MHB What Are the Minimal Polynomials of Matrix Powers?

[evinda]

Hello! (Wave)

If the matrix $A \in M_n(\mathbb{C})$ has $m_A(x)=(x^2+1)(x^2-1)$ as its minimal polynomial, then I want to find the minimal polynomials of the matrices $A^2$ and $A^3$.

($M_n(k)$=the $n \times n$ matrices with elements over the field $k=\mathbb{R}$ or $k=\mathbb{C}$)

Is there a relation that connects the minimal polynomial of a matrix $B$ with the minimal polynomial of the powers of $B$ ? (Thinking)

 

[I like Serena]

Hello! (Wave)

If the matrix $A \in M_n(\mathbb{C})$ has $m_A(x)=(x^2+1)(x^2-1)$ as its minimal polynomial, then I want to find the minimal polynomials of the matrices $A^2$ and $A^3$.

($M_n(k)$=the $n \times n$ matrices with elements over the field $k=\mathbb{R}$ or $k=\mathbb{C}$)

Is there a relation that connects the minimal polynomial of a matrix $B$ with the minimal polynomial of the powers of $B$ ? (Thinking)

Hey evinda! (Wave)

Don't we have that $\lambda$ is an eigenvalue of $A$ iff $\lambda$ is a root of the minimal polynomial?

Now suppose $\lambda$ is an eigenvalue of $A$ with eigenvector $\mathbf v$.
Then what can we say about $A^2\mathbf v$? (Wondering)

 

[evinda]

Hey evinda! (Wave)

Don't we have that $\lambda$ is an eigenvalue of $A$ iff $\lambda$ is a root of the minimal polynomial?

(Nod)

Now suppose $\lambda$ is an eigenvalue of $A$ with eigenvector $\mathbf v$.
Then what can we say about $A^2\mathbf v$? (Wondering)

$A^2\mathbf v= A(A \mathbf v)=A(\lambda \mathbf v)=\lambda(A \mathbf v)=\lambda(\lambda \mathbf{v})=\lambda^2 \mathbf v$

and

$A^3 \mathbf v=A(A^2 \mathbf v)=A(\lambda^2 \mathbf v)=\lambda^2(A \mathbf v)=\lambda^3 \mathbf v$.

So the eigenvalues of $A^2$ are $\lambda^2$ and the eigenvalues of $A^3$ are $\lambda^3$.

In our case, the eigenvalues of $A$ are $\pm i$ and $\pm 1$.

Thus the eigenvalues of $A^2$ are $-1,1$ and the eigenvalues of $A^3$ are $-i, i, -1,1$.

So the minimal polynomials of $A^2$ and $A^3$ are $(x+1)(x-1)=x^2-1$ and $(x+i)(x-i)(x+1)(x-1)=(x^2+1)(x^2-1)$,respectively.

Right? (Thinking)

 

[I like Serena]

(Nod)

$A^2\mathbf v= A(A \mathbf v)=A(\lambda \mathbf v)=\lambda(A \mathbf v)=\lambda(\lambda \mathbf{v})=\lambda^2 \mathbf v$

and

$A^3 \mathbf v=A(A^2 \mathbf v)=A(\lambda^2 \mathbf v)=\lambda^2(A \mathbf v)=\lambda^3 \mathbf v$.

So the eigenvalues of $A^2$ are $\lambda^2$ and the eigenvalues of $A^3$ are $\lambda^3$.

In our case, the eigenvalues of $A$ are $\pm i$ and $\pm 1$.

Thus the eigenvalues of $A^2$ are $-1,1$ and the eigenvalues of $A^3$ are $-i, i, -1,1$.

So the minimal polynomials of $A^2$ and $A^3$ are $(x+1)(x-1)=x^2-1$ and $(x+i)(x-i)(x+1)(x-1)=(x^2+1)(x^2-1)$,respectively.

Right?

Right! (Nod)

 

[evinda]

Right! (Nod)

Nice... Thanks a lot! (Happy)