I am given that $f(x)$, $g(x)$ and $h(x)$ are polynomials such that $f(x)=g(x)h(x)$.
I am then asked to show using induction on the degree of $f$ that $f(A)$ would be an $n \times n$-matrix if $A$ is an $n \times n$ matrix. I am told any constant term of $f(x), c,$ will be replaced by $cI$ in the matrix polynomial $f(A)$, where $I$ is the $n \times n$ identity matrix - same size as A.
Wouldn’t this be trivial as if $I$ is the $n \times n$ identity matrix, A would need to be the same size in order to add to I and thus any linear combination of powers of A would be the same size? I don’t know what induction has to do with this.
I am also asked to show that if $f(x)=g(x) h(x)$ and $f(A)=0$, then for any vector $v$ and any polynomial $b(A)$, then $b(A) h(A) v \in\ker(g(A))$ and, similarly, $b(A) g(A) v \in\ker(h(A))$. I don’t know how to start because I think that $b(A)h(A)v$ and $b(A)g(A)v$ are both vectors and we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.
Any hints or help would be greatly appreciated!
To show that $f(A)$ is an $n\times n$-matrix, you could prove by induction that $A^k$ is an $n\times n$-matrix for all $k\in\Bbb{N}$, and then note that any linear combination of $n\times n$-matrices is again an $n\times n$-matrix. I agree that it's a rather trivial thing to prove though.
To show that $b(A)h(A)v\in\ker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)v\in\ker g(A)$. You claim that
Indeed you can't take powers of vectors, but that's not what is being claimed.
You have just shown (by induction) that $g(A)$ is an $n\times n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $n\times n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)v\in\ker g(A)$ it suffices to show that the product is $0$.