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I am new to Linear Algebra and would like some feedback regarding my answer to the following question

A and B are two square matrices of order n. Prove or refute (with a counterexample) the following statements:

  1. If A is real and the algebraic multiplicity of each one of its eigenvalues is 1, then A is diagonalizable over R.
  2. If A and B have the same characteristic polynomial, then they have the same rank as well.
  3. If A and B are row-equivalent, then they have the same characteristic polynomial.
  4. If A and B are similar, then $A^k$ and $B^k$ are similar (k is a natural number).

My answer:

1.

$1 \leq$ geometric multiplicity $\leq$ algebraic multiplicity

If the algebraic multiplicity of each eigenvalue is 1, then $1 \leq$ geometric multiplicity $\leq$ 1, therefore the geometric multiplicity is 1. This means that the geometric multiplicity and the algebraic multiplicity are equal, and therefore the matrix is diagonalizable over R.

2.

The statement is incorrect. Counterexample:

A=$$\begin{bmatrix}0&1\\0&0 \end{bmatrix}$$ A's rank is 1, and its characteristic polynomial is $λ^2$.

B= $$\begin{bmatrix} 0&0\\0&0 \end{bmatrix}$$ B's rank is 0, and its characteristic polynomial is $λ^2$.

3.

The statement is incorrect. Counterexample:

A= $$\begin{bmatrix} 2&0\\0&2 \end{bmatrix}$$

B= $$\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$$

These two matrices are row equivalent, but A's characteristic polynomial is (λ-2)(λ-2), while B's characteristic polynomial is (λ-1)(λ-1).

4.

Correct: by definition A and B are similar if and only if there is an invertible matrix P, which satisfies $P^{-1}AP=B$

If $P^{-1}AP=B$, then $A=PBP^1$

$A^k=PDP^{-1}·PDP^{-1}…PDP^{-1}$, which repeats itself k times. As $P^{-1}·P$ cancel out, $A^k=PB^kP^{-1}$. Therefore $P^{-1}A^kP=P^{-1}PB^kP^{-1}P=B^k$

Thank you!

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  • $\begingroup$ This all looks fine to me ! $\endgroup$ – J.F Jan 19 at 17:41
  • $\begingroup$ Your answer to 1 is correct if we assume that all eigenvalues of $A$ are real. Otherwise, it is false. All the rest is correct (minor issue: your argument for 4 requires $k > 0$, but of course the $k = 0$ case is trivial). $\endgroup$ – darij grinberg Jan 19 at 18:19
  • $\begingroup$ @darijgrinberg Thank you! But in the question it is given that A is real and it asks if A is diagonalizable over R, so it should not be a problem, right? $\endgroup$ – dalta Jan 19 at 18:26
  • $\begingroup$ All seems good to me! $\endgroup$ – Yuval Gat Jan 19 at 18:44

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