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I am taking a signals processing class online and we are covering some basic linear algebra. Here is a question that I'm struggling with from the class.

Find a matrix $A$ which satisfies all of the following characteristics:

  • $\text{dim }\text{col}(A) = 2$
  • $\begin{bmatrix} 1 & 1 & 2 & -1\end{bmatrix}^T \in \text{null}(A)$
  • $\begin{bmatrix} 3 & 3 & 6 \end{bmatrix}^T \in \text{col}(A)$
  • $\begin{bmatrix} 3 & 0 & 0 & 3\end{bmatrix}^T \in \text{col}(A^T)$

Attempt

So far, I've started by making the first two columns of my matrix $A$ independent.

$$\begin{bmatrix} 1 & 0 & \alpha_1 & \beta_1 \\ 1 & 0 & \alpha_2 & \beta_2 \\ 0 & 1 & \alpha_3 & \beta_3 \end{bmatrix}$$

From this I can clearly fulfill the requirement that $\begin{bmatrix} 3 & 3 & 6 \end{bmatrix}^T \in \text{col}(A)$. Then I start filling in the $\alpha_i$ and $\beta_i$ values to fulfill the other characteristics.

The problem is, whenever I manage to satisfy a couple of the characteristics, I break the third one. I know how to individually satisfy these characteristics, but not how to satisfy them together. How can I go about constructing a matrix that satisfies all these characteristics?

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Here's an idea. Construct a suitable matrix of the form $$ \pmatrix{3&0&0&3\\3&0&0&3\\6&?&?&6}. $$ The first row and first column ensure that conditions 2 and 4 are satisfied. The fact that the middle row is the same as the first ensures that the rank is at most 2. In order to make sure that the rank is exactly $2$, it suffices to choose non-zero entries for the unknown entries. Note, however, that these must be selected in such a way that condition 2 holds.

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The last condition says that the matrix has a row $$ [3 \; 0\; 0 \; 3]. $$ Notice that this row's dot product with the generator of your nullspace, i.e. with the vector $$ \nu = [1 \; 1 \; 2 \; −1] $$ is zero. Furthermore, the same is true of a row that is either a scalar multiple of the above row, or has the form $$ [3\; \alpha\; \beta \; 3]. $$ where $[\alpha \; \beta] \cdot [1 \; 2] = 0$.

Furthermore, the requirement that the rank of the matrix is 2 can be met by making two of the three rows linearly dependent. So, let's make the rows of our matrix: $$ [3 \; 0\; 0 \; 3], $$ $$ [3 \; \alpha\; \beta \; 3], $$ and, finally, the row that equals twice the first row: $$ [6 \; 0\; 0 \; 6]. $$ All that is left to do is choose suitable nonzero values for $\alpha, \beta$.

Each of these rows, when dotted with our desired nullspace generator $\nu$, gives zero, hence $\nu$ is indeed in the nullspace, as required. The rank of the resulting matrix is 2 as long as $\alpha, \beta$ are not both zero. And the first (and also the last) column satisfies the 3rd bulleted requirement.

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