So am trying to find signature of a bilinear form that satisfies $ f\mathrm (e_i\mathrm , e_j \mathrm)$ for all $i, j$ where $\mathrm e_i$ are the standard unit vectors.

So the only way I can think how to do this is to form a matrix $\mathrm A$ that is an $n \times n$ real symmetric matrix with $1$ in every entry and try find the number of positive, negative, and zero eigenvalues.

But now I'm stuck after constructing this matrix as its $n \times n$ and the usual method of $det\mathrm(A\mathrm - cI\mathrm)=0$ isn't possible, or is it? I don't know how to find all the eigenvalues for $n \times n$ matrix as have only computed for $2 \times 2$, $3 \times 3$, etc. No idea how to do it for $n \times n$ even if I do know all the entries.

As A is real symmetric matrix I know there should be $n$ real eigenvalues not necessarily distinct. Using definition of eigenvalues I think $c=0$ and $c=n$ may be two and $n$ could have multiplicity $n-1$ but that's not exactly proving it. Can anyone tell me what methods are available for computing eigenvalues for $n \times n$ if you know all the entries?

Or if there is a different way I could approach this problem? Just to clarify, I'm only a first year undergrad so have not come across a lot of the theory and possible methods. Oh and this is homework but I'm looking for advice, not just entire solution, as I want to know how I can do this when faced with similar situation for arbitrarily sized matrices.

Thanks for any hints or advice.

  • $\begingroup$ What do you mean "that satisfies $f(e_i,e_j)$ for all $i, j$ "? What is $f$? $\endgroup$ – BaronVT Feb 17 '15 at 19:32

Hint: If all entries are 1 then the rank (dimension of column span) of your matrix is just 1 because all columns are the same, and so the rank-nullility theorem says that the dimension of the kernel of your matrix (hence the multiplicity of eigenvalue $0$) is $n-1$. So there is just one non-zero eigenvalue with multiplicity one. By looking at your matrix, can you figure out what the one non-zero eigenvalue and associated eigenvector are?

  • $\begingroup$ hi sorry for late reply, I went to read in library. Is the one non-zero eigenvalue is n ? and the eigenvector is n(1 1 1 . . .1) with this being a column, I don't know how to write column vector of this kind. Is this correct? $\endgroup$ – user24907 Feb 17 '15 at 21:05
  • $\begingroup$ @user24907 Yes you got it. So you have eigenvalue $n$ of multiplicity 1 and eigenvalue $0$ of multiplicity $n-1$. $\endgroup$ – user2566092 Feb 17 '15 at 21:14
  • $\begingroup$ Thank you, so is this the main method for finding eigenvalues when you cant put them in solvable polynomial? I'm still unsure on what method I can use when dealing with arbitrary sized matrices such as $n \times n$ . Now I need to do same again except now its all 0 on diagonal and 1 everywhere else. $\endgroup$ – user24907 Feb 17 '15 at 21:27
  • $\begingroup$ @user24907 This is just a trick based on inspection of the matrix. There are many tricks. If you'd like a trick for your next matrix, let that matrix be $B$ and note $B = A - I$ where $A$ is the matrix in this problem and $I$ is the identity matrix. If $A$ has eigenvalue $\lambda$ with multiplicity $k$ the $B$ has eigenvalue $\lambda - 1$ with multiplicity $k$. That's easy to show based on definition of eigenvectors and eigenvalues. $\endgroup$ – user2566092 Feb 17 '15 at 21:37
  • $\begingroup$ Thanks, yes from reading your answer and other similar problems I can see that it really does come down to using different tricks for different situations and knowing the different ways you can think about eigenvalues. I was just too focused on the usual way of constructing polynomial and solving for roots. I'm going to look for other problems which will force me to find eigenvalues using different methods/tricks. Thanks a lot for your time and wisdom, I really appreciate it. $\endgroup$ – user24907 Feb 17 '15 at 21:56

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