# Is the Matrix Diagonalizable if $A^2=4I$

I have two question:

Let $$A$$ be a non-scalar matrix, $$A_{k \times k} \in \Bbb R$$, and $$A^2=4I$$. Is the matrix A, always diagonalizable in $$\Bbb R$$?

I know that the answer is yes:

$$A^2=4I \rightarrow A^2-4I=0$$

Then by the Cayley Hamilton theorem I know that the matrix satisfies the equation above.

Now I don't know how to explain the fact that the characteristic polynomial is $$P_A=(\lambda-2)(\lambda+2)$$, then the characteristic polynomial has two different roots, and no more. what's the reason for it?

Second Question - Irrelevant to the first question

When we say that a matrix is diagonalizable if it has different linear roots, it means that for if I have the following characteristic polynomial (for example) $$(t-1)^2(t-2)$$ then the matrix is not diagonalizable since the root 1 appears twice in the characteristic polynomial?

• If we are to be strict, the answer is yes, not true. – Mariano Suárez-Álvarez Oct 12 '15 at 20:19
• Is $k=2$? The characteristic polynomial has degree $k$. – copper.hat Oct 12 '15 at 20:20
• Your explanation of the answer is not correct. As $A^2-AI=0$, you know that the matrix is a root of the polynomial $X^2-4$, so that the minimal polynomial of $A$ divides $X^2-4$. Cayley-Hamilton has nothing to do with this. – Mariano Suárez-Álvarez Oct 12 '15 at 20:21
• $A^2-4I=0$ means that the chracterisctic polynomial is $\lambda^2-4= (\lambda-2)(\lambda+2)=0$ – Emilio Novati Oct 12 '15 at 20:21
• @EmilioNovati, no, it does not. – Mariano Suárez-Álvarez Oct 12 '15 at 20:21

1. There are only two roots since a polynomial of degree $n$ has exactly $n$ roots (in $\mathbb{C}$). Here, $n=2$, so there are only $2$ roots.

2. The statement in your second question is not correct. For example, consider the identity matrix of dimension $n$. Clearly the identity matrix is diagonalizable (as it is diagonal), but it has characteristic polynomial $(t-1)^n$.

• Maybe the second statement is true for the minimal polynomial? I think I was confused. – Alan Oct 12 '15 at 20:21
• @Alan Yes, it is true for the minimal polynomial, but not for the characteristic polynomial. – Ben Sheller Oct 12 '15 at 20:22

The matrix $$A$$ is a root of the polynomial $$t^2-4$$, hence the minimal polynomial of $$A$$, a divisor of $$t^2-4$$, has only simple roots. This is is equivalent to $$A$$ being diagonalisable.

Answer to the second question: a matrix over a field $$K$$ is diagonalisable over $$K$$ if and only if its minimal polynomial splits over $$K$$ into a product of distinct linear factors.

I'll try to answer without using the minimal and characteristic polynomials.

Consider the Jordan normal form of $A$, $$A = Q J Q^t,$$ with $Q$ orthogonal, and split $J$ into its diagonal part $D$ plus its strictly upper triangular part $U$, namely $$J = D + U.$$ Then, $$4I = A^2 = Q J Q^t QJ Q^t = Q J^2 Q^t \Longrightarrow J^2 = 4 I.$$ Exploiting the splitting, we obtain $$4I = (D + U)(D+U) = D^2 + DU + UD + U^2.$$ Notice that $D^2$ is diagonal, whereas $DU + UD + U^2$ is upper triangular, with zeros on the main diagonal. Hence, $$D^2 = 4I \Longrightarrow D = \pm2I.$$ Moreover, the upper triangular part of $4I$ is zero, hence $$0 = DU + UD + U^2 = \pm 4 U + U^2 = U ( U \pm 4 I).$$ This is possible only for $U = 0$.