Null Space, Nullity, Range, Rank of a Projection Linear Transformation

Ohio State University exam problems and solutions in mathematics

Problem 450

Let $\mathbf{u}=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}$ and $T:\R^3 \to \R^3$ be the linear transformation
\[T(\mathbf{x})=\proj_{\mathbf{u}}\mathbf{x}=\left(\, \frac{\mathbf{u}\cdot \mathbf{x}}{\mathbf{u}\cdot \mathbf{u}} \,\right)\mathbf{u}.\]

(a) Calculate the null space $\calN(T)$, a basis for $\calN(T)$ and nullity of $T$.

(b) Only by using part (a) and no other calculations, find $\det(A)$, where $A$ is the matrix representation of $T$ with respect to the standard basis of $\R^3$.

(c) Calculate the range $\calR(T)$, a basis for $\calR(T)$ and the rank of $T$.

(d) Calculate the matrix $A$ representing $T$ with respect to the standard basis for $\R^3$.

(e) Let
\[B=\left\{\, \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
-1 \\
1
\end{bmatrix} \,\right\}\] be a basis for $\R^3$.
Calculate the coordinates of $\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}$ with respect to $B$.

(The Ohio State University, Linear Algebra Exam Problem)

 
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Proof.

(a) Calculate the null space $\calN(T)$, a basis for $\calN(T)$ and nullity of $T$.

Let $\mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}\in \R^3$ and suppose that we have $\mathbf{x}\in \calN(T)$.
This means that $T(\mathbf{x})=\mathbf{0}$.
Thus we have
\[\left(\, \frac{\mathbf{u}\cdot \mathbf{x}}{\mathbf{u}\cdot \mathbf{u}} \,\right)\mathbf{u}=0.\] Since $\mathbf{u}\neq \mathbf{0}$, we must have
\begin{align*}
0=\mathbf{u}\cdot \mathbf{x}=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}\cdot \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}=x+y.
\end{align*}
This yields that the null space $\calN(T)$ consists of the vectors of the form
\[\mathbf{x}=\begin{bmatrix}
x \\
-x \\
z
\end{bmatrix}=x\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}+z\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \tag{*}\] for any real numbers $x, z$.
Thus, we have
\[\calN(T)=\Span\left\{\, \begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix} \,\right\}.\]

It follows from (*) that the vectors $\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}$ are linearly independent, hence they form a basis of $\calN(T)$.
The nullity of $T$ is the dimension of $\calN(T)$ by definition, and hence the nullity of $T$ is $2$.

 

(b) Find the determinant $\det(A)$

Note that the nullity of $A$ is the same as the nullity of $T$. So by part (a), the nullity of $A$ is $2$. This implies that the equation
\[A\mathbf{x}=\mathbf{0}\] has a nonzero solution $\mathbf{x}$.
Hence the matrix $A$ is singular, and thus the determinant $\det(A)=0$.

 

(c) Calculate the range $\calR(T)$, a basis for $\calR(T)$ and the rank of $T$.

Note that by the formula of the linear transformation $T:\R^3\to \R^3$, every vector $\mathbf{x}\in \R^3$ is mapped to a scalar multiple of the vector $\mathbf{u}$.
Also, since $\mathbf{u}$ is mapped to $\mathbf{u}$ itself, the linear transformation $T$ is nonzero.
Hence the range is
\[\calR(T)=\Span(\mathbf{u}),\] and a basis for $\calR(T)$ is the set $\{\mathbf{u}\}$.
The rank of $T$ is the dimension of the range $\calR(T)$ by definition.
Thus, the rank of $T$ is $1$.

 

(d) Calculate the matrix $A$ representing $T$ with respect to the standard basis for $\R^3$.

Let
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}, \mathbf{e}_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\] be the standard basis vectors of $\R^3$.

We calculate
\begin{align*}
T(\mathbf{e}_1)&=\left(\, \frac{\mathbf{u}\cdot \mathbf{e}_1}{\mathbf{u}\cdot \mathbf{u}} \,\right)\mathbf{u}
=\frac{1}{2}\mathbf{u}=\begin{bmatrix}
1/2 \\
1/2 \\
0
\end{bmatrix}\\[6pt] T(\mathbf{e}_2)&=\left(\, \frac{\mathbf{u}\cdot \mathbf{e}_2}{\mathbf{u}\cdot \mathbf{u}} \,\right)\mathbf{u}
=\frac{1}{2}\mathbf{u}=\begin{bmatrix}
1/2 \\
1/2 \\
0
\end{bmatrix}\\[6pt] T(\mathbf{e}_3)&=\left(\, \frac{\mathbf{u}\cdot \mathbf{e}_3}{\mathbf{u}\cdot \mathbf{u}} \,\right)\mathbf{u}
=\frac{0}{2}\mathbf{u}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.
\end{align*}
Thus, the matrix representation $A$ of $T$ is
\begin{align*}
A&=\begin{bmatrix}
T(\mathbf{e}_1) & T(\mathbf{e}_2) & T(\mathbf{e}_3) \\
\end{bmatrix}\\
&=\begin{bmatrix}
1/2 & 1/2 & 0 \\
1/2 &1/2 &0 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}

 

(e) Calculate the coordinate vector

Let $\mathbf{v}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}$ and let
\[ [\mathbf{v}]_B=\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}\] be the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
This means that we have
\begin{align*}
\mathbf{v}=a \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}+b \begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+c\begin{bmatrix}
0 \\
-1 \\
1
\end{bmatrix}.
\end{align*}
To determine $a, b, c$, we solve the system
\begin{align*}
\begin{bmatrix}
1 & -1 & 0 \\
0 &1 &-1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}=\mathbf{v}.
\end{align*}
The augmented matrix of the system is
\begin{align*}
\left[\begin{array}{rrr|r}
1 & -1 & 0 & x \\
0 &1 & -1 & y \\
0 & 0 & 1 & z
\end{array} \right].
\end{align*}

Applying the elementary row operations, we obtain
\begin{align*}
\left[\begin{array}{rrr|r}
1 & -1 & 0 & x \\
0 &1 & -1 & y \\
0 & 0 & 1 & z
\end{array} \right] \xrightarrow{R_1+R_2}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & x+y \\
0 &1 & -1 & y \\
0 & 0 & 1 & z
\end{array} \right]\\[6pt] \xrightarrow{\substack{R_1+R_3\\ R_2+R_3}}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & x+y+x \\
0 &1 & 0 & y+z \\
0 & 0 & 1 & z
\end{array} \right].
\end{align*}
Hence we have
\begin{align*}
a=x+y+z, b=y+z, c=z,
\end{align*}
and the coordinate vector of $\mathbf{v}$ with respect to the basis $B$ is
\[[\mathbf{v}]_B=\begin{bmatrix}
x+y+z \\
y+z \\
z
\end{bmatrix}.\]


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