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For $V = sum_s=1^t A_s A_s^T$ to be non-singular $(A_s)_s=1^t$ needs to span $R^d$

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0

$begingroup$

I am reading a book on bandits algorithm and inside a proof it says the following:

Let $(A_s)_s=1^t$ be sequence of vectors in $R^d$. Construct a matrix $V$ such that:

$$ V = sum_s=1^t A_s A_s^T$$

Now for $V$ to be non-singular, $(A_s)_s=1^t$ must span $R^d$ and $t$ must be at least $d$.

I can see why $t$ needs to be at least $d$ but I am not sure how to prove that $(A_s)_s=1^t$ must span $R^d$.

Any help in understanding this would be much appreciated.

linear-algebra matrices vector-spaces

share|cite|improve this question

asked Mar 21 at 14:05

hi15hi15

1586

$endgroup$

add a comment | 

0

$begingroup$

I am reading a book on bandits algorithm and inside a proof it says the following:

Let $(A_s)_s=1^t$ be sequence of vectors in $R^d$. Construct a matrix $V$ such that:

$$ V = sum_s=1^t A_s A_s^T$$

Now for $V$ to be non-singular, $(A_s)_s=1^t$ must span $R^d$ and $t$ must be at least $d$.

I can see why $t$ needs to be at least $d$ but I am not sure how to prove that $(A_s)_s=1^t$ must span $R^d$.

Any help in understanding this would be much appreciated.

linear-algebra matrices vector-spaces

share|cite|improve this question

asked Mar 21 at 14:05

hi15hi15

1586

$endgroup$

add a comment | 

$begingroup$

I am reading a book on bandits algorithm and inside a proof it says the following:

Let $(A_s)_s=1^t$ be sequence of vectors in $R^d$. Construct a matrix $V$ such that:

$$ V = sum_s=1^t A_s A_s^T$$

Now for $V$ to be non-singular, $(A_s)_s=1^t$ must span $R^d$ and $t$ must be at least $d$.

I can see why $t$ needs to be at least $d$ but I am not sure how to prove that $(A_s)_s=1^t$ must span $R^d$.

Any help in understanding this would be much appreciated.

linear-algebra matrices vector-spaces

share|cite|improve this question

asked Mar 21 at 14:05

hi15hi15

1586

$endgroup$

share|cite|improve this question

asked Mar 21 at 14:05

hi15hi15

1586

share|cite|improve this question

asked Mar 21 at 14:05

hi15hi15

1586

asked Mar 21 at 14:05

hi15hi15

1586

asked Mar 21 at 14:05

hi15hi15

1586

1 Answer
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1

$begingroup$

For $V$ to be non-singular, it has to have rank $d$. That means that the rows (or columns, whichever you prefer) span all of $R^d$. Taking a close look at these columns, you will see that every single column is a linear combination of the $A_s$. Hence, the rows of $V$ span at most the same space as the $A_s$.

share|cite|improve this answer

answered Mar 21 at 14:29

DirkDirk

4,488219

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1

$begingroup$

For $V$ to be non-singular, it has to have rank $d$. That means that the rows (or columns, whichever you prefer) span all of $R^d$. Taking a close look at these columns, you will see that every single column is a linear combination of the $A_s$. Hence, the rows of $V$ span at most the same space as the $A_s$.

share|cite|improve this answer

answered Mar 21 at 14:29

DirkDirk

4,488219

$endgroup$

add a comment | 

1

$begingroup$

For $V$ to be non-singular, it has to have rank $d$. That means that the rows (or columns, whichever you prefer) span all of $R^d$. Taking a close look at these columns, you will see that every single column is a linear combination of the $A_s$. Hence, the rows of $V$ span at most the same space as the $A_s$.

share|cite|improve this answer

answered Mar 21 at 14:29

DirkDirk

4,488219

$endgroup$

add a comment | 

$begingroup$

For $V$ to be non-singular, it has to have rank $d$. That means that the rows (or columns, whichever you prefer) span all of $R^d$. Taking a close look at these columns, you will see that every single column is a linear combination of the $A_s$. Hence, the rows of $V$ span at most the same space as the $A_s$.

share|cite|improve this answer

answered Mar 21 at 14:29

DirkDirk

4,488219

$endgroup$

share|cite|improve this answer

answered Mar 21 at 14:29

DirkDirk

4,488219

answered Mar 21 at 14:29

DirkDirk

4,488219

answered Mar 21 at 14:29

DirkDirk

4,488219