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I have been given a diagonalizable matrix $A \in K^{n \times n}$ and a polynomial $f \in K[X]$ for a field $K$. I need to prove that $f(A)$ is diagonalizable. Because the matrix $A$ is a arbitrary matrix, it follows: $$A = \left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array}\right) \in K^{n \times n} \, for \, a_1,\dots,a_n \in K.$$ Assuming that $f$ is a normalized polynomial, it follows: $$f = X^n + a_{n-1}X^{n-1} + \dots + a_1X + a_0$$

My first idea was to simply insert the given the matrix $A$ in the polynomial $f$ which results in the following: $$f(A) = \left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array}\right)^n + a_{n-1} \left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array}\right)^{n-1} + \dots + a_1\left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array}\right)^1 + a_0 \left(\begin{array}{cccc} 1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 \end{array}\right)$$ I know that in order to determine the matrix $A^n$ I can use the following: $$A^n = TB^nT^{-1} \text{$\,$ where B is a diagonal matrix}.$$ I know such matrix $B$ exists because matrix $A$ is diagonalizable, so $A$ is similar to a diagonal matrix. I don't know if this is the right approach because from this point on I am stuck.

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    $\begingroup$ Yes, you are on the right track. Can you show $f(A)=Tf(B)T^{-1}$? $\endgroup$ – Thomas Shelby Jun 28 at 14:39
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I think you can do it as this. Let $$f(X) = X^n + a_{n-1}X^{n-1} + \cdots +a_1 X + a_0, \quad \in K[X]$$ be your polynomial and let $A$ diagonalize as $$A = TBT^{-1}.$$ Then you have \begin{align*} f(A) &=A^n + a_{n-1}A^{n-1} + \cdots + a_1A + a_0I \\ &=TB^nT^{-1} + a_{n-1}TB^{n-1}T^{-1} + \cdots + a_1 TBT^{-1} + a_0 I \\ &=T \left(B^nT^{-1} + a_{n-1}B^{n-1}T^{-1} + \cdots + a_1 BT^{-1} + a_0 T^{-1} \right) \\ &=T \left(B^n + a_{n-1}B^{n-1} + \cdots + a_1 B + a_0 I \right)T^{-1} \end{align*} where we use the property, that for scalars $\lambda \in K$ and Matrices $A,B$, $$\lambda (AB) = (\lambda A)B = A(\lambda B)$$ Now we have to show, that the matrix $B^n + a_{n-1}B^{n-1} + \cdots + a_1 B + a_0 I$is a diagonal one. Let $\lambda_1,\dots, \lambda_n$ be the different eigenvalues, then $$B^i = \begin{pmatrix} \lambda_1^i &0 &0 &\cdots &0 \\ 0 &\lambda_2^i& 0 &\cdots &0 \\ 0 &0&\lambda_3^i &\cdots &0\\\vdots&&&\ddots\\0&0&0&\cdots & \lambda_n^i\end{pmatrix},\quad \text{ for } 1 \le i \le n.$$ Since the sum of diagonal matrices is diagonal, and multiplication by scalar gives also a diagonal matrix, it should be fine.

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You observed that $T B^n T^{-1} = A$ for some diagonal matrix $B$, and we know $f(A)$ is simply going to be some linear combinations of the powers of $A$. Let $f(A) = a_0 I + a_1 A + a_2 A^2 + \cdots + a_n A^n$. What can we say about each of the individual terms?

We can actually rewrite $$f(A) = a_0 TT^{-1} + a_1 T B T^{-1} + \cdots + T B^n T^{-1} = T(a_0 I + a_1 B + \cdots + a_n B^n) T^{-1} = T f(B) T^{-1}$$ Aha! so $f(A)$ is similar to the matrix $f(B)$. Now we know that, if $f(B)$ is a diagonal matrix, then we will have shown that $f(A)$ is diagonalizable (by definition, since it will be similar to a diagonal matrix). Can you explain why $f(B)$ is necessarily a diagonal matrix? Hint: $B$ is a diagonal matrix.

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  • $\begingroup$ $f(B)$ is necessarily a diagonal matrix because the sum of two diagonal matrices is again a diagonal matrix. Furthermore the power of a diagonal matrix is again a diagonal matrix. Would this be the right explanation? $\endgroup$ – Nick Jun 28 at 18:17
  • $\begingroup$ Correct. The two facts that you have just stated show that any linear combination of powers of a diagonal matrix is diagonal, so any polynomial evaluated on a diagonal matrix is still diagonal. $\endgroup$ – paulinho Jun 28 at 18:18
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You were certainly on the right track. Also, there is no need to assume that the polynomial $f(x)$ is monic. To prove the question, write the polynomial $f(x) \in K[x]$ as \begin{equation} f(x) = \sum_{i=0}^r a_ix^i \end{equation} Since $A$ is diagonalizable by assumption, there is an invertible matrix $T$ and a diagonal matrix $B$ such that \begin{equation} A = TBT^{-1} \end{equation} So, we have that \begin{align} f(A) &:= \sum_{i=0}^ra_iA^i \\ &= \sum_{i=0}^ra_i (TBT^{-1})^i \\ &= \sum_{i=0}^r T\cdot (a_iB^i)\cdot T^{-1} \\ &= T \cdot \left( \sum_{i=0}^ra_iB^i \right) \cdot T^{-1} \\ &:= T \cdot f(B) \cdot T^{-1} \end{align}

(I leave it to you to see why each equal sign above is valid.)

Since $B$ is diagonal, it is easy to verify that powers of $B$ are diagonal; it follows that $f(B)$ is a diagonal matrix. This shows that $f(A)$ is similar to $f(B)$, and hence is diagonalizable.

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