Finding missing dual basis vectors

Question

I have two dual basis vectors $b_1^*,b_2^* \in (\mathbb R^{4\times1})^*$,whereas the star should denote that I am working in the dual vector space, and $E$ stands for the canonical basis. My dual basis vectors in respect to the canonical basis coordinates are:

$$\langle b_1^*,E\rangle = (1,1,1,1)$$ $$\langle b_2^*,E\rangle = (1,0,0,1)$$

Now I would like to find two additional dual basis vectors, so that they form a complete basis of $(\mathbb R^{4\times1})^*$. After that, I would like to find the coordinates of my "normal" basis. (I think its called bidual but I am not sure). I think once I found my dual basis, I could find an inverse matrix which should give me my regular basis coordinates, but how do I find those two additional basis vectors first of all?

Edit: After thinking for a bit: Couldn't I just arbitrarily choose linearly independent vectors, such as:

$$\langle b_3^*,E\rangle = (1,0,0,0)$$ $$\langle b_4^*,E\rangle = (0,1,0,0)$$

Answer 1

My previous answer is overly constrained. If we identify elements of $V^*=(\mathbb R^{n\times1})^*$ with row vectors of coordinates relative to the dual of a basis of $V=\mathbb R^{n\times1}$, then applying an element of $V^*$ to a vector in $V$ becomes matrix multiplication. The dual basis condition $b_i^*b_j=\delta_{ij}$ then says that the matrices formed from the respective coordinate vectors relative to the standard bases (or any pair of dual bases, for that matter) are inverses.

So, we can take for $b_3^*$ and $b_4^*$ any extension to a complete basis of $V^*$ and find the corresponding basis of $V$ by inverting the matrix with these dual basis vectors as its rows. For example, taking $b_3^*=(1,0,0,0)$ and $b_4^*=(0,1,0,0)$, we form the matrix $$\pmatrix{1&1&1&1\\1&0&0&1\\1&0&0&0\\0&1&0&0}$$ and invert it to get $$\pmatrix{0&0&1&0\\0&0&0&1\\1&-1&0&-1\\0&1&-1&0}.$$ The basis vectors $b_i$ are the columns of this matrix.

Answer 2

Let $V=\mathbb R^{4\times1}$ and let $W\subseteq V$ be a subspace. If $T:V\to V$ is a linear operator with kernel $W$, we know that the kernel of $T^*$ annihilates the image of $T$ and the image of $T^*$ annihilates $W$.

Let $W=\operatorname{span}\{b_1,b_2\}$. Since $b_1^*$ and $b_2^*$ are dual to these vectors, $W^*=\operatorname{span}\{b_1^*,b_2^*\}$. Choose for $T$ orthogonal projection onto $W^\perp$. Its dual $T^*$ projects onto the orthogonal complement of $W^*$, so we can choose for $b_3^*$ and $b_4^*$ any basis of $(W^*)^\perp$. The dual basis to this can then be found by inverting the matrix which has these vectors as rows: the resulting matrix will have the dual basis for its columns.

Update: Why this particular choice for $T$? If we identify $V^*$ with $\mathbb R^{n\times1}$ in the obvious way, the dual basis conditions become orthogonality conditions among the vectors. This means that we will be looking at intersections of orthogonal complements of various spaces to find these vectors, which is equivalent to examining the image and kernel of various orthogonal projection operators. This also provides a succinct reason why one can’t select just any vectors outside of the span of $b_1^*$ and $b_2^*$ for $b_3^*$ and $b_4^*$: they also have to be in the correct subspace of $V^*$, specifically, the annihilator of the span of $b_1$ and $b_2$, which is the orthogonal complement of the span of $b_1^*$ and $b_2^*$.