I have a homework for my linear algebra class at my university the thing is that we get a 4×4 matrix A then we have to find it's Transpose which is pretty easy and then find the matrix B=(A^T)*A also A=(A^T).The problem is that the matrix B's elements are all three figures making it very difficult to find the characteristic polyonymal, the eigenvalues and eigenvectors.no question actualy asks to find them but i can't find ways to answear them without doing so.things i try to prove are
(1) if the eigenvectors are linearly independent(i tried to do that by saing that it has 4 real distinct eigenvalues i can prove that the eigenvalues are real cause I found that B=B^T but i have no idea how to prove that it has 4 distinct eigenvalues)
(2) if the eigenvectors are rectangular(not sure if rectangular is the right terminology for what i want to ask i am greek) with each other (i would try to prove that the inner product of the eigenvectors equals zero by calculating them but as stated above finding the eigenvectors are hard to find with this matrix) a fraind of mine that found the eigenvectors by using some program then tried to calcute the inner products of the eigenvectors what he found was that the inner product was very close to 0 but since the eigen vectors were found by some program we can't be certain.
every and all advice is welcomed
Best Answer
If $B=A^TA$, then $B$ is symmetric, because $B^T=(A^TA)^T=A^T(A^T)^T=A^ TA=B$. You should have a theorem that says symmetric matrices have real eigenvalues and the eigenvectors corresponding to different eigenvalues are orthogonal (which is the term you want instead of rectangular). If any eigenvalues do match, the eigenvectors can be chosen to be orthogonal as well.
Added: here is a simple example to show what I mean about repeated eigenvalues and eigenvectors. Let $M=\begin {pmatrix} 1&0&0\\0&1&0\\0&0&2 \end {pmatrix}$ It has eigenvalues $1,1,2$. The eigenvector corresponding to $2$ is $(0,0,1)^T$ Any vector of the form $(x,y,0)^T$ will be an eigenvector corresponding to $1$. Note that all of these are orthogonal to $(0,0,1)^T$. We can choose two orthonormal eigenvectors to span the space (the $xy$ plane), for instance $(\frac 35,\frac 45,0)^T$ and $(\frac 45,-\frac 35,0)^T$ (or, of course, the more mundane one).