Let $T: \R^n \to \R^m$ be a linear transformation.
Suppose that the nullity of $T$ is zero.

If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.

Proof.

Suppose that we have a linear combination
\[c_1T(\mathbf{x}_1)+c_2T(\mathbf{x}_2)+\cdots+c_k T(\mathbf{x}_k)=\mathbf{0}_m,\] where $\mathbf{0}_m$ is the $m$ dimensional zero vector in $\R^m$.
To show that the set $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is linearly independent, we need to show that $c_1=c_2=\cdots=c_k=0$.

Using the linearity of $T$, we have
\[T(c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k)=\mathbf{0}_m.\] Then the vector $c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k$ is in the nullspace $\calN(T)$ of $T$. Since the nullity, which is the dimension of the nullspace, is zero, we have $\calN(T)=\{\mathbf{0}_n\}$. This yields
\[c_1\mathbf{x}_1+c_2\mathbf{x}_2+\cdots+c_k \mathbf{x}_k=\mathbf{0}_n.\]

Since the vectors $\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k$ are linearly independent, we must have $c_1=c_2=\dots=c_k=0$ as required.

Thus we conclude that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.

The post If the Nullity of a Linear Transformation is Zero, then Linearly Independent Vectors are Mapped to Linearly Independent Vectors first appeared on Problems in Mathematics.

Define two functions $T:\R^{2}\to\R^{2}$ and $S:\R^{2}\to\R^{2}$ by
\[
T\left(
\begin{bmatrix}
x \\ y
\end{bmatrix}
\right)
=
\begin{bmatrix}
2x+y \\ 0
\end{bmatrix}
,\;
S\left(
\begin{bmatrix}
x \\ y
\end{bmatrix}
\right)
=
\begin{bmatrix}
x+y \\ xy
\end{bmatrix}
.
\] Determine whether $T$, $S$, and the composite $S\circ T$ are linear transformations.

Solution.

We will prove that $T$ and $S\circ T$ are linear transformations, but $S$ is not.

$T$ is alinear transformation

To prove that $T$ is a linear transformation, note that for any $\mathbf{x},\mathbf{y}\in\R^{2}$, if we write
\[
\mathbf{x}
=
\begin{bmatrix}
x_{1} \\ x_{2}
\end{bmatrix}
,\;
\mathbf{y}
=
\begin{bmatrix}
y_{1} \\ y_{2}
\end{bmatrix}
,
\] then we have
\begin{align*}
T\left(\mathbf{x}+\mathbf{y}\right)
&=
T\left(
\begin{bmatrix}
x_{1}+y_{1} \\ x_{2}+y_{2}
\end{bmatrix}
\right)
=
\begin{bmatrix}
2(x_{1}+y_{1})+(x_{2}+y_{2}) \\ 0
\end{bmatrix}
\\
&=
\begin{bmatrix}
2x_{1}+x_{2} \\ 0
\end{bmatrix}
+
\begin{bmatrix}
2y_{1}+y_{2} \\ 0
\end{bmatrix}
=
T(\mathbf{x})+T(\mathbf{y})
.
\end{align*}

Next, for any scalar $r$, we have
\[
T(r\mathbf{x})
=
T\left(r
\begin{bmatrix}
x_{1} \\ x_{2}
\end{bmatrix}
\right)
=
T\left(
\begin{bmatrix}
rx_{1} \\ rx_{2}
\end{bmatrix}
\right)
=
\begin{bmatrix}
2rx_{1}+rx_{2} \\ 0
\end{bmatrix}
=r
\begin{bmatrix}
2x_{1}+x_{2} \\ 0
\end{bmatrix}
=
rT(\mathbf{x})
.
\] Hence $T$ is a linear transformation.

$S$ is not a linear transformation

To prove that $S$ is not a linear transformation, observe that
\[
S\left(
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
\right)
=
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
,
\quad
S\left(
\begin{bmatrix}
0 \\ 1
\end{bmatrix}
\right)
=
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
,
\quad
S\left(
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
\right)
=
\begin{bmatrix}
2 \\ 1
\end{bmatrix}
.
\] Therefore,
\begin{align*}
S\left(
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
+
\begin{bmatrix}
0 \\ 1
\end{bmatrix}
\right)
&=
S\left(
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
\right)
=
\begin{bmatrix}
2 \\ 1
\end{bmatrix}
% \\
% &
\neq
\begin{bmatrix}
2 \\ 0
\end{bmatrix}
=
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
+
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
\\
&=
S\left(
\begin{bmatrix}
1 \\ 0
\end{bmatrix}
\right)
+
S\left(
\begin{bmatrix}
0 \\ 1
\end{bmatrix}
\right)
.
\end{align*}
Thus it is not the case that $S(\mathbf{x}+\mathbf{y})=S(\mathbf{x})+S(\mathbf{y})$ for all $\mathbf{x},\mathbf{y}\in\R^{2}$. It follows that $S$ cannot be a linear transformation.

The composite $S\circ T$ is a lineawr transformation

To prove that $S\circ T$ is linear, note that for any $\mathbf{x}\in\R^{2}$,
\[
S\circ T(\mathbf{x})
=
S\left(
T\left(
\begin{bmatrix}
x \\ y
\end{bmatrix}
\right)\right)
=
S\left(
\begin{bmatrix}
2x+y \\ 0
\end{bmatrix}
\right)
% =
% \begin{bmatrix}
% 2x+y+0 \\ (2x+y)\cdot 0
% \end{bmatrix}
=
\begin{bmatrix}
2x+y \\ 0
\end{bmatrix}
=
T(\mathbf{x})
.
\] Therefore, $S\circ T=T$. Since $T$ is a linear transformation, we can immediately conclude that $S\circ T$ is a linear transformation. Hence $T$ and $S\circ T$ are linear, while $S$ is not.

The post Are These Linear Transformations? first appeared on Problems in Mathematics.

Let $\mathrm{P}_2$ denote the vector space of polynomials of degree $2$ or less, and let $T : \mathrm{P}_2 \rightarrow \mathrm{P}_2$ be the derivative linear transformation, defined by
\[ T( ax^2 + bx + c ) = 2ax + b . \]

Is $T$ diagonalizable? If so, find a diagonal matrix which represents $T$. If not, explain why not.

Solution.

The standard basis of the vector space $\mathrm{P}_2$ is the set $B = \{ 1 , x , x^2 \}$. The matrix representing $T$ with respect to this basis is
\[ [T]_B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} . \]

The characteristic polynomial of this matrix is
\[ \det ( [T]_B – \lambda I ) = \begin{vmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 2 \\ 0 & 0 & -\lambda \end{vmatrix} = \, – \lambda^3 . \] We see that the only eigenvalue of $T$ is $0$ with algebraic multiplicity $3$.

On the other hand, a polynomial $f(x)$ satisfies $T(f)(x) = 0$ if and only if $f(x) = c$ is a constant. The null space of $T$ is spanned by the single constant polynomial $\mathbb{1}(x) = 1$, and thus is one-dimensional. This means that the geometric multiplicity of the eigenvalue $0$ is only $1$.

Because the geometric multiplicity of $0$ is less than the algebraic multiplicity, the map $T$ is defective, and thus not diagonalizable.

The post Is the Derivative Linear Transformation Diagonalizable? first appeared on Problems in Mathematics.

Let $\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $ \langle \mathbf{v} , \mathbf{w} \rangle = \mathbf{v}^\trans \mathbf{w}$. A linear transformation $T : \R^2 \rightarrow \R^2$ is called an orthogonal transformation if for all $\mathbf{v} , \mathbf{w} \in \R^2$,
\[\langle T(\mathbf{v}) , T(\mathbf{w}) \rangle = \langle \mathbf{v} , \mathbf{w} \rangle.\]

For a fixed angle $\theta \in [0, 2 \pi )$ , define the matrix
\[ [T] = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \] and the linear transformation $T : \R^2 \rightarrow \R^2$ by
\[T( \mathbf{v} ) = [T] \mathbf{v}.\]

Prove that $T$ is an orthogonal transformation.

Solution.

Suppose we have vectors $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ and $\mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} $ . Then,
\[T(\mathbf{v}) = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} \cos(\theta) v_1 – \sin (\theta) v_2 \\ \sin(\theta) v_1 + \cos (\theta) v_2 \end{bmatrix},\] and
\[ T(\mathbf{w}) = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} \cos(\theta) w_1 – \sin (\theta) w_2 \\ \sin(\theta) w_1 + \cos (\theta) w_2 \end{bmatrix}.\]

Then we find the inner product for these two vectors:
\begin{align*}
&\langle T(\mathbf{v} ) , T( \mathbf{w} ) \rangle \\
&= \begin{bmatrix} \cos(\theta) v_1 – \sin (\theta) v_2 & \sin(\theta) v_1 + \cos (\theta) v_2 \end{bmatrix} \begin{bmatrix} \cos(\theta) w_1 – \sin (\theta) w_2 \\ \sin(\theta) w_1 + \cos (\theta) w_2 \end{bmatrix} \\[6pt] &= \biggl( \cos(\theta) v_1 – \sin(\theta) v_2 \biggr) \biggl( \cos(\theta) w_1 – \sin ( \theta) w_2 \biggr) \\[6pt] &\qquad + \biggl( \sin (\theta) v_1 + \cos (\theta) v_2 \biggr) \biggl( \sin (\theta) w_1 + \cos(\theta) w_2 \biggr) \\[6pt] &= \cos^2(\theta) ( v_1 w_1 + v_2 w_2 ) + \sin(\theta) \cos(\theta) ( – v_1 w_2 – v_2 w_1 + v_1 w_2 + v_2 w_1 ) \\ &\qquad + \sin^2 (\theta) ( v_2 w_2 + v_1 w_1 ) \\[6pt] &= \left( \cos^2 ( \theta) + \sin^2 ( \theta ) \right) ( v_1 w_1 + v_2 w_2 ) \\
&= v_1 w_1 + v_2 w_2 \\
&= \langle \mathbf{v} , \mathbf{w} \rangle .
\end{align*}

This proves that $T$ is an orthogonal transformation. For the second-to-last equality, we used the Pythagorean identity $\sin^2 ( \theta ) + \cos^2 ( \theta ) = 1$.

The post The Rotation Matrix is an Orthogonal Transformation first appeared on Problems in Mathematics.

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 $C([-1, 1])$ denote the vector space of real-valued functions on the interval $[-1, 1]$. Define the vector subspace
\[W = \{ f \in C([-1, 1]) \mid f(0) = 0 \}.\]

Define the map $T : C([-1, 1]) \rightarrow W$ by $T(f)(x) = f(x) – f(0)$. Determine if $T$ is a linear map. If it is, determine its nullspace and range.

Proof.

$T$ is a lineawr transformation

First we must check that $T$ takes values in $W$, as the problem implies but does not prove. Consider $f \in C([-1 , 1])$. The function $T(f)$ lies in $W$ if and only if $T(f)(0) = 0$. We quickly check that it satisfies this condition:in $C([-1, 1])$ lies
\[ T(f)(0) = f(0) – f(0) = 0 . \]

Now that we know $T$ takes values in $W$, we show that it is a linear transformation. We will prove this by showing that $T$ satisfies both of the axioms for linear transformations. First, suppose that $f, g \in C([-1, 1])$. Then
\begin{align*}
T(f+g)(x) &= (f+g)(x) – (f+g)(0) \\
&= f(x) + g(x) – f(0) – g(0) \\
&= f(x) – f(0) + g(x) – g(0) \\
&= T(f)(x) + T(g)(x) . \end{align*}

Now for a scalar $c \in \mathbb{R}$ we have
\begin{align*}
T( cf )(x) &= (cf)(x) – (cf)(0) \\
&= c f(x) – c f(0) \\
&= c ( f(x) – f(0) ) \\
&= c T(f)(x) . \end{align*}

Thus we have proven that $T$ is a linear transformation.

The nullspace of $T$

Next, we will prove that the nullspace of $T$ is
\[\mathcal{N}(T) = \{ f \in C([-1 , 1]) \mid f(x) \mbox{ is a constant function } \}.\] Suppose that $f \in \calN(T)$, that is, $f$ satisfies
\[0 = T(f)(x) = f(x) – f(0).\] Then $f(x) = f(0)$ for all $x \in [-1, 1]$. This means that $f$ is a constant function. On the other hand, if $f(x)$ is a constant function, then $T(f)(x) = f(x) – f(0) = 0$. We see that $f$ lies in the nullspace of $T$ if and only if it is a constant function.

The range of $T$

Next, we want to find the range of $T$. We claim that
\[\mathcal{R}(T) = W.\]

Suppose that $f \in W$, that is, it is a function such that $f(0) = 0$. Then
\[T(f)(x) = f(x) – f(0) = f(x),\] and so we see that $f \in \mathcal{R}(T)$.
Conversely, suppose that $f \in \mathcal{R}(T)$, so that $f = T(g)$ for some $g \in C([-1, 1])$. Then
\[f(0) = T(g)(0) = g(0) – g(0) = 0,\] and so $f \in W$. Thus every $f \in C([-1, 1])$ lies in the range of $T$ if and only if $f(0) = 0$.

\end{proof}

The post Find the Nullspace and Range of the Linear Transformation $T(f)(x) = f(x)-f(0)$ first appeared on Problems in Mathematics.

For an integer $n > 0$, let $\mathrm{P}_n$ denote the vector space of polynomials with real coefficients of degree $2$ or less. Define the map $T : \mathrm{P}_2 \rightarrow \mathrm{P}_4$ by
\[ T(f)(x) = f(x^2).\]

Determine if $T$ is a linear transformation.

If it is, find the matrix representation for $T$ relative to the basis $\mathcal{B} = \{ 1 , x , x^2 \}$ of $\mathrm{P}_2$ and $\mathcal{C} = \{ 1 , x , x^2 , x^3 , x^4 \}$ of $\mathrm{P}_4$.

Solution.

$T$ is a linear transformation

To prove that $T$ is a linear transformation, we must show that it satisfies both axioms for linear transformations. For $f, g \in \mathrm{P}_2$, we have
\[T( f+g )(x) = (f+g)(x^2) = f(x^2) + g(x^2) = T(f)(x) + T(g)(x)\] while for a scalar $c \in \mathbb{R}$, we have
\[ T( c f )(x) = (cf)(x^2) = c f(x^2) = c T(f)(x).\] We see that $T$ is a linear transformation.

The matrix representation for $T$

To find its matrix representation, we must calculate $T(f)$ for each $f \in \mathcal{B}$ and find its coordinate vector relative to the basis $\mathcal{C}$. We calculate
\[T(1) = 1 , \quad T(x) = x^2 , \quad T(x^2) = x^4.\] Each of these is an element of $\mathcal{C}$. Their coordinate vectors relative to $\mathcal{C}$ are thus
\[[ T(1) ]_{\mathcal{B}} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} , \quad [ T(x) ]_{\mathcal{B}} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} , \quad [ T(x^2) ]_{\mathcal{C}} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}.\]

The matrix representation of $T$ is found by combining these columns, in order, into one matrix:

\[ [T]_{\mathcal{B}}^{\mathcal{C}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\] Click here if solved 26

The post Find the Matrix Representation of $T(f)(x) = f(x^2)$ if it is a Linear Transformation first appeared on Problems in Mathematics.

Let $C ([0, 3] )$ be the vector space of real functions on the interval $[0, 3]$. Let $\mathrm{P}_3$ denote the set of real polynomials of degree $3$ or less.

Define the map $T : C ([0, 3] ) \rightarrow \mathrm{P}_3 $ by
\[T(f)(x) = f(0) + f(1) \cdot x + f(2) \cdot x^2 + f(3) \cdot x^3.\]

Determine if $T$ is a linear transformation. If it is, determine its nullspace.

$T$ is a linear transformation

We will show that $T$ satisfies both of the axioms for linear transformations. Suppose $f, g \in C([0, 3])$. Then
\begin{align*}
&T(f+g)(x)\\
&= (f+g)(0) + (f+g)(1) \cdot x + (f+g)(2) \cdot x^2 + (f+g)(3) \cdot x^3 \\
&= f(0) + f(1) \cdot x + f(2) \cdot x^2 + f(3) \cdot x^3 + g(0) + g(1) \cdot x + g(2) \cdot x^2 + g(3) \cdot x^3 \\
&= T(f)(x) + T(g)(x) . \end{align*}

Thus we see that $T(f+g) = T(f) + T(g)$.

Next, suppose $f \in C([0, 3])$ and $c \in \mathbb{R}$. Then,
\begin{align*}
&T(cf)(x)\\
&= (cf)(0) + (cf)(1) \cdot x + (cf)(2) \cdot x^2 + (cf)(3) \cdot x^3 \\
&= c f(0) + c f(1) \cdot x + c f(2) \cdot x^2 + c f(3) \cdot x^3 \\
&= c ( f(0) + f(1) \cdot x + f(2) \cdot x^2 + f(3) \cdot x^3 ) \\
&= c T(f)(x) . \end{align*}

This shows that $T$ is a linear transformation.

The nullspace of $T$

Next, we will find the nullspace of $T$. Suppose that $f \in C([0, 3])$ such that $T(f) = 0$. That means that
\[f(0) + f(1) \cdot x + f(2) \cdot x^2 + f(3) \cdot x^3 = 0.\]

This implies that $f(0) = f(1) = f(2) = f(3) = 0$. On the other hand, if $f(0)=f(1)=f(2)=f(3)=0$, then $T(f)(x) = 0$. Thus, the nullspace is
\[\mathcal{N}(T) = \{ f \in C([0, 3]) \mid f(0) = f(1) = f(2) = f(3) = 0 \}.\] Click here if solved 23

The post Is the Map $T(f)(x) = f(0) + f(1) \cdot x + f(2) \cdot x^2 + f(3) \cdot x^3$ a Linear Transformation? first appeared on Problems in Mathematics.

Let $C (\mathbb{R})$ be the vector space of real functions. Define the map $T$ by $T(f)(x) = (f(x))^2$ for $f \in C(\mathbb{R})$.

Determine if $T$ is a linear transformation or not. If it is, determine the range of $T$.

Solution.

We claim that $T$ is not a linear transformation.

We can see this by seeing
\[ T(f+g)(x) = ( f(x) + g(x) )^2 = f^2(x) + 2 f(x) g(x) + g^2 (x) \] while
\[ T(f)(x) + T(g)(x) = f^2(x) + g^2(x) . \] Because $T(f+g) \neq T(f) + T(g)$, we see that $T$ is not a linear transformation.

For a specific example, consider $f(x) = g(x) = x$. Then
\[ T( f+g)(x) = (2x)^2 = 4x^2 , \] while
\[ T(f)(x) + T(g)(x) = x^2 + x^2 = 2x^2 . \] Clearly $T(f+g) \neq T(f) + T(g)$.

The post Is the Map $T(f)(x) = (f(x))^2$ a Linear Transformation from the Vector Space of Real Functions? first appeared on Problems in Mathematics.

Let $V$ be the vector space of $2 \times 2$ matrices with real entries, and $\mathrm{P}_3$ the vector space of real polynomials of degree 3 or less. Define the linear transformation $T : V \rightarrow \mathrm{P}_3$ by
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = 2a + (b-d)x – (a+c)x^2 + (a+b-c-d)x^3.\]

Find the rank and nullity of $T$.

Solution.

To find the rank of $T$, we will use its matrix representation relative to the standard bases for $V$ and $\mathrm{P}_3$. Define the matrices
\[\mathbf{e}_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} , \, \mathbf{e}_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\] and the polynomials
\[p_1(x) = 1 , \quad p_2(x) = x , \quad p_3(x) = x^2 , \quad p_4(x) = x^3.\]

Then the standard bases are
\[B = \{ \mathbf{e}_1 , \mathbf{e}_2 , \mathbf{e}_3 , \mathbf{e}_4 \} \subset V \mbox{ and } C = \{ p_1 , p_2 , p_3 , p_4 \} \subset \mathrm{P}_3.\]

We now find the matrix of representation of $T$ relative to these bases. This will be a $4 \times 4$ matrix whose $i$-th row is the column vector of $T(e_i)$ relative to the basis $C$. For example, we have
\[ T(e_1) = 2 – x^2 + x^3 = 2p_1 – p_3 + p_4 , \] and so the first column of the matrix is the column vector
\[ [T(e_1)]_{C} = \begin{bmatrix} 2 \\ 0 \\ -1 \\ 1 \end{bmatrix} . \]

This vector becomes the first column of the matrix. After performing this process for the other three basis elements $e_2 , e_3 , e_4$, we get the matrix
\[ [T]_{B}^{C} = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & -1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix}.\]

To find the rank of $T$, we row-reduce this matrix:
\begin{align*}
\begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & -1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix} \xrightarrow[-R_3]{ \frac{1}{2} R_1 } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix} \xrightarrow{ R_3 – R_1 } \\[6pt] \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix}
\xrightarrow{R_4 – R_1 – R_2 + R_3} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}.
\end{align*}

This reduced matrix has three non-zero rows, and so has rank 3. Thus the orginal linear transformation $T$ also has rank 3. The nullity can be computed using the equation
\[ \mathtt{rank}(T) + \mathtt{nullity}(T) = \dim (V).\] Using the values $\mathtt{rank}(T) = 3$ and $\dim (V) = 4$, the nullity of $T$ must be 1.

The post The Rank and Nullity of a Linear Transformation from Vector Spaces of Matrices to Polynomials first appeared on Problems in Mathematics.