Question

anyone know how to proof matrix?
A is nxn matrix and Ax= x
show that A = I

Answer 1

I guess is it given that A is an invertible matrix?

Answer 2

yes

Answer 3

If A is invertible then there is some matrix B in which AB equals the identity. I am not sure if multiplying both sides by a Matrix B will get you there or not. But then A must equal A inverse or in this case B, and therefore A must Be equal to the identity... it seems so trivial trying to prove this. Ughh.

Answer 4

if \(Ax=x\) for every \(x\) then let's consider the cannonical basis of \(\mathbb{R}^n\)
\(\{\varepsilon_1,\varepsilon_2,\dots,\varepsilon_n\}\).

Answer 5

ok

Answer 6

then what happens?

Answer 7

if we denote by \(A^{(i)}\) the i-th column of A. Then
\[ \large A\varepsilon_i=\varepsilon_i \]
by hypothesis. But also
\[ \large A\varepsilon=A^{(i)} \]
therefore
\[ \large A^{(i)}=\varepsilon_i \]
That is
\[ \large A=[\varepsilon_1\quad\varepsilon_2\quad\dotsb\quad
\varepsilon_n]=I \]

Answer 8

sorry. it should be
\[ \large A\varepsilon_i=A^{(i)} \]
in the fourth line.

Answer 9

got it?