# Given a diagonalizable matrix $A$ and polynomial $f$, prove $f(A)$ is diagonalizable

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.

• Yes, you are on the right track. Can you show $f(A)=Tf(B)T^{-1}$? – Thomas Shelby Jun 28 at 14:39

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.

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.

• $f(B)$ 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? – Nick Jun 28 at 18:17
• 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. – paulinho Jun 28 at 18:18

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 $$$$f(x) = \sum_{i=0}^r a_ix^i$$$$ Since $$A$$ is diagonalizable by assumption, there is an invertible matrix $$T$$ and a diagonal matrix $$B$$ such that $$$$A = TBT^{-1}$$$$ 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.