end{bmatrix}\\[6pt] \xrightarrow{\substack{\frac{1}{2}R_3 \\ -\frac{1}{7} R_4}}
\begin{bmatrix}
1 & 0 & -1 & 3 \\
0 &1 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1
\end{bmatrix}
\xrightarrow{\substack{R_1+R_3 \\ R_4-R_3}}
\begin{bmatrix}
1 & 0 & 0 & 4 \\
0 &1 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form and the first three columns contain the leading 1.
Thus the first three vectors of $T$ is a basis of $\Span(T)$.
Hence
\[\{[A_1]_B, [A_2]_B, [A_3]_B\}\] is a basis of $\Span(T)$.
This yields that
\[\{A_1, A_2, A_3\}\] is a basis of $\Span(S)$ by the correspondence of coordinate vectors.

Comment.

These are Quiz 9 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

• Quiz 1. Gauss-Jordan elimination / homogeneous system.
• Quiz 2. The vector form for the general solution / Transpose matrices.
• Quiz 3. Condition that vectors are linearly dependent/ orthogonal vectors are linearly independent
• Quiz 4. Inverse matrix/ Nonsingular matrix satisfying a relation
• Quiz 5. Example and non-example of subspaces in 3-dimensional space
• Quiz 6. Determine vectors in null space, range / Find a basis of null space
• Quiz 7. Find a basis of the range, rank, and nullity of a matrix
• Quiz 8. Determine subsets are subspaces: functions taking integer values / set of skew-symmetric matrices
• Quiz 9. Find a basis of the subspace spanned by four matrices
• Quiz 10. Find orthogonal basis / Find value of linear transformation
• Quiz 11. Find eigenvalues and eigenvectors/ Properties of determinants
• Quiz 12. Find eigenvalues and their algebraic and geometric multiplicities
• Quiz 13 (Part 1). Diagonalize a matrix.
• Quiz 13 (Part 2). Find eigenvalues and eigenvectors of a special matrix

The post Quiz 9. Find a Basis of the Subspace Spanned by Four Matrices first appeared on Problems in Mathematics.