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In going over some example problems from class notes, I came across the following problem:

"Suppose we have a regression model $y=\beta_0+\beta_1x_1+\beta_2x_2+\beta_3x_4+\beta_4x_4+\epsilon$. We wish to test the null hypothesis that $H_0: \beta_1=\beta_2$. Express the hypothesis in terms of $A\beta=c$, provide $A$ and $c$ which yield the correct formulation of $H_0$. What is the rank of $A$?"

So, unless I am misunderstanding the problem, I would go about answering it thusly:

1) Restate the null hypothesis as $H_0: \beta_1-\beta_2=0$.

2) Set $c=0$.

3) Then, for $A\beta=c$, we have $A=[0,1,-1,0,0]$ and (no idea how to typeset a column vector) $\beta'=[\beta_0,\beta_1,\beta_2,\beta_3,\beta_4]$. Do the vector multiplication out, and you end up with $\beta_1-\beta_2=0$, which is our null hypothesis.

But, then, how do I answer the last part of the question? The rank of $A$? I've never even heard anyone talk about rank in the context of a vector. Did I do something wrong here and $A$ is supposed to be some $m$ x $n$ matrix for which I could calculate a rank? Or was this just a bad question (the professor in question is notorious for asking questions that don't make any sense)?

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  • $\begingroup$ It looks to me like $A$ is a $1\times 5$ matrix. $\endgroup$ – whuber Aug 6 '14 at 14:08
  • $\begingroup$ This may help you resolve one part of your confusion - en.wikipedia.org/wiki/… $\endgroup$ – Glen_b Aug 6 '14 at 14:11
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Your procedure is correct.

The rank can be very well interpreted for a single vector as well: the (row) rank will be the number of linearly independent rows. Given that a row vector has a single row it translates to whether the row will be linearly independent itself. Considering now the the definition of linear dependence (''has a linear combination with not every coefficient being zero that is the zero vector'') applied to a single vector (''multiplied with a non-zero scalar it is the zero vector'') it is immediately obvious that a non-zero vector is linearly independent (there is no scalar with which we can multiply a non-zero vector and the result will be the zero vector) and a zero vector is linearly dependent (the multiplier can be any scalar).

To sum up: the rank is 0 if the vector is the zero vector, 1 otherwise. In your case, the rank is thus 1.

In that specific context, the question makes some sense (at least generally speaking...) as it is customary to check prior to such parameter-restricion testings whether matrix $A$ is of full row rank so that the equations are not redundant. (See Hayashi Econometrics, p. 39.)

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  • $\begingroup$ Thanks Tamas! That is more or less the answer I would have guessed, I just had no logical explanation for why I would choose it. $\endgroup$ – Ryan Simmons Aug 6 '14 at 15:32
  • $\begingroup$ @RyanSimmons: You're welcome! $\endgroup$ – Tamas Ferenci Aug 6 '14 at 15:34

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