# help needed!!! proof of invertible matrices

• November 11th 2012, 01:14 PM
kaya2345
help needed!!! proof of invertible matrices
The question has been attached as an image.

any help or light on this question would be hugely appreciated!!

thank you http://static1.tsrfiles.co.uk/4.7/im...lies/smile.png
• November 11th 2012, 01:29 PM
Plato
Re: help needed!!! proof of invertible matrices
Quote:

Originally Posted by kaya2345
The question has been attached as an image.
any help or light on this question would be hugely appreciated!!
thank you http://static1.tsrfiles.co.uk/4.7/im...lies/smile.png

If $\delta _{j,k} = \left\{ {\begin{array}{rl} {1,} & {j = k} \\ {0,} & {j \ne k} \\ \end{array} } \right.$ and $I_n=\left[\delta _{j,k}\right]$.

Is $I_n$ invertible?

Can we take away any of those ones?
• November 11th 2012, 01:32 PM
kaya2345
Re: help needed!!! proof of invertible matrices
Quote:

Originally Posted by Plato
If $\delta _{j,k} = \left\{ {\begin{array}{rl} {1,} & {j = k} \\ {0,} & {j \ne k} \\ \end{array} } \right.$ and $I_n=\left[\delta _{j,k}\right]$.

Is $I_n$ invertible?

Can we take away any of those ones?

thank you. and yes In is invertible
• November 11th 2012, 01:57 PM
HallsofIvy
Re: help needed!!! proof of invertible matrices
That shows, I think, that the minimum number of 1s in such an invertible n by n matrix is n. But the question was about the maximum number. We do know that if a matrix has two columns the same, it is not invertible. I think we can avoid that by having a 0, at a different row, in each column. That implies that the maximum number of 1s in such an invertible n by n matrix is $n^2- n$.
• November 11th 2012, 02:02 PM
kaya2345
Re: help needed!!! proof of invertible matrices
Quote:

Originally Posted by HallsofIvy
That shows, I think, that the minimum number of 1s in such an invertible n by n matrix is n. But the question was about the maximum number. We do know that if a matrix has two columns the same, it is not invertible. I think we can avoid that by having a 0, at a different row, in each column. That implies that the maximum number of 1s in such an invertible n by n matrix is $n^2- n$.

Thank you, for that. Could I prove this using proof by induction? or contradiction?
• November 11th 2012, 02:21 PM
Plato
Re: help needed!!! proof of invertible matrices
Quote:

Originally Posted by HallsofIvy
That shows, I think, that the minimum number of 1s in such an invertible n by n matrix is n. But the question was about the maximum number.

That mistake was once caused by my laptop not doing well with those attachments.
I wish we could outlaw the posting of questions in images.