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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 $\Bbb R$?

Answer

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?

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    $\begingroup$ If we are to be strict, the answer is yes, not true. $\endgroup$ – Mariano Suárez-Álvarez Oct 12 '15 at 20:19
  • $\begingroup$ Is $k=2$? The characteristic polynomial has degree $k$. $\endgroup$ – copper.hat Oct 12 '15 at 20:20
  • $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Oct 12 '15 at 20:21
  • $\begingroup$ $ A^2-4I=0$ means that the chracterisctic polynomial is $\lambda^2-4= (\lambda-2)(\lambda+2)=0$ $\endgroup$ – Emilio Novati Oct 12 '15 at 20:21
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    $\begingroup$ @EmilioNovati, no, it does not. $\endgroup$ – Mariano Suárez-Álvarez Oct 12 '15 at 20:21
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  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 $(t-1)^n$.

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

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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 $U = 0$.

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