The way you showed is wrong because with your hypothesis we may have .
Maybe there is a typo in the hint given by your prof, since for example if , is not an eigenvalue.
here is nxn matrices . (A)=1 and tr(A)=n
find the representative polinomial of A and tell if A is diagonisable.
there is a way which my prof said is wronggg:
i said that this A could be a representation matrices for some transformation
and if (A)=1 then dim ImT=1 too,so dim KerT=n-1
so if we take a basis sfor this kernel and expand it
to .
so the first n-1 columns are zero columns exsept the last n'th column
and from the law that the traces of similar matrices are equal i get
that the diagonal of the representative matrices is all zero except there is 'n' in the corner
and from here its very easy to prove that its diagonisable.
but my prof said that its wrong because even if the representative matrice is diagonisable
it doesnt say anything about A
this is how my prof recommended to start to solve it:
if (A)=1 then we have eigen value 0 with n-1 eigen vectors.
dont know how to continue from here
??
i believe transgalactic means: ρ(A) = rank(A).
rank(A) = 1 --> nullity(A) = n-1.
the nullity of A is the dimension of the null space of A, which is also the eigenspace corresponding to the eigenvalue 0.
this tells us the geometric multiplicity of 0 is n-1, so the algebraic multiplicity of 0 is at least n-1.
this, in turn, tells us that where is the characteristic polynomial of A.
now A has to have at least 1 non-zero eigenvalue since nullity(A) < n. for the time being, let's call this eigenvalue λ, and let v be an eigenvector corresponding to λ.
then if is a basis for null(A), then
is an eigenbasis for V, so A is diagonalizable.
if P is the matrix whose columns are the eigenbasis vectors, then is the matrix with all 0's, and λ in the lower-right hand corner.
and as you pointed out .
so the characteristic polynomial is: .
"now A has to have at least 1 non-zero eigenvalue since nullity(A) < n"
why is that?
before you say
" is an eigenbasis for V, so A is diagonalizable"
you must first prive that this basis consists only from eigen vectors
you havent prove that "v" eigen vector in this step
if nullity(A) = n, then A is the 0-matrix. the 0-matrix (and only the 0-matrix) has rank 0.
A does not have rank 0, so A is not the 0-matrix. rank(A) = 1, so every vector in the column space of A,
that is every vector in the image A(V), is a multiple of just one vector in V. let's call this basis of im(A), w.
so for EVERY vector v in V, A(v) = cw, for some scalar c (which scalar depends on which vector v we choose).
but w is in V, so A(w) = cw. thus w is an eigenvector for A, with eigenvalue c. because w is a BASIS vector, it is non-zero.
can the eigenvalue c be 0? no. because if A(w) = 0w = 0, then every vector in the image of A would have to be 0 as well
(w is a basis for im(A)), contradicting the fact that A is NOT the 0-matrix.
so, A has an eigenvector with a non-zero eigenvalue (we can in fact, use any non-zero column of A as this eigenvector,
and we know A HAS a non-zero column, because A is not the 0-matrix).
in fact, A has a LOT of eigenvectors with non-zero eigenvalue, but they are all scalar multiples of each other. that is what rank(A) = 1 means.
for example, suppose A is the matrix:
[3 0 0]
[1 0 0]
[0 0 0]. then (3,1,0) is an eigenvector for A with eigenvalue 3. indeed, A(x,y,z) = (3x,x,0) = x(3,1,0),
so {(3,1,0)} is a basis for the image of A. we get the same thing if A =
[2 2 0]
[1 1 0]
[0 0 0]. here A(x,y,z) = (2x+2y,x+y,0) = (x+y)(2,1,0). either non-zero column works as an eigenvector.
to find the eigenvalue, just find A(2,1,0) = (2+1)(2,1,0) = 3(2,1,0), so the eigenvalue is 3.