Let $A \in M_{n\times n}(\mathbb{C})$. Which of the following statement(s) is/are true?

Question

Let $A \in M_{n\times n}(\mathbb{C})$. Which of the following statement(s) is/are true?

$(A)$ There exists $B \in M_{n\times n} (\mathbb{C})$ such that $B^2 = A.$

$(B)$ A is diagonalizable

$(C)$ There exists an invertible matrix P such that $P AP^ {−1}$ is upper-triangular.

$(D)$ A has an eigenvalue.

My works : i know that every real number is a complex number so we xan assume real as a complex,...so option A) is True

For $A=\begin{bmatrix} 1 &0 \\ 0& 1\end{bmatrix}$ i take $B = \begin{bmatrix} 0 &1 \\ 1& 0\end{bmatrix}$

Option B is not true take $A=\begin{bmatrix} 1 &0 \\ 0& 0\end{bmatrix}$

Option C is True by definition

option D is also true by definition

Is my answer is correct or not

thanks u

Answer 1

No, statement (A) is not true in general. It is not enough to find a $B$ just for one example of $A$. Here is a counterexample. Take $$ A=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}. $$ Assume that $B^2=A$ for some $B$. Then $B$ is nilpotent, since $B^4=A^2=0$. Since $B\in M_2(\Bbb{C})$, we conclude that $B^2=0$. So $A=B^2=0$, a contradiction.

(C) is true, because the characteristic polynomial can be factored into linear factors over $\Bbb{C}$ by the fundamental theorem of algebra. This has been asked and answered at MSE, e.g., here:

When is a matrix triangularisable?

Answer 2

Your answers are not correct.

For A): there does not necessarily exist such a $B$. For example, there is no matrix $B$ such that $B^2 = A$ when $$ A = \pmatrix{0&1\\0&0} $$ Interestingly, there necessarily exists such a $B$ if $A$ is invertible or diagonalizable.

B) Indeed, not every matrix is diagonalizable. However, the example you chose is a diagonal matrix, which is of course diagonalizable. The matrix I've written above would work as a counterexample for this option as well.

C) This statement is true. However, "by definition" is not a valid justification in this case.

D) This statement is true. Again, I think that "by definition" is insufficient justification. Ultimately, this fact (in the general case) is a consequence of the fundamental theorem of algebra.

Answer 3

(A) is actually false. For instance, $\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$ has no square root in $M_{2\times2}(\mathbb{C})$.

(B) is also false, but the matrix that you mentioned is diagonalizable. Use the one I mentioned above instead.

(C) That's true, but not by definition. You can prove it using (D) and induction.

(D) It's true but, again, not by definition. It's a consequence of the Fundamental Theorem of Algebra.