Proving that $A+A^2+A^3$ has eigenvalue $\lambda+\lambda^2+\lambda^3$ where $\lambda$ is an eigenvalue of $A$

Question

I know that $$P= \begin{pmatrix} 4 & 5 & 1 \\ -9 & -6 & 0 \\ 8 & 4 & 2 \end{pmatrix}$$

---

I also know that

$$D= \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}$$

Where $a,b,c$ are eigenvalues of $A+A^2+A^3$

---

In this case the eigenvalues of :

$$A+A^2+A^3= \lambda + \lambda^2 + \lambda^3$$

Where $\lambda$ is an eigenvalue of A. Can someone prove this statement for me.

Answer 1

We know that a matrix $M^n = PD^nP^{-1}$. Then $A = PDP^{-1}$ so $$ A + A^2 + A^3 = PDP^{-1} + PD^2P^{-1} + PD^3P^{-1} = P(D+D^2+D^3)P^{-1} $$ You have already found $P$ so once you have $D$ you just add.

Answer 2

$$(A+B)x=Ax+Bx=\lambda x + \eta x= (\lambda+ \eta)x$$ Finally $$A^2x=(AA)x=A(Ax)=A(\lambda x)=\lambda (Ax)=\lambda (\lambda x)= \lambda^2x$$

Answer 3

Let $\lambda$ be an eigenvalue of $A$ and $v$ be the corresponding eigen vector. Then we have $Av=\lambda v$.

Multiplying both sides by $A$ from left we have $A^2 v=A\lambda v= \lambda Av=\lambda^2 v$, where we have used the fact that $Av=\lambda v$. Therefore $\lambda^2$ is an eigenvalue of $A^2$ and $v$ is the corresponding eigen vector. Similarly $A^3v=\lambda^3 v$.

Therefore $(A^3+ A^2+ A) v=(\lambda^3+\lambda^2+\lambda) v$ I.e., $ \lambda^3+\lambda^2+\lambda$ is an eigenvalue of $ A^3+ A^2+ A$.

In general, $p(A) v= p(\lambda) v$, where $p$ is a polynomial.

Answer 4

This is general. If $\lambda$ is an eigenvalue of the matrix $M$ with coefficients in $\mathbf{R}$ let's say, then, if $P(T)$ is a polynomial with real coefficient, the matrix $P(M)$ has $P(\lambda)$ as eigenvalue.

Indeed, let's write $P(T) = \sum_{i=0}^n a_i T^i$ where the $a_i$'s are real. Then, if $X$ is an eigenvector of $M$ for the eigenvalue $\lambda$ (that is, if $X\not=0$ and $M X = \lambda X$) we have :

$P(M) X = \left( \sum_{i=0}^n a_i M^i \right) X = \sum_{i=0}^n a_i ( M^i X ) = \sum_{i=0}^n a_i ( \lambda^i X ) = \left( \sum_{i=0}^n a_i \lambda^i \right) X = P(\lambda) X$.

Of course, the fact that $M^i X = \lambda^i X$ can be shown by a trivial induction : true for $i = 0$ because $M^{0} = Id$, and if $M^i X = \lambda^i X$ then $M^{i+1} X = M ( M^i X ) = M (\lambda^i X) = \lambda^i MX = \lambda^{i+1} X$.