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)

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 $1$.

The post Projection to the subspace spanned by a vector first appeared on Problems in Mathematics.

Let $A$ be an $m \times n$ real matrix.
Then the kernel of $A$ is defined as $\ker(A)=\{ x\in \R^n \mid Ax=0 \}$.

The kernel is also called the null space of $A$.
Suppose that $A$ is an $m \times n$ real matrix such that $\ker(A)=0$. Prove that $A^{\trans}A$ is invertible.

(Stanford University Linear Algebra Exam)

We will give two proofs.

Both proofs try to prove $\ker(A^{\trans}A)=0$. The method in the 1st proof is more or less direct computation.

For the second proof, you need to remember the relation between the transpose and the orthogonal complement of a vector space.

1st proof

Since the size of the transpose $A^{\trans}$ is $n\times m$, the product $A^{\trans}A$ is a $n\times n$ square matrix.
To show that it is invertible, we show that the kernel of $A^{\trans}A$ is trivial.
Then the result follows since $A^{\trans}A$ is an injective linear transformation from $\R^n$ to $\R^n$, thus an isomorphism. Hence $A^{\trans}A$ is invertible.

To show that the kernel of $A^{\trans}A$ is trivial, denote $A=[A_1 A_2\dots A_n]$, where $A_i$ is the $i$-th column vector of $A$.
Note that the column vectors $A_i$ are linearly independent since $\ker(A)=0$.

Then
\[A^{\trans}=\begin{bmatrix}
A_1^{\trans} \\
A_2^{\trans} \\
\vdots \\
A_n^{\trans}
\end{bmatrix}.\]

Now we suppose that $A^{\trans}Ax=0$ for $x=\begin{bmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{bmatrix}\in R^n$.
Then we have
\begin{align*}
0=&A^{\trans}A x\\[6pt] &=\begin{bmatrix}
A_1^{\trans}A_1 & A_1^{\trans}A_2 & \cdots & A_1^{\trans}A_n \\
A_2^{\trans}A_1 &A_2^{\trans}A_2 & \cdots & A_2^{\trans}A_n \\
\vdots & \vdots & \vdots & \vdots \\
A_n^{\trans}A_1 & A_n^{\trans}A_2 & \cdots & A_n^{\trans}A_n
\end{bmatrix}x\\[6pt] &=\begin{bmatrix}
A_1^{\trans}(x_1 A_1+\cdots +x_n A_n) \\
A_2^{\trans}(x_1 A_1+\cdots+x_n A_n \\
\vdots \\
A_n^{\trans}(x_1 A_1+\cdots+x_n A_n)
\end{bmatrix}.
\end{align*}

Therefore we have $A_i^{\trans}(x_1 A_1+\cdots+x_n A_n)=0$ for all $i=1,\dots, n$.
Equivalently, we have the inner product $A_i\cdot Ax=0$ for all $i=1,\dots,n$.

If $Ax=x_1 A_1+\cdots+x_n A_n\neq 0$, then the vectors $A_1, A_2, \dots, A_n, Ax$ are linearly independent vectors in $\R^n$ since $A_1,\dots, A_n$ are linearly independent and the inner products with $A_i$ and the vector $Ax$ are all zero, hence they are orthogonal.

However, this is a contradiction since there are $n+1$ linearly independent vectors in $\R^n$ of dimension $n$ vector space. Thus, we must have $Ax=0$. Since the kernel of $A$ is trivial, this implies $x=0$. Therefore the kernel of $A^{\trans}A$ is trivial.

2nd proof

In the second proof, we still want to prove the kernel of $A^{\trans}A$ is trivial.

Let $x\in \ker(A^{\trans}A)$.
Then we have $A^{\trans}(Ax)=0$ and thus $Ax \in \ker(A^{\trans})=\im(A)^{\perp}$.

On the other hand, clearly $Ax \in \im(A)$.
Thus $Ax \in \im(A)\cap \im(A)^{\perp}=\{0\}$.
So we have $Ax=0$, and $x=0$ since $\ker(A)=0$.

Therefore $\ker(A^{\trans}A)=0$, and since any injective homomorphism from $\R^n$ to itself is an isomorphism, we conclude that $A^{\trans}A$ is invertible.

The post If the Kernel of a Matrix $A$ is Trivial, then $A^T A$ is Invertible first appeared on Problems in Mathematics.