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.
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.
Show Hint
If \(v\in \text{nullspace}(A)\cap \text{rangespace}(A^T)\), then \(v^Tv=0\) can be used directly.
Updated On: Jun 1, 2026
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Correct Answer: 3
Solution and Explanation
Step 1: Use the given condition.
Since \(v\in \text{nullspace}(A)\), we have
\[ Av=0 \]
Step 2: Use the second condition.
Since \(v\in \text{rangespace}(A^T)\), there exists a vector \(y\) such that
\[ v=A^T y \]
Step 3: Multiply \(Av=0\) by \(y^T\).
\[ y^TAv=0 \]
Step 4: Rewrite using transpose.
\[ y^TAv=(A^Ty)^Tv \]
Since \(A^Ty=v\), we get
\[ v^Tv=0 \]
Step 5: Compute \(v^Tv\).
\[ v^Tv=(\sqrt{5}i)^2+(2i)^2+x^2 \]
\[ v^Tv=-5-4+x^2 \]
Step 6: Equate to zero.
\[ x^2-9=0 \]
\[ x^2=9 \]
Step 7: Find \(|x|\).
\[ |x|=3 \] \[ \boxed{3.0} \]
Since \(v\in \text{nullspace}(A)\), we have
\[ Av=0 \]
Step 2: Use the second condition.
Since \(v\in \text{rangespace}(A^T)\), there exists a vector \(y\) such that
\[ v=A^T y \]
Step 3: Multiply \(Av=0\) by \(y^T\).
\[ y^TAv=0 \]
Step 4: Rewrite using transpose.
\[ y^TAv=(A^Ty)^Tv \]
Since \(A^Ty=v\), we get
\[ v^Tv=0 \]
Step 5: Compute \(v^Tv\).
\[ v^Tv=(\sqrt{5}i)^2+(2i)^2+x^2 \]
\[ v^Tv=-5-4+x^2 \]
Step 6: Equate to zero.
\[ x^2-9=0 \]
\[ x^2=9 \]
Step 7: Find \(|x|\).
\[ |x|=3 \] \[ \boxed{3.0} \]
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