Question

I don't understand this proof. Can someone explain it?

Answer 1

Let \(A\) be a \(n\times n\) matrix. How does \(A\mathbf{x}=\mathbf{b},\,\mathbf{x},\mathbf{b}\in \mathbb{R}^n\) has a unique solution for every \(\mathbf{b}\) implies a matrix \(C_{n\times n}\) such that \(AC=I_n\)?

Proof:
Let \(\mathbf{e}_n\) be the n-th standard basis of \(\mathbb{R}^n\). By the assumption, there exist \(\mathbf{c}_i\) such that \(A\mathbf{c}_i=\mathbf{e}_i\). This implies \(A\underbrace{\begin{bmatrix}\mathbf{c}_1 & \mathbf{c}_2 & \dots & \mathbf{c}_n\end{bmatrix}}_C=\underbrace{\begin{bmatrix}\mathbf{e}_1 & \mathbf{e}_2 & \dots & \mathbf{e}_n\end{bmatrix}}_{I_n}\)

Answer 2

I don't understand how the individual vector \(\mathbf{c}_i\) transforms into the matrix \(C\).

Answer 3

\(\large [ e_i]\) is a basis of \(\large \mathbb{R}^n\)

that means, all the combinations of n vectors in \(\large [e_i]\) span the vector space

Answer 4

so, we can find vectors \(c_i\) such that :
\(\large Ac_1 = e_1\)
\(\large Ac_2 = e_2\)
\(\cdots \)
\(\large Ac_n = e_n\)

Answer 5

well Ax=b is a system of equation which one of the following is true
1- has no solution
2_has exactly one solution
3_ has infinitly many solution

Ax=b has exactly one solution for every nX1 matrix b
which applies there exist c1,c2,c3,...
such that
Ac1- e1=0
Ac2-e2=0

ect
then
A(c1+c2+...cn)- (e1+e2+...)=0
then
AC=I
and as Oops said before its the same to A^-1

Answer 6

I get the part \(Ac_1=e_1,\,Ac_2=e_2,\,\dots,\,Ac_n=e_n\) but I don't get why \(C=\begin{bmatrix}c_1&c_2&\dots&c_n\end{bmatrix}\) is a matrix such that \(AC=I_n\). I think I have a problem on understand how matrix multiplication works.

Answer 7

*understanding

Answer 8

multiplying a matrix A by a column vector is same as taking linear combnations of columns of A

Answer 9

@ganeshie8 can u help me with a question when your done here

Answer 10

\[\large [ Ac_1 ~ Ac_2 ~ Ac_3 \cdots Ac_n] = [e_1~e_2~e_3~\cdots e_n] \]

Answer 11

lets take an example maybe ?

Answer 12

Is this the definition of matrix multiplication?

Answer 13

nope, its one of the several ways of multiplying matrices

Answer 14

consider below multiplication by a column vector :
\[

\left[
\begin{array}{ccc}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9

\end{array} \right]


\left[
\begin{array}{}
a \\
b \\
c

\end{array} \right]


\]

Answer 15

all it says is this :
take \(a\) number of first column,
take \(b\) number of second column,
take \(c\) nujmber of third column

add them

thats clearly a linear combination of columns of left matrix, right ?

Answer 16

\[

\left[
\begin{array}{ccc}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9

\end{array} \right]


\left[
\begin{array}{}
a \\
b \\
c

\end{array} \right]

=
a \left[
\begin{array}{}
1 \\
2 \\
3

\end{array} \right]
+b \left[
\begin{array}{}
4\\
5 \\
6

\end{array} \right]
+
c\left[
\begin{array}{}
7 \\
8 \\
9

\end{array} \right]

\]

Answer 17

adding one more vector for the second matrix on left side just adds one more column on the right side result

Answer 18

Can you represent the following multiplication as sum of matrices or vectors?
\[
\begin{bmatrix}
1&4&7\\
2&5&8\\
3&6&9
\end{bmatrix}
\begin{bmatrix}
a&d\\
b&e\\
c&f
\end{bmatrix}
\]

Answer 19

the result just two different columns, each column is a linear combination of columns of the left matrix

Answer 20

column1 of result :

\[
a \left[
\begin{array}{}
1 \\
2 \\
3

\end{array} \right]
+b \left[
\begin{array}{}
4\\
5 \\
6

\end{array} \right]
+
c\left[
\begin{array}{}
7 \\
8 \\
9

\end{array} \right]

\] \]

Answer 21

column2 of result :

\[d\left[
\begin{array}{}
1 \\
2 \\
3

\end{array} \right]
+e \left[
\begin{array}{}
4\\
5 \\
6

\end{array} \right]
+
f\left[
\begin{array}{}
7 \\
8 \\
9

\end{array} \right]

\]

Answer 22

Is this correct?
\[
\begin{align}
\begin{bmatrix}
1&4&7\\
2&5&8\\
3&6&9
\end{bmatrix}
\begin{bmatrix}
a&d\\
b&e\\
c&f
\end{bmatrix}=&\
a
\begin{bmatrix}
1&0\\
2&0\\
3&0\\
\end{bmatrix}
+
b
\begin{bmatrix}
4&0\\
5&0\\
6&0\\
\end{bmatrix}
+
c
\begin{bmatrix}
7&0\\
8&0\\
9&0\\
\end{bmatrix}+\\
&\ d
\begin{bmatrix}
0&1\\
0&2\\
0&3\\
\end{bmatrix}
+
e
\begin{bmatrix}
0&4\\
0&5\\
0&6\\
\end{bmatrix}
+
f
\begin{bmatrix}
0&7\\
0&8\\
0&9\\
\end{bmatrix}
\end{align}
\]

Answer 23

Yep ! this representation looks much better :)