# How to find eigenvalues of $n \times n$ real symmetric matrix?

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.

• What do you mean "that satisfies $f(e_i,e_j)$ for all $i, j$ "? What is $f$? – 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?
• @user24907 Yes you got it. So you have eigenvalue $n$ of multiplicity 1 and eigenvalue $0$ of multiplicity $n-1$. – user2566092 Feb 17 '15 at 21:14
• 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. – user24907 Feb 17 '15 at 21:27
• @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. – user2566092 Feb 17 '15 at 21:37