Matrix multiplication of an nxn matrix is the scaling

chaoseverlasting

Geometrically, matrix multiplication of an nxn matrix is the scaling, and rotation of a vector in n dimensions true? So when you find the inverse of a matrix, what you're actually doing is finding a transformation such that in the 'transformed space' the vector is a unit vector.

If the inverse matrix ($$A^{-1}$$)is plotted in the original space, then does it have any relation to the original matrix($$A$$)?

What I mean by that is, if you have a function $$y=f(x)$$ in 2 D space, and you find the inverse function $$x=f^{-1}(y)$$ the inverse function is a reflection of the function $$y=f(x)$$ about the line $$y=x$$. Does the inverse matrix ($$A^{-1}$$) have any such relation to the original matrix($$A$$)?

roam

Re: Matrices

The relation between $$A^{-1}$$ & $$A^$$... the inverse of $$A^$$, where it exists, is denoted by $$A^{-1}$$ and $$AA^{-1}$$ is equal to I ie. identity matrix = $$A^{-1}A$$

HallsofIvy

Homework Helper
Re: Matrices

The relation between $$A^{-1}$$ & $$A^$$... the inverse of $$A^$$, where it exists, is denoted by $$A^{-1}$$ and $$AA^{-1}$$ is equal to I ie. identity matrix = $$A^{-1}A$$
True, but I don't believe that was the question he was asking.

Geometrically, matrix multiplication of an nxn matrix is the scaling, and rotation of a vector in n dimensions true? So when you find the inverse of a matrix, what you're actually doing is finding a transformation such that in the 'transformed space' the vector is a unit vector.

If the inverse matrix ($$A^{-1}$$)is plotted in the original space, then does it have any relation to the original matrix($$A$$)?

What I mean by that is, if you have a function $$y=f(x)$$ in 2 D space, and you find the inverse function $$x=f^{-1}(y)$$ the inverse function is a reflection of the function $$y=f(x)$$ about the line $$y=x$$. Does the inverse matrix ($$A^{-1}$$) have any such relation to the original matrix($$A$$)?
What do you mean by "plotted in the original space"? An n by n matrix operates on n dimensional vectors and itself exist in an n2 vector space. In order to "graph" A and A-1 you would need an n4 dimensional graph.

chaoseverlasting

Re: Matrices

True, but I don't believe that was the question he was asking.

What do you mean by "plotted in the original space"? An n by n matrix operates on n dimensional vectors and itself exist in an n2 vector space. In order to "graph" A and A-1 you would need an n4 dimensional graph.
Why would you need an n4 dimensional graph? Say for a 3x3 matrix, you would need a 9 D space then. Its kind of hard for me to imagine that, but say if we took each row or column to be a plane in 3D (where I suppose each plane corresponds to a projection of 9D space on 3D), would there be a relation between the planes of $$A$$ and $$A^{-1}$$ ?

uman

Re: Matrices

A plane? Wouldn't each row or column of a three by three matrix simply be a vector in $$F^3$$?

chaoseverlasting

Re: Matrices

Yeah, each row or column of a 3x3 matrix would be a vector, but if you assume that vector to be the normal vector passing through a given point then you get three planes depending on whatever matrix equation you use. I dont think a standalone 3x3 matrix could represent planes by itself.

Even so, I'm guessing here, the projection would give you three vectors instead of three planes. So you would have a set of three vectors corresponding to $$A$$ and three vectors corresponding to $$A^{-1}$$, would the vectors of A and A-1 be co-related in any way?

Re: Matrices

olliemath

Re: Matrices

What I mean by that is, if you have a function $$y=f(x)$$ in 2 D space, and you find the inverse function $$x=f^{-1}(y)$$ the inverse function is a reflection of the function $$y=f(x)$$ about the line $$y=x$$. Does the inverse matrix ($$A^{-1}$$) have any such relation to the original matrix($$A$$)?
In general, any function (including e.g. matrices) $$f:R^n\rightarrow R^n$$ has a graph in $$R^{2n}$$ which is the set $$\{(\textbf{x},f(\textbf{x})):\textbf{x}\in R^n\}$$. If f is invertible then the graph of $$f^{-1}$$ is $$\{(f(\textbf{x}),\textbf{x}):\textbf{x}\in R^n\}$$. You get one from the other by reflection through a hyperplane and you can easily write out the matrix for this.

For a function on R this is very usefull as a visual aid. In general, not so. I think what you want to know is if we can read off the value that A^{-1} assigns to x using the graph of A? Because we can't visualise the graph we can't.

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chaoseverlasting

Re: Matrices

In general, any function (including e.g. matrices) $$f:R^n\rightarrow R^n$$ has a graph in $$R^{2n}$$ which is the set $$\{(\textbf{x},f(\textbf{x})):\textbf{x}\in R^n\}$$. If f is invertible then the graph of $$f^{-1}$$ is $$\{(f(\textbf{x}),\textbf{x}):\textbf{x}\in R^n\}$$. You get one from the other by reflection through a hyperplane and you can easily write out the matrix for this.

For a function on R this is very usefull as a visual aid. In general, not so. I think what you want to know is if we can read off the value that A^{-1} assigns to x using the graph of A? Because we can't visualise the graph we can't.
Thats exactly it. How would you go about writing a matrix for this? Perhaps we cant use it as a visual aid, but it would simplify a lot of other calculations.

olliemath

Re: Matrices

Use

$$B:=\left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right]$$

where I is the identity on $$R^n$$. The coordinates of a general point on the graph are $$(Ix,Ax)$$, so the coords of a general point on the graph of the inverse are $$B(Ix,Ax)=(Ax,Ix)$$. One usually writes out the matrix $$(A,I):R^n\rightarrow R^{2n}$$ and uses gaussian elimination to convert it to $$(I,A^{-1})$$, which again gives points on the graph of the inverse, but in a more useful fashion.

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