Let \(A\) be an \(n \times n\) matrix with rank \(k\). Consider the following statements:

- If \(A\) has real entries, then \(AA^t\) necessarily has rank $k$
- If \(A\) has complex entries, then \(AA^t\) necessarily has rank \(k\).

- 1 and 2 are true
- 1 and 2 are false
- 1 is true and 2 is false
- 1 is false and 2 is true

TIFR 2019 GS Part A, Problem 7

Linear algebra

Easy

Do you really need a hint? Try it first!

\(x\) is a null vector of \(AA’\) implies that \(x\) is a null vector of \(A’\)
Let \(q_i\) be the null vector of \(AA’\) i.e \(AA’q_i=0 \Rightarrow q_i’AA’q_i=0 \Rightarrow ||A’q_i||_2=0 \Rightarrow A’q_i=0 \) [\(A’q_i \in \Bbb R^n \) ]
\(q_i\) is a null vector of \(A’\)

\(x\) is a null vector of \(A’\) implies that $\(x\) is a null vector of \(AA’\)

\(Rank(AA’)=Rank(A’)=Rank (A)\)

Consider \(A=

\begin{pmatrix}

1 & i \\

0 & 0 \\

\end{pmatrix}\) then \(AA’=

\begin{pmatrix}

0 & 0 \\

0& 0 \\

\end{pmatrix}\)

\begin{pmatrix}

1 & i \\

0 & 0 \\

\end{pmatrix}\) then \(AA’=

\begin{pmatrix}

0 & 0 \\

0& 0 \\

\end{pmatrix}\)

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