Question

Let \(A\in M_3(\mathbb{C})\). Suppose the column vector \[ v=\begin{pmatrix} \sqrt{5}i\\ 2i\\ x \end{pmatrix} \] in \(\mathbb{C}^3\) belongs to the intersection of nullspace\((A)\) and rangespace\((A^T)\). Then \(|x|=\underline{}\) rounded off to one decimal place.

Hint:

If \(v\in \text{nullspace}(A)\cap \text{rangespace}(A^T)\), then \(v^Tv=0\) can be used directly.

Answer 1

Correct Answer: 3

Step 1: Use both memberships.
The vector $v$ lies in the null space of $A$, so $Av=0$. It also lies in the range of $A^T$, so $v=A^T y$ for some $y$.

Step 2: A neat orthogonality fact.
The null space of $A$ and the range of $A^T$ are orthogonal complements, so a vector in both must be orthogonal to itself.

Step 3: Write the condition.
That gives $v^T v=0$. Plugging in the entries, $(\sqrt5\,i)^2+(2i)^2+x^2=0$.

Step 4: Simplify.
\[ -5-4+x^2=0 \]

Step 5: Solve.
\[ x^2=9 \quad\Rightarrow\quad |x|=3 \]
\[ \boxed{3.0} \]