Projection to the subspace spanned by a vector

Johns Hopkins University math exam problems and solutions

Problem 60

Let $T: \R^3 \to \R^3$ be the linear transformation given by orthogonal projection to the line spanned by $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$.

(a) Find a formula for $T(\mathbf{x})$ for $\mathbf{x}\in \R^3$.

(b) Find a basis for the image subspace of $T$.

(c) Find a basis for the kernel subspace of $T$.

(d) Find the $3 \times 3$ matrix for $T$ with respect to the standard basis for $\R^3$.

(e) Find a basis for the orthogonal complement of the kernel of $T$. (The orthogonal complement is the subspace of all vectors perpendicular to a given subspace, in this case, the kernel.)

(f) Find a basis for the orthogonal complement of the image of $T$.

(g) What is the rank of $T$?

(Johns Hopkins University Exam)

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Proof.

(a) Find a formula for $T(\mathbf{x})$ for $\mathbf{x}\in \R^3$

For any vector $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\in \R^3$,
we have $\mathbf{x}=T(\mathbf{x})+\mathbf{v}$, where $\mathbf{v}=\mathbf{x}-T(\mathbf{x})$, which is perpendicular to the vector $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$.
Since $T(\mathbf{x})\in W$, we have $T(\mathbf{x})=t\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$ for some number $t$.
Thus $\mathbf{x}=t\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}+\mathbf{v}$.
To determine the number $t$, we take the inner product with $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$ and obtain
\begin{align*}
\mathbf{x}\cdot \begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}
=t\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}\cdot \begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix} + \mathbf{v} \cdot \begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix} \\
\Leftrightarrow \,\,\,\, x_1+2x_2+2x_3=9t
\end{align*}
Here $\mathbf{v} \cdot \begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}=\mathbf{0}$ since $\mathbf{v}$ is perpendicular to $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$.
Therefore we have $t=\frac{1}{9}(x_1+2x_2+2x_3)$, and the formula is
\[T(\mathbf{x})=\frac{1}{9}(x_1+2x_2+2x_3)\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}.\]

 

(b) Find a basis for the image subspace of $T$

Let $V$ be the subspace spanned by $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$ in $\R^3$.
Since $T$ is a projection to the subspace $W$, the image is $W$ itself.
(Any vector in $W$ is mapped to itself by the projection $T$.) Since $W$ is spanned by just one vector $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$, it is one-dimensional and a basis is $\left\{\, \begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix} \,\right\}$.

(c) Find a basis for the kernel subspace of $T$

If $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \in \ker T$, then by the formula in part (a), we have $x_1+2x_2+2x_3=0$.
Thus the vectors in the kernel of $T$ can be written as
\[\mathbf{x}=\begin{bmatrix}
-2s-2t \\
s \\
t
\end{bmatrix}=
\begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}s+
\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}t,\] where $s$ and $t$ are free variables.
Since the vectors $\begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}$ and $\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}$ are linearly independent and they span the kernel, the basis of $\ker T$ is
\[ \left \{\, \begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}
\,\right \}.\]

(d) Find the $3 \times 3$ matrix for $T$ with respect to the standard basis for $\R^3$

 The matrix for $T$ is given by $[T(\mathbf{e}_1), T(\mathbf{e}_2), T(\mathbf{e}_3)]$, where $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ are the standard basis unit vectors for $\R^3$.
By the formula in part (a) we compute
\[T(\mathbf{e}_1)=\frac{1}{9}\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}, T(\mathbf{e}_2)=\frac{2}{9}\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}, T(\mathbf{e}_3)=\frac{2}{9}\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}.\] Therefore the matrix for $T$ with respect to the standard basis is
\[\frac{1}{9}\begin{bmatrix}
1 & 2 & 2 \\
2 &4 &4 \\
2 & 4 & 4
\end{bmatrix}.\]

(e) Find a basis for the orthogonal complement of the kernel of $T$

Note that the kernel consists of vectors in $\R^3$ that are perpendicular to $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$.
Therefore the vectors perpendicular to the vectors in the kernel is parallel to $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$.
Thus a basis for the orthogonal complement is $\left\{\, \begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix} \,\right \}$.

(f) Find a basis for the orthogonal complement of the image of $T$

The image of $T$ is the subspace $W$ spanned by $\begin{bmatrix}
1 \\
2 \\
2
\end{bmatrix}$.
Thus the orthogonal complement of the image is the same as the kernel of $T$. Thus a basis is
\[ \left \{\, \begin{bmatrix}
-2 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}
\,\right\}\] as we saw in part (c).

(g) What is the rank of $T$?

The rank of $T$ is the dimension of the image of $T$. The image is $T$ and it is one-dimensional since it is spanned by only one vector. Thus the rank of $T$ is $1$.


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