1. ## proof question about positive semidefinite matrix(important 4 regression analysis)

this is a proof I encounter in William Greene econometric analysis appendix section.

If A is a n*k with full column rank and n > k, then (A')(A)is positive definite and (A)(A') is positive semi-definite.

proof given by author is as follows,

By assumption, Ax is not equal to zero. So, x'A'Ax = (Ax)'(Ax) = y'y = summation y^2 > 0

for the latter case, because A has more rows then columns, then there is an X such that A'x = 0, thus we can only have y'y >= 0

What I dont understand is the the bold and underline part.

P/S: this question comes from pg 835, of William Greene Econometric Analysis 5th edition textbook.

2. Originally Posted by phoenicks
this is a proof I encounter in William Greene econometric analysis appendix section.

If A is a n*k with full column rank and n > k, then (A')(A)is positive definite and (A)(A') is positive semi-definite.

proof given by author is as follows,

By assumption, Ax is not equal to zero. So, x'A'Ax = (Ax)'(Ax) = y'y = summation y^2 > 0
$\displaystyle A$ has full column rank means that the columns of $\displaystyle A$ are linearly independent. see that if $\displaystyle v_1, \cdots , v_k$ are the columns of $\displaystyle A$ and $\displaystyle \bold{x}=[x_1 \ x_2 \cdots \ x_k]^T,$ then $\displaystyle A \bold{x}=x_1v_1 + \cdots + x_kv_k.$

so if $\displaystyle A \bold{x}=\bold{0},$ then $\displaystyle x_1v_1 + \cdots + x_k v_k = 0,$ and thus $\displaystyle x_j = 0,$ for all $\displaystyle j,$ because $\displaystyle v_1, \cdots , v_k$ are linearly independent. hence the only solution of $\displaystyle A \bold{x}=\bold{0}$ is $\displaystyle \bold{x}=\bold{0}.$

for the latter case, because A has more rows then columns, then there is an X such that A'x = 0, thus we can only have y'y >= 0

What I dont understand is the the bold and underline part.

P/S: this question comes from pg 835, of William Greene Econometric Analysis 5th edition textbook.

i think by $\displaystyle A'$ you mean $\displaystyle A^T,$ the transpose of $\displaystyle A.$ assuming that the entries of $\displaystyle A$ come from a field $\displaystyle F,$ the matrix $\displaystyle A^T$ has $\displaystyle n$ colums and these columns are in $\displaystyle F^k.$ we know that $\displaystyle \dim F^k = k.$
thus if $\displaystyle n > k,$ then any $\displaystyle n$ vectors in $\displaystyle F^k$ are linearly dependent. now $\displaystyle A^T$ has $\displaystyle n$ columns, say $\displaystyle w_1, \cdots , w_n,$ and $\displaystyle n > k.$ so they are linearly dependent, i.e. there exists $\displaystyle \bold{0} \neq \bold{x}=[x_1 \ x_2 \cdots \ x_n]^T$
such that $\displaystyle x_1w_1 + \cdots + x_nw_n = \bold{0}.$ this also can be written as: $\displaystyle A^T \bold{x} = \bold{0}.$