Let \(T:\mathbb{R}^2\rightarrow \mathbb{R}^3\) be a linear transformation
such that \(T\left(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\right) =
\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}\)
and \(T\left(\begin{bmatrix} 1 \\ 1\end{bmatrix}\right) =
\begin{bmatrix} 0 \\ -1 \\ 1\end{bmatrix}\).
What is \(T\left(\begin{bmatrix} 0 \\ -1\end{bmatrix}\right)\)?
We first attempt to write \(\begin{bmatrix} 0\\-1\end{bmatrix}\)
as a linear combination of \(\begin{bmatrix} 2\\1\end{bmatrix}\)
and \(\begin{bmatrix} 1\\1\end{bmatrix}\). Otherwise, there is not enough
information to answer the question.
Solving for \(\alpha\) and \(\beta\) in
\(\begin{bmatrix} 0\\-1\end{bmatrix}
= \alpha \begin{bmatrix} 2 \\ 1\end{bmatrix} + \beta \begin{bmatrix}
1 \\ 1\end{bmatrix},\)
we obtain \(\alpha = 1\) and \(\beta = -2\).
Consider the linear transformation \(T:\mathbb{R}^4 \rightarrow
\mathbb{R}^{2\times 2}\) given by
\(T \left( \left[{\begin{array}{c}
a \\
b \\
c \\
d\\
\end{array} } \right] \right) = \left[ \begin{array}{cc} 2a+b+c & a-d \\ a+b+c-d & 0 \\ \end{array} \right].\)
Determine a basis for the range of \(T\).
We see from the definition of \(T\) that the range of \(T\) is
the set of all matrices given by
\(a \begin{bmatrix}
2 & 1 \\
1 & 0
\end{bmatrix}+
b \begin{bmatrix}
1 & 0 \\
1 & 0
\end{bmatrix}+
c
\begin{bmatrix}
1 & 0 \\
1 & 0
\end{bmatrix}+
d
\begin{bmatrix}
0 & -1 \\
-1 & 0
\end{bmatrix}\)
where \(a,b,c,d \in \mathbb{R}\).
Since the second and third matrix in the above expression are the same,
the range of \(T\) is spanned by the set
\(\left\{\begin{bmatrix}
2 & 1 \\
1 & 0
\end{bmatrix},
\begin{bmatrix}
1 & 0 \\
1 & 0
\end{bmatrix},
\begin{bmatrix}
0 & -1 \\
-1 & 0
\end{bmatrix}\right\}.\)
We now show that the set is linearly independent, thus proving that it is
a basis. Suppose that \(x,y,z\in \mathbb{R}\) are such that
\[x \begin{bmatrix}
2 & 1 \\
1 & 0
\end{bmatrix}+
y \begin{bmatrix}
1 & 0 \\
1 & 0
\end{bmatrix}+
z
\begin{bmatrix}
0 & -1 \\
-1 & 0
\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}.\]
Comparing entries, we have the system
\begin{eqnarray*}
2x + y & = & 0 \\
x - z & = & 0 \\
x + y - z& = & 0
\end{eqnarray*}
Subtracting the third equation from the first, we obtain \(x = 0\).
Hence, we must have \(y = 0\) and \(z = 0\) as well. So the only solution
is to have \(x = y = z = 0\). So the set of matrices is linearly independent.
Give a basis for the kernel of \(T\).
The kernel of \(T\) is the set of all
\(\begin{bmatrix} a\\b\\c \\d\end{bmatrix}\) such that
\(T\left(\begin{bmatrix} a\\b\\c \\d\end{bmatrix}\right) =
\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.\)
This is true if and only if
\begin{eqnarray*}
2a + b + c & = & 0 \\
a - d & = & 0 \\
a+b+c-d & = & 0.
\end{eqnarray*}
Note that this is a homogenous system.
The coefficient matrix is
\(\begin{bmatrix} 2 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \\
1 & 1 & 1 & -1 \end{bmatrix}\).
Row-reducing this matrix gives
\(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \end{bmatrix}\).
Note that only the third column is not a pivot column.
Thus, the set of solutions is given by
\(\begin{bmatrix} a\\b\\c \\d\end{bmatrix} =
\begin{bmatrix} 0 \\ -s \\ s \\ 0\end{bmatrix}
= s\begin{bmatrix} 0 \\ -1 \\ 1 \\ 0\end{bmatrix}\)
for all \(s \in \mathbb{R}\).
Hence, a basis for the kernel is given by
\(\left\{\begin{bmatrix} 0 \\ -1 \\ 1 \\ 0\end{bmatrix}\right\}\).
Is \(T\) surjective? Injective?
Since we found a basis for the range of \(T\)
with three elements, the dimension of the range of \(T\) is three.
But the dimension
of the codomain of \(T\) is \(4\). So \(T\) is not surjective.
\(T\) is also not injective because its kernel contains the nonzero vector
\(\begin{bmatrix} 0 \\ -1 \\ 1 \\ 0\end{bmatrix}\).
Example 3
Let \(\mathbb{F}\) denote the field \(GF(2)\).
Let \(T:\mathbb{F}^4 \rightarrow \mathbb{F}^3\) be a linear transformation
given by \(T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4
\end{bmatrix}\right) = \begin{bmatrix} x_1 + x_2 + x_4 \\ x_1 + x_3 + x_4
\\ x_2 + x_3 + x_4 \end{bmatrix}\).
Give a basis for the kernel of \(T\).
Note that \(T\) can be written as
\(T(x) = Ax\) with \(A = \begin{bmatrix} 1 & 1 & 0 & 1\\
1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1\end{bmatrix}\). Hence, the kernel of \(T\) is given
by the nullspace of \(A\). We now row-reduce \(A\).
Let \(T:\mathbb{R}^3\rightarrow \mathbb{R}^3\) be a linear transformation
given by \(T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) =
\begin{bmatrix} x_1 + x_2 + x_3 \\ -x_1 -x_2 + x_3 \\ x_2 - x_3\end{bmatrix}\).
Determine if \(T\) is invertible.
Note that \(T\) can be written as
\(T(x) = Ax\) with \(A = \begin{bmatrix} 1 & 1 & 1\\
-1 & -1 & 1 \\
0 & 1 & -1\end{bmatrix}\). Hence, the linear transformation
\(T\) is invertible
if and only if \(A\) is invertible.
We now row-reduce \(A\).
We could continue with the row reduction until we get to a matrix in RREF
and see if there is any nonpivot column.
However, we can take a shortcut by noting that
the last matrix is upper triangular with a nonzero determinant, namely
\(1\cdot 1 \cdot 2 = 2\). Hence, the matrix is nonsingular,
implying that \(A\) is also nonsingular. Thus, \(A\) is invertible
and so \(T\) is invertible.
Example 5
Let \(T:\mathbb{R}^3\rightarrow \mathbb{R}^2\) be a linear transformation
given by \(T\left(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\right) =
\begin{bmatrix} x_1 + x_2 \\ -x_1 -x_2 + x_3\end{bmatrix}\).
Let \(\Gamma =
\left (\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix},
\begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix},
\begin{bmatrix} 0 \\ 1 \\ 1\end{bmatrix}
\right )\) be an ordered basis for
\(\mathbb{R}^3\) and
let \(\Omega =
\left ( \begin{bmatrix} 1 \\ 1\end{bmatrix},
\begin{bmatrix} -1 \\ 1\end{bmatrix} \right )\) be an ordered basis for
\(\mathbb{R}^2\).
Find \([T]_\Gamma^\Omega\).
Let \(P_2\) denote the vector space of polynomials in \(x\)
with real coefficients having degree at most \(2\).
Let \(T:P_2\rightarrow \mathbb{R}^3\) be a linear transformation
given by \(T (ax^2 + bx + c) =
\begin{bmatrix} 2a-b \\ b+2c \\ 0\end{bmatrix}\).
Determine the dimension of the kernel of \(T\).
We first find a basis for the kernel of \(T\).
For \(ax^2 + bx + c\) to be in the kernel of \(T\), we need
\(2a-b = 0\) and \(b+2c= 0\).
Solving for \(b\) and \(c\) in terms of \(a\), we obtain
\(b = 2a\) and \(c = -a\).
Thus, the kernel is the set of polynomials of the form
\(ax^2 + 2ax - a = a(x^2 + 2x - 1)\).
Hence, the kernel of \(T\) is spanned by
a single vector, implying that the dimension of the kernel of \(T\)
is 1.
Example 7
Let \(P_1\) denote the vector space of polynomials in \(x\)
with real coefficients having degree at most \(1\).
Let \(\Gamma = (x-1, 1)\) and \(\Omega = (x+1, x-2)\) be ordered
bases for \(P_1\).
Let \(T:P_1\rightarrow P_1\) be a linear transformation such that
\([T]_\Gamma^\Omega = \begin{bmatrix} 1 & 2 \\ 1 & -1 \end{bmatrix}\).
What is \(T(ax + b)\)?
Let \(u = ax + b\).
Then \([u]_\Gamma = \begin{bmatrix} a \\ a+b\end{bmatrix}\) since
\(ax + b = a(x-1) + (a+b)\).
It follows that
\([T(u)]_\Omega = [T]_\Gamma^\Omega [u]_\Gamma
= \begin{bmatrix} 1 & 2 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} a \\ a+b
\end{bmatrix} = \begin{bmatrix} 3a+2b \\ -b\end{bmatrix}\).