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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 $b(A) h(A) v inker(g(A))$ and, similarly, $b(A) g(A) v inker(h(A))$. 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 $ntimes n$-matrix, you could prove by induction that $A^k$ is an $ntimes n$-matrix for all $kinBbbN$, and then note that any linear combination of $ntimes n$-matrices is again an $ntimes n$-matrix. I agree that it's a rather trivial thing to prove though.

To show that $b(A)h(A)vinker g(A)$ for any polynomial $b$ and any vector $v$, it suffices to show that $h(A)vinker g(A)$. You claim that

...we can’t insert vectors into $g(A)$ or $h(A)$ because you can’t take powers of vectors.

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 $ntimes n$-matrix. So you can multiply it by an $n$-vector. Because $h(A)$ is also an $ntimes n$-matrix $h(A)v$ is an $n$-vector. So you can multiply $g(A)$ by $h(A)v$. To show that $h(A)vinker g(A)$ it suffices to show that the product is $0$.