# Thread: Span of a vector space

1. ## Span of a vector space

Here is a question i got for one of my exercises and I'm not too sure how i would approach such a problem.

Is it possible to span the space of n×n matrices M(n,n) using the powers
of a single matrix A, i.e. I, A, A^2, . . ., A^n, . . .?

First thing i thought about was if there were matrices with the property such that as u keep multiplying it by itself you would get another matrix independent of the previous ones but that turned out to be a confusing. I also tried thinking about the opposite where if all matrices eventually become linearly dependent at some point therefore disproving that there exists n*n linearly independent matrices hence proving that it is not possible to span the space. It does seem a bit complex though doing it the way i was thinking about, so any insight to how i can solve this problem simpler would be great.

2. taking complex matrices makes it easier to understand. any complex matrix is similar to a upper traingular one. so wlog u may assume that u started with an upper traingular one (why?).

any power of an upper traingular matrix is also upper traingular... so all the powers will at most span only upper traingular matrices, leaving out the rest.

for real matrices you may argue the same way using jordan forms. in this case too the answer seems to be 'no'.

3. looking at it another way ...

if powers of a matrix spanned the whole space, then the smallest degree of a polynomial which is satisfied by that matrix woudld be greater than $\displaystyle n^2$ ... but you can see by the characteristic eqn that there must be a polynomial of degree n which is satisfied by that matrix.

4. Originally Posted by nirax
looking at it another way ...

if powers of a matrix spanned the whole space, then the smallest degree of a polynomial which is satisfied by that matrix woudld be greater than $\displaystyle n^2$ ... but you can see by the characteristic eqn that there must be a polynomial of degree n which is satisfied by that matrix.
can u elaborate on this please. i dont understand what u mean

5. Absolutely not. If it is true, then every matrix B will be a polynomial of A. That implies that the multiplication between matrix is commutative, which is obviously impossble.

6. consider $\displaystyle A, A^2, A^3, ..., A^{n^2}, A^{{n^2}+1}, ...$ if this set spans the full space then you must get $\displaystyle {n^2}$ lin independent ones among these. the highest power occuring in such a set must be at least $\displaystyle {n^2}$ if you try to write an expression of the form $\displaystyle a_1.A + a_2.A^2 + a_3. A^3 + ... + a_i.A^{i}$, this sum would never be zero unless you take $\displaystyle i$ to be atleast $\displaystyle {n^2}$ by the earlier argument of lin independence.

so the minimal polynomial of this matrix has degree at least $\displaystyle {n^2}$, contradicting the fact that minimal poly cannot be of degree higher than $\displaystyle {n}$.

which step you do not follow ??

7. Originally Posted by nirax
consider $\displaystyle A, A^2, A^3, ..., A^{n^2}, A^{{n^2}+1}, ...$ if this set spans the full space then you must get $\displaystyle {n^2}$ lin independent ones among these. the highest power occuring in such a set must be at least $\displaystyle {n^2}$ if you try to write an expression of the form $\displaystyle a_1.A + a_2.A^2 + a_3. A^3 + ... + a_i.A^{i}$, this sum would never be zero unless you take $\displaystyle i$ to be atleast $\displaystyle {n^2}$ by the earlier argument of lin independence.

so the minimal polynomial of this matrix has degree at least $\displaystyle {n^2}$, contradicting the fact that minimal poly cannot be of degree higher than $\displaystyle {n}$.

which step you do not follow ??
why is it the fact that minimal polynomial cannot be of degree higher than n? if u look in my OP u would see that i wrote:
"Is it possible to span the space of n×n matrices M(n,n) using the powers
of a single matrix A, i.e. I, A, A^2, . . ., A^n, . . .?"
so it is not limited to n

8. Originally Posted by ah-bee
why is it the fact that minimal polynomial cannot be of degree higher than n? if u look in my OP u would see that i wrote:
"Is it possible to span the space of n×n matrices M(n,n) using the powers
of a single matrix A, i.e. I, A, A^2, . . ., A^n, . . .?"
so it is not limited to n
By the Cayley Hamilton Theorem,the eigenpolynomial of any matrix A is a annihilator of A. And any annihilator of A is a multiple of the minimal polynomial. But the degree of eigenpolynomial of A is n, which implies that the degree minimal polynimial cannot be greater than n.

9. Originally Posted by ynj
By the Cayley Hamilton Theorem,the eigenpolynomial of any matrix A is a annihilator of A. And any annihilator of A is a multiple of the minimal polynomial. But the degree of eigenpolynomial of A is n, which implies that the degree minimal polynimial cannot be greater than n.
i see. thanks for that, never seen that theorem before.

10. Actually, you can refer #5, which is much simpler..

11. thanks.. figured it out