- #1
skrat
- 748
- 8
Hi there,
I have a homework where I have to do this:
Prove that square matrix is invertible if the columns of the matrix are linearly independent.
There is also a hint: You can help with the following statement: Linear transformation L: U->V is bijection when a vector space basis N for U, it's "picture" (- i don't know the right english word) L(N) is a basis for vector space V. This applies for any space basis for U.
I tried several ways but I just can't get to the end. I don't want anybody here to write me the whole prove - I will hate you for that. However, I would be really happy if somebody could put me on the right path, show me the way and give me clues obvious enough to make it to the end.
If necessary I can write it down how I tried to solve this but I seriously doubt it would be any help to you!
Thanks in advance!
EDIT: OK, I changed my mind. I would be happy to know where I got it wrong. My procedure:
Let [itex]A:U\rightarrow V[/itex] linear transformation and
[itex]A=\begin{bmatrix}
a_{11} & a_{12}^{} &\cdots &a_{1n} \\
a_{21}& a_{22} &\cdots &a_{2n} \\
\vdots && &\vdots\\
a_{n1}&a_{n2} & \cdots &a_{nn}
\end{bmatrix}[/itex]
if [itex]N=\begin{Bmatrix}
u_{1} & \cdots & u_{n}
\end{Bmatrix}[/itex] basis for vector space U and
[itex]M=\begin{Bmatrix}
A(u_{1}) & \cdots & A(u_{n})
\end{Bmatrix}=\begin{Bmatrix}
v_{1} & \cdots & v_{n}
\end{Bmatrix}[/itex] basis for vector space V.
Than I decided to take one vector x from space U:
[itex]x=\alpha _{1}u_{1}+\alpha _{2}u_{2}+\cdots +\alpha _{n}u_{n}[/itex]
What linear transformation A does with x is:
[itex]Ax=\begin{bmatrix}
a_{11} & a_{12}^{} &\cdots &a_{1n} \\
a_{21}& a_{22} &\cdots &a_{2n} \\
\vdots && &\vdots\\
a_{n1}&a_{n2} & \cdots &a_{nn}
\end{bmatrix}\cdot \begin{bmatrix}
x_{1} \\
x_{2}\\
\vdots \\
x_{n}
\end{bmatrix}=\begin{bmatrix}
b_{1} \\
b_{2}\\
\vdots \\
b_{n}
\end{bmatrix}[/itex]
Where I am not sure what exactly [itex]x_{1}... x_{n}[/itex] are (yes, components, but combination of what?) and the same goes for b's..
Hmmm, Is this even close to right way?
I have a homework where I have to do this:
Prove that square matrix is invertible if the columns of the matrix are linearly independent.
There is also a hint: You can help with the following statement: Linear transformation L: U->V is bijection when a vector space basis N for U, it's "picture" (- i don't know the right english word) L(N) is a basis for vector space V. This applies for any space basis for U.
I tried several ways but I just can't get to the end. I don't want anybody here to write me the whole prove - I will hate you for that. However, I would be really happy if somebody could put me on the right path, show me the way and give me clues obvious enough to make it to the end.
If necessary I can write it down how I tried to solve this but I seriously doubt it would be any help to you!
Thanks in advance!
EDIT: OK, I changed my mind. I would be happy to know where I got it wrong. My procedure:
Let [itex]A:U\rightarrow V[/itex] linear transformation and
[itex]A=\begin{bmatrix}
a_{11} & a_{12}^{} &\cdots &a_{1n} \\
a_{21}& a_{22} &\cdots &a_{2n} \\
\vdots && &\vdots\\
a_{n1}&a_{n2} & \cdots &a_{nn}
\end{bmatrix}[/itex]
if [itex]N=\begin{Bmatrix}
u_{1} & \cdots & u_{n}
\end{Bmatrix}[/itex] basis for vector space U and
[itex]M=\begin{Bmatrix}
A(u_{1}) & \cdots & A(u_{n})
\end{Bmatrix}=\begin{Bmatrix}
v_{1} & \cdots & v_{n}
\end{Bmatrix}[/itex] basis for vector space V.
Than I decided to take one vector x from space U:
[itex]x=\alpha _{1}u_{1}+\alpha _{2}u_{2}+\cdots +\alpha _{n}u_{n}[/itex]
What linear transformation A does with x is:
[itex]Ax=\begin{bmatrix}
a_{11} & a_{12}^{} &\cdots &a_{1n} \\
a_{21}& a_{22} &\cdots &a_{2n} \\
\vdots && &\vdots\\
a_{n1}&a_{n2} & \cdots &a_{nn}
\end{bmatrix}\cdot \begin{bmatrix}
x_{1} \\
x_{2}\\
\vdots \\
x_{n}
\end{bmatrix}=\begin{bmatrix}
b_{1} \\
b_{2}\\
\vdots \\
b_{n}
\end{bmatrix}[/itex]
Where I am not sure what exactly [itex]x_{1}... x_{n}[/itex] are (yes, components, but combination of what?) and the same goes for b's..
Hmmm, Is this even close to right way?
Last edited: