Loomis and Sternberg Problem 1.56

Question

Hints please! Also, what are the authors talking about in the second part of the question? Aren’t the two functionals in the first part also linear transformations whose null space is the above span?

Note: I realise that this question has already been asked here - Finding linear functionals on $\mathbb R^4$ the intersection of whose null spaces is a linear span. I’m posting this again as the only answer to the linked question does not explain how the solution was obtained or give any hints for solving the problem.

Answer 1

Hopefully these hints work for the OP. I have also provided full solutions, hidden inside spoilers, for other readers.

Constructing the linear functionals

Let's call the two linear functionals $$ f_1:\mathbb{R}^4\to\mathbb{R}\,,~~~~~~~~~~ f_2:\mathbb{R}^4\to\mathbb{R}\,. $$ Take $v_1=(1,1,1,1)$, $v_2=(1,0,-1,0)$, and extend this to a basis by finding $v_3$ and $v_4$ such that $\{v_1,v_2,v_3,v_4\}$ are linearly independent. Using the fact that the dimension of the kernel for each of these linear functionals is

$\dim\ker f_j=3$,

construct $f_1$ and $f_2$ by defining the action of these operators on the basis elements above so that the intersection of the kernels is the span of $v_1$ and $v_2$. Here is a more detailed hint:

You know that three of the four basis vectors must be sent to zero by each linear functional. How do you choose which vectors get sent to zero so that any linear combination of $v_1$ and $v_2$ is sent to zero by both linear functionals, but still have the linear functionals be distinct operators?

and here is the actual construction:

Take $f_1(v_1)=f_1(v_2)=f_1(v_3) = 0$ and $f_1(v_4) = A\neq0$ and $f_2(v_1)=f_2(v_2)=f_2(v_4) = 0$ and $f_1(v_3)=B\neq0$. You can choose whatever you want for the values of $A$ and $B$, although there are more or less natural choices. Then, is is straight-forward to show that if $x=a_1v_1+a_2v_2+a_3v_3+a_4v_4$ is in the intersection of the kernels, then $a_3$ and $a_4$ must be zero.

Constructing the linear transformation

I'm not entirely sure that this is what the second question is getting at, but here's my take on it. Note that $v_1$ and $v_2$ are not chosen entirely arbitrary, i.e., they are "related" in a way, i.e., they are

orthogonal to each other.

Based on that observation, what is the most natural way to choose $v_3$ and $v_4$ to extend to a full basis for $\mathbb{R}^4$? Well,

Choose $v_3$ and $v_4$ to be orthogonal to each other and to the original vectors so that the entire basis is orthogonal.

Once you have that special basis in hand, there is a "natural" choice of linear transformation that has $v_1$ and $v_2$ as an (orthogonal) basis for the kernel.

Hint 1:

You just "strip" away the $v_1$ and $v_2$ parts of a vector.

Hint 2:

This is a projection of the vector onto the subspace spanned by $v_3$ and $v_4$. The linear functionals can actually be chosen to be defined via the inner product of the input vector with these vectors, and that's the direct relationship, as far as I can tell.

Example

Here is I think the most natural way of doing this construction

Basis:

Choose $v_3 = (0, -1, 0, 1 )$ and $v_4 = (1, -1, 1, -1 )$, which is pretty natural way to choose the new vectors so that the entire basis is orthogonal.

Construction of the linear functionals:

Then define $f_1(v) = \langle v, v_4 \rangle$ and $f_2(v) = \langle v, v_3 \rangle$, where $\langle v, w \rangle$ is the inner product between the vectors $v$ and $w$. Alternatively, define $f_1(v) = \langle v, v_4 \rangle / \lVert v_4 \rVert$ and $f_2(v) = \langle v, v_3 \rangle / \lVert v_3 \rVert$, where $\lVert w \rVert $ is the norm of $w$. This alternative definition will render the induced linear transformation as a projection operator.

Construction of the linear functionals in terms of components:

Once we define $f_1(v) = \langle v, v_4 \rangle$, the matrix of the transformation is just the row matrix $f_1 = v_4^T = \begin{bmatrix} 1 & -1 & 1 & -1 \end{bmatrix}$, and similarly, $f_2 = v_3^T = \begin{bmatrix} 0 & -1 & 0 & 1 \end{bmatrix}$. If we want to use the alternative definitions, just divide these by the norms of the corresponding vectors.

Construction of the associated linear transformation:

We can construct the projection operator $P$ onto the space spanned by $v_3$ and $v_4$ by defining\begin{align}P = \langle \cdot, v_4 \rangle v_4/\lVert v_4 \rVert^2 + \langle\cdot, v_3 \rangle v_3/\lVert v_3 \rVert^2\,.\end{align}By a direct calculation, you can see that maps $a_1v_1+a_2v_2+a_3v_3+a_4v_4$ to $a_3v_3+a_4v_4$.

Construction of the associated linear transformation as a matrix:

Without showing the details, here's how you do it. Construct $\langle \cdot, v_4 \rangle v_4$ by doing the matrix multiplication $v_4v_4^T$, then multiply squared-norm of $v_4$. Do the same for $v_3$ and add. The result is\begin{align}\frac{1}{4}\begin{bmatrix}1 & -1 & 1 & -1\\-1 & 3 & -1 & -1\\1 & -1 & 1 & -1 \\-1 & -1 & -1 & 3\end{bmatrix}\end{align}