Left invertible mapping left inverse of matrix

[fabbi007]

Homework Statement

relation from R^2-->R^2 ( R is real line)

(y1) [0 1] (x1)
(y2) =[-1 1] (x2)

is this left invertible?

if so what is the left inverse?

y1,y2 are element in a 2by 1 matrix, same with x1, x2. the elemenst 0,1,-1,1 are in a 2x2 matrix. I did no know how to represent a matrix here.

Homework Equations

A mapping is left invertible only if it is one-to-one

The Attempt at a Solution

The above relation is one-to-one because x is getting mapped into y with a 2 dimensional relation into a 2 dimensional space. If we solve this we get y1=x2; y2=-x1+x2. Hence I concluded it is left invertible. Also the mapping is onto, since the Range of the mapping is yielding Y.

I am not sure how I can find the left inverse now. Any ideas

 

[Mark44]

Is this what you're trying to represent? \left(\begin{array}{c}y_1 \\ y_2 \end{array}\right) =<br /> \left( \begin{array}{c c} 0 &amp; 1 \\ -1 &amp; 1 \end{array}\right)<br /> \left(\begin{array}{c}x_1 \\ x_2 \end{array}\right)

Click on equation above to see the LaTeX code I used.

If we write your matrix equation as y = Ax, you can apply the left inverse to both sides to get A^-1y = A^-1Ax = x. In other words, you can use the left inverse to solve for x.

BTW, you should be posting these questions in the Calculus & Beyond section. The questions you are asking are math questions, not engineering.

 

[fabbi007]

I
If we write your matrix equation as y = Ax, you can apply the left inverse to both sides to get A^-1y = A^-1Ax = x. In other words, you can use the left inverse to solve for x.

BTW, you should be posting these questions in the Calculus & Beyond section. The questions you are asking are math questions, not engineering.

Thanks for the latex code Mark. This is from my electrical engineering course. Hence I posted here. Also, the above mapping is right invertible because from the definition the range=Y. Would there be different left inverse and right inverse for a mapping if it is both one-to-one and onto? My guess from left inverse is for the mapping would be

<br /> \left( \begin{array}{c c} 1 &amp; -1 \\ 1 &amp; 0 \end{array}\right)<br />

and since the mapping is right invertible too. the right inverse would be the same as left inverse, right?

<br /> \left( \begin{array}{c c} 1 &amp; -1 \\ 1 &amp; 0 \end{array}\right)<br />

 

[Mark44]

Yeah, I think they would both be the same in this problem. I'm not certain on this, but I think that if you're dealing with square matrices, the left-inverse and right-inverse will be the same. Where I think this comes into play is when you're dealing with non-square matrices.

 

[fabbi007]

Yeah, I think they would both be the same in this problem. I'm not certain on this, but I think that if you're dealing with square matrices, the left-inverse and right-inverse will be the same. Where I think this comes into play is when you're dealing with non-square matrices.

Thanks Mark. I get it now. It is indeed a square matrix and there is only one inverse. Also since from definitions if a mapping is both left and right invertible then it has an inverse, meaning only one inverse.

 

[fabbi007]

Yeah, I think they would both be the same in this problem. I'm not certain on this, but I think that if you're dealing with square matrices, the left-inverse and right-inverse will be the same. Where I think this comes into play is when you're dealing with non-square matrices.

How can you prove that if a mapping F:X->Y is both left and right invertible that there exists only one left inverse and one right inverse. I am trying to understand the theory, I could understand the example though. Can you give me a hint?