Show it is impossible to have solution for $AX-XA=I$?

Question

For each matrix $A_{n×n}$, explain why it is impossible to find a solution for $X_{n×n}$ in the matrix equation

$$ AX-XA=I $$

Hint: Consider the trace function.

My question: I can understand that by taking the trace from both sides we get zero in the left and $n$ in the right hand side. However, I do not understand why manipulating both sides and getting unreasonable result means the true proof.

Do you understand how a proof by contradiction works in general? — angryavian, Feb 11 '19 at 03:57
@angryavian: I know how it works, but what assumption should I take to get the contradiction? Assuming the equality holds? — , Feb 11 '19 at 04:12
Yes, exactly: assume there exists a matrix $X$ satisfying $AX-XA = I$. — angryavian, Feb 11 '19 at 04:12

score 2 · Accepted Answer · answered Feb 11 '19 at 05:50

An implication $A \implies B$ is an assertion of the form : "If $A$ is true, then $B$ is true".

In particular, suppose $A \implies B$ is true. If it happens that $B$ is a statement like $1 = 0$ or something i.e. which is always false, then, $A$ also has to be false, because if it were true, then because of the implication being true we would get that $B$ is true, which can never happen.

Now, taking the trace says that "$AX - XA = I$ for two matrices $A,X$" implies "$n = 0$" i.e. the implication is a true statement. Since the latter statement is always false, but the implication is true, the first statement must also be false.

score 1 · Answer 2 · answered Feb 11 '19 at 06:39

1

Functions cannot map an input to more than one output, i.e. $a=b \implies f(a) = f(b)$ (the converse means the function is injective).

The trace of a matrix is a function and so we derive our contradiction.

answered Feb 11 '19 at 06:39

qwr

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Show it is impossible to have solution for $AX-XA=I$?

2 Answers2