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Let $A=\begin{pmatrix} 1&2\\ 2&4\\ 0&0\\ \lambda&1 \end{pmatrix}$. Find the rank of $(AA^T)$ with respect to $\lambda$.

$A^T=\begin{pmatrix} 1&2&0&\lambda\\ 2&4&0&1 \end{pmatrix}$

$AA^T=\begin{pmatrix} 5&10&0&\lambda+2\\ 10&20&0&2\lambda+4\\ 0&0&0&0\\ \lambda+2&2\lambda+4&0&\lambda^2+1 \end{pmatrix}$

What should I do now? The first 3 columns are linearly dependent, right?

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    $\mathrm{rank}(AA^{\top})=\mathrm{rank}(A)$. – 2017-01-26
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    Yes, they're linearly dependent since they're all multiples of the first column. – 2017-01-26
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    $\text{rank} (AA^T) = \text{rank}(A) = 1$ for $\lambda=1/2$, and the rank is $2$ for $\lambda \neq 1/2$. – 2017-01-26
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    @StubbornAtom The OP hasn't said that the underlying field is real (although I do believe that it is real). – 2017-01-26
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    @user1551 Whenever there is lack of specification/rigour in the question, you tend to assume things. – 2017-01-26
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    The usage of numbers other than $0$ and $1$ indicates that the underlying field does not have characteristic $2$. Therefore the multiplicative inverse of $2$ exists and Alex' solution holds. – 2017-01-26
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    @ReinhardMeier $AA^T$ and $A$ need not have the same rank over finite fields – 2017-01-26
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    @ReinhardMeier for example, take this matrix over $\Bbb F_5$ with $\lambda = -2$ – 2017-01-26
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    You are right, I focused too much on the rank of $A$ and forgot to consider the matrix multiplication. – 2017-01-26

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