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Math Help - Rank of the given Matrix

  1. #1
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    Rank of the given Matrix

    Let \bf{x} and \bf{y} \in \mathbb{R}^{n} be 2 non zero column vectors. Let \bf{y}^{T} denote the transpose of \bf{y}. Let A=\bf{x} \bf{y}^{T}. What is the rank of A?
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  2. #2
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    What do you mean by y^T, because we don't know what "y" is! I think maybe you're referring to the case where x=y. And I think in this case xx^T will have the full column rank if det(x^Tx) \neq 0, and it will have the full row rank if det(xx^T) \neq 0.

    Since x^Tx is square, and it is an nxn matrix, it follows that x^Tx is invertible if and only if x^Tx has rank n. However x^Tx has the same rank as x itself, so x^Tx is invertible if rank(x)=n (this means if x has full column rank).

    Again I'm assuming x=y.
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  3. #3
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    I see no reason to assume that x= y. The problem makes sense without that. For example, if x= \begin{bmatrix}a \\ b\end{bmatrix} and y= \begin{bmatrix}c \\ d\end{bmatrix}, then y^T= \begin{bmatrix}c & d\end{bmatrix} and
    xy^T= \begin{bmatrix}a \\ b\end{bmatrix}\begin{bmatrix}c & d\end{bmatrix}= \begin{bmatrix}ac & ad \\ bc & bd\end{bmatrix}. The determinant of that is is acbd- adbc= 0 so it does NOT have rank 2. It is not, unless at least one of x and y is the 0 vector, all "0"s so it does not have rank 0. It has rank 1.
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